# When Does A Quadratic Have Real Roots? (3 Ways To Tell)

When working with quadratic equations, the roots (solutions) can be either real or complex.  It is helpful to know what signs to look for to tell when the solutions are real.

So, when does a quadratic have real roots?  A quadratic equation has real roots when the discriminant is positive or zero (not negative).  From an algebra standpoint, this means b2 >= 4ac.  Visually, this means the graph of the quadratic (a parabola) touches the x axis at least once.

Of course, a quadratic that touches the x axis only once, at the vertex, has one repeated real root, instead of two distinct real roots.

In this article, we’ll talk about how you can tell that a quadratic has real solutions.  We’ll also look at some examples, along with how to write a quadratic equation given its real roots.

There are a few ways to tell when a quadratic equation has real roots:

• Look at the discriminant – if it is positive or zero, the roots are real.
• Look at the graph – if the parabola touches the x-axis, then the roots are real.
• Look at the coefficients – there are some special cases that will tell you when there are real solutions to the quadratic (more on this later in the article!)

### Look At The Discriminant

The first way to tell if a quadratic has real roots is to look at the discriminant.  If the discriminant is positive or zero, then the quadratic equation has real roots.

Remember that for the quadratic equation:

• ax2 + bx + c = 0

the discriminant is given by the expression:

• b2 – 4ac

To get a positive or zero discriminant, we need:

• b2 – 4ac >= 0
• b2 >= 4ac

There are really two distinct cases when a quadratic has real roots:

• The first case is when the discriminant is positive (b2 – 4ac > 0) – this gives us two distinct real roots.
• The second case is when the discriminant is zero (b2 – 4ac = 0) – this gives us one repeated real root.

#### Discriminant Is Positive (Two Distinct Real Roots)

Here is one example of a quadratic equation with two real roots:

• 2x2 + 10x + 8 = 0

In this case, a = 2, b = 10, and c = 8.  This gives us:

• = (10)2
• = 100
• 4ac
• = 4(2)(8)
• = 64

So, b2 – 4ac > 0 (since 100 – 64 = 36 > 0), and thus the discriminant is positive.  This means that the quadratic has two distinct real roots.

The two real roots are -1 and -4. You can see this by factoring the quadratic as 2(x + 1)(x + 4) = 0.

(If you wish, you can verify these solutions by using the quadratic formula with a = 2, b = 10, and c = 8).

#### Discriminant Is Equal To Zero (One Repeated Real Root)

Here is one example of a quadratic equation with one repeated real root:

• x2 + 4x + 4 = 0

In this case, a = 1, b = 4, and c = 4.  This gives us:

• = (4)2
• = 16
• 4ac
• = 4(1)(4)
• = 16

So, b2 – 4ac = 0 (since 16 – 16 = 0), and thus the discriminant is zero.  This means that the quadratic has one repeated real root.

The repeated real root is x = -2.  You can see this by factoring x2 + 4x + 4 as (x + 2)(x + 2).

(You can also use the quadratic formula with a = 1, b = 4, and c = 4 to verify this solution.)

### Look At The Graph

Another way to tell if a quadratic has real solutions is to look at its graph.  For any quadratic equation, the graph will be a parabola.

Remember that one of the key features of a parabola is its vertex.  The vertex of a parabola is sort of like the “mountain top” (for negative values of a) or “valley bottom” (for positive values of a).

If the parabola touches the x axis, then the quadratic has real solutions.  There are two possible cases:

• The parabola touches the x axis twice (this gives us two real roots)
• The parabola touches the x axis once, at the vertex (this gives us one repeated real root)

#### Graph of the Quadratic Touches The X Axis Twice (Two Real Roots)

As you can see in the graph pictured above, the parabola touches the x axis in two different places. This means the quadratic equation x2 – 6x + 8 has two real roots, x = 2 and x = 4 (that is, both of the x values where the parabola and x axis intersect).

Be careful: for a quadratic equation to have two real roots, its graph must touch the x axis twice.

If the graph touches the x axis once, then the quadratic has one repeated real root (see the case below).  If the graph does not touch the x axis at all, then the quadratic has no real roots (instead, it has two distinct complex roots).

#### Graph of the Quadratic Touches The X Axis Once (One Repeated Real Root)

As you can see in the graph pictured above, the vertex (valley bottom) of this parabola lies on the x axis. This means the quadratic equation x2 – 4x + 4 has one real solution.

Be careful: for a quadratic equation to have one solution, its graph must touch the x axis exactly once.

If the graph touches the x axis twice, then it has two real solutions.

If the graph does not touch the x axis at all, then it has two complex solutions (and no real solutions).

### Look At The Coefficients

You can also look at the coefficients of a quadratic equation in standard form to tell if it has real roots.  Remember that the standard form of a quadratic equation has zero on one side, and terms in descending order on the other:

• ax2 + bx + c = 0

There are a few different cases to consider here.

#### When The Coefficient of x2 Is Equal To 1 (a = 1)

In this case, we look for b2 >= 4c.  If this is true, then the quadratic has real roots.

For example, the quadratic equation x2 + 4x + 3 = 0 has a = 1, b = 4, and c = 3.

This gives us:

• = 42
• = 16
• 4ac
• = 4*1*3
• = 12

Then b2 > 4ac (since 16 > 12), and so there are two distinct real roots for this quadratic: x = -1 and x = -3.

#### When The Coefficient of x2 Is Not Equal To 1 (a is not equal to 1)

In this case, divide the entire quadratic equation by a.  Then, you are in the first case, when the coefficient of x2 is equal to 1.

For example, take the equation 3x2 + 24x + 48 = 0.  Since a = 3, we will divide both sides by 3.

This leaves us with x2 + 8x + 16 = 0.  For this new equation, a = 1, b = 8, and c = 16.

Then we have:

• = 82
• = 64
• 4ac
• = 4*1*16
• = 64

This means that b2 = 4ac, so there is one repeated real root, which is x = -4.

#### When The Coefficient Of X Is Equal To 0 (b = 0)

In this case, our discriminant simplifies to -4ac.  To get real roots, the discriminant must be positive or zero.

This means -4ac >= 0.  There are two cases to consider:

• If a and c have opposite signs (one is positive and the other is negative), then the quadratic equation has real roots.
• If a and c have the same signs (both are positive or both are negative), then the quadratic equation has two complex roots.

For example, the quadratic equation x2 – 4 = 0 has two real roots.  In this case, a = 1, b = 0, and c = -4.

Since b = 0 and also a and c have opposite signs (a is positive, c is negative), we know there are real roots.  In fact, the solutions are 2 and -2.

You can verify this with the quadratic formula, or by factoring as a difference of squares: (x+ 2)(x – 2) = 0.

Remember that you can always use a calculator to help you verify the solutions to a quadratic equation.  You can also use a quadratic equation solver, such as this one from WolframAlpha.

For WolframAlpha’s calculator, remember that:

• The quadratic coefficient (x2 coefficient) means a
• The linear coefficient (x coefficient) means b
• The constant coefficient means c

## Examples of Quadratic Equations With Real Roots

Here are some examples of quadratic equations with real roots.  Look at them to see if you notice a pattern before reading further.

• x2 + x – 2 = 0 (solutions: x = 1 and x = -2)
• x2 + 2x – 3 = 0 (solutions: x = 1 and x = -3)
• x2 + 3x – 4 = 0 (solutions: x = 1 and x = -4)
• x2 + 4x – 5 = 0 (solutions: x = 1 and x = -5)
• x2 + 5x – 6 = 0 (solutions: x = 1 and x = -6)

One thing you might notice is that the x2 coefficients (a values) are equal to 1.  This simplifies the discriminant to b2 – 4c.

Another thing you might notice is that the x coefficients (b values) are all whole numbers:

• 1, 2, 3, 4, 5

You might also notice is the constant terms (c values) are negative integers:

• -2, -3, -4, -5, -6

This guarantees that the discriminant will be positive (which gives us two distinct real roots for the quadratic).

One last thing you might notice is that the solutions are 1 and c (where c is the constant term).

After we see the pattern, we can use it to create quadratic equations with real roots.  We can pick any whole number n and create a quadratic with real roots 1 and –n.

All we need to do is FOIL (x – 1)(x + n) to find our quadratic.  This would look like:

• (x + 1)(x + n) = 0
• x2 + (n-1)x – n = 0

For instance, let’s say we choose n = 10.  Then we would need to FOIL (x – 1)(x + 10), which comes out to x2 + 9x – 10.

Then the quadratic x2 + 9x – 10 = 0 has two real roots: x = 1 and x = -10.

We can also multiply a quadratic equation by any real number (except zero) to get a new equation with the same roots.  For example, if we multiply x2 + 9x – 10 = 0 by 2, we get 2x2 + 18x – 20 = 0, which also has roots of 1 and -10.

## How Do You Write A Quadratic Equation With Given Real Roots?

Now it’s time to think about working backwards.  That means taking real numbers and writing a quadratic equation with those numbers as its solutions.

This is easy to do.  Even better, there are infinitely many such equations for a given set of roots.

Let’s say we want to write a quadratic equation with two real solutions r and s.  All we need to do is FOIL (x – r)(x – s) to find the quadratic.

This would give us the quadratic equation x2 – (r+s)x + rs.

### Example: Writing A Quadratic With Given Distinct Real Roots

Let’s say we want to find a quadratic equation with real roots 4 and 7.  Then we have r = 4 and s = 7.

The equation would be a multiple of (x – r)(x – s) = 0.  Substituting r = 4 and s = 7 gives us:

• (x – r)(x – s) = 0
• (x – 4)(x – 7) = 0

Using FOIL on the left side gives us x2 – 11x + 28 = 0. We can take any multiple of this equation (for example, we could multiply both sides by 2 without changing the solutions of x = 4 and x = 7).

## Conclusion

Now you know when a quadratic equation has real roots.  You also know what to look out for in terms of the discriminant, the graph, and the coefficients.

You might also want to read my article on when to use the quadratic equation or my article about how to factor a quadratic binomial (you can use a special case of the quadratic formula).

This article goes into detail on how to use a quadratic to find the nature of the solutions (real or complex) of a cubic function.

Don’t forget to subscribe to my YouTube channel & get updates on new math videos!

When dealing with quadratic equations, we often see either two real solutions or two complex solutions.  However, it is also possible that a quadratic will have exactly one solution.

So, when does a quadratic have one solution?  A quadratic equation has one solution when the discriminant is zero.  From an algebra standpoint, this means b2 = 4ac.  Visually, this means the graph of the quadratic (a parabola) will have its vertex resting on the x-axis.

Of course, we can also look at the coefficients in a quadratic equation to tell if it has one solution.

In this article, we’ll talk about how you can tell that a quadratic has one solution.  We’ll also look at some examples, along with how to write a quadratic equation given that it only has one solution.

Let’s get started.

There are a few different ways to tell when a quadratic equation has one solution:

• Look at the discriminant – if it is zero, there is only one solution to the quadratic.
• Look at the graph – if the vertex of the parabola rests on the x-axis, there is only one solution to the quadratic.
• Look at the coefficients – there is a special pattern that will tell you when there is only one solution to the quadratic (more on this later in the article!)

### Look At The Discriminant

The first way to tell if a quadratic has one solution is to look at the discriminant.  If the discriminant is zero, then the quadratic equation has only one real solution.

Remember that for the quadratic equation given by

• ax2 + bx + c = 0

To get a discriminant of zero, we need to set b2 – 4ac equal to zero.  This gives us:

• b2 – 4ac = 0
• b2 = 4ac

Here is one example of a quadratic equation with only one solution:

•  x2 – 6x  + 9

In this case, a = 1, b = -6, and c = 9.  This gives us:

• b2 = (-6)2 = 36
• 4ac = 4(1)(9) = 36

So, b2 = 4ac, and thus the discriminant is zero.  This means that the quadratic has only one solution: x = 3.

You can find this solution by using the quadratic formula.  You can also find it by factoring the quadratic:

• x2 – 6x  + 9 = (x – 3)(x – 3)
Дополнительно:  Прошивка root irbis

### Look At The Graph

Another way to tell if a quadratic has one solution is to look at the graph of the quadratic.  For any quadratic equation, its graph will be a parabola.

Remember that a key feature of a parabola is its vertex.  The vertex of a parabola is sort of like the “mountain top” (for negative values of a) or “valley bottom” (for positive values of a).

As you can see in the graph pictured above, the vertex (valley bottom) of this parabola lies on the x axis. This means the quadratic equation x2 – 4x + 4 has one real solution (at x = 2).

Be careful: for a quadratic equation to have only one real solution (a double root), its graph must touch the x axis exactly once.

If the graph touches the x axis twice, then it has two distinct real solutions.

If the graph does not touch the x axis at all, then it has two complex solutions (and no real solutions).

### Look At The Coefficients

You can also look at the coefficients of a quadratic equation in standard form to tell if it has one real solution.  Remember that the standard form of a quadratic equation has zero on one side, and terms in descending order on the other:

• ax2 + bx + c = 0

There are two possible cases to deal with here: either a = 1, or a is not equal to 1.

#### When The Coefficient of x2 Is Equal To 1 (a = 1)

In this case, we look for c = b2 / 4.  If this is true, then the quadratic has one solution, which is x = -b / 2.

For example, the quadratic equation x2 + 8x + 16 has one solution.  In this case, a = 1, b = 8, and c = 16. Then we have:

• b2 / 4
• = 82 / 4
• = 64 / 4
• = 16

which is the same as the value of c.  The one solution is:P

• x = -b / 2
• x = -8/2
• x = -4

We can also find this solution by factoring the quadratic as:

• x2 + 8x + 16 = (x + 4)(x + 4)

which gives us a solution of x = -4 (a real repeated root). We can also use the quadratic formula with a = 1, b = 8, and c = 16 to get the same result of x = -4.

#### When The Coefficient of x2 Is Not Equal To 1 (a is not equal to 1)

In this case, divide the entire quadratic equation by a.  Then, you are in the first case, when the coefficient of x2 is equal to 1.

For example, consider the equation:

• 2x2 + 12x + 18 = 0

Since a = 2, we will divide both sides by 2.

This leaves us with:

• x2 + 6x + 9 = 0

For this new quadratic equation, we have a = 1, b = 6, and c = 9.

Then we have:

• b2 / 4
• = 62 / 4
• = 36 / 4
• = 9

which is the same as the value of c.  The one solution is x = -b / 2 = -6 / 2 = -3.

We can also find this solution by factoring the quadratic as:

• x2 + 6x + 9 = (x + 3)(x + 3)

which gives us a solution of x = -3 (a real repeated root). We can also use the quadratic formula with a = 1, b = 6, and c = 9 to get the same result of x = -3.

Remember that you can always use a calculator to help you verify the solutions to a quadratic equation.  You can also use a quadratic equation solver, such as this one from WolframAlpha.

For WolframAlpha’s calculator, remember that:

• The quadratic coefficient (x2 coefficient) means a
• The linear coefficient (x coefficient) means b
• The constant coefficient means c

## Examples of Quadratic Equations With One Solution

Here are some examples of quadratic equations with one solution.  Look at them to see if you notice a pattern before reading further.

• x2 + 2x + 1 = 0 (solution: x = -1)
• x2 + 4x + 4 = 0 (solution: x = -2)
• x2 + 6x + 9 = 0 (solution: x = -3)
• x2 + 8x + 16 = 0 (solution: x = -4)
• x2 + 10x + 25 = 0 (solution: x = -5)

One thing you might notice is that the x coefficients (b values) are all even numbers:

• 2, 4, 6, 8, and 10

These come from doubling the numbers 1, 2, 3, 4, and 5.

Another thing you might notice is that the constant terms (c values) are all perfect squares:

• 1, 4, 9, 16, and 25

These come from squaring the numbers 1, 2, 3, 4, and 5.

One last thing you might notice is that the solutions are the negatives of the numbers 1, 2, 3, 4, and 5.

After we notice the pattern, we see the beauty of it: you can pick any whole number n and create a quadratic with a single solution (the solution is n).

Here’s how to do it.

• First, choose your whole number n. Remember that a whole number has no fractions or decimals.  For this example, I choose n = 7.
• Next, calculate 2n.  This will be the value of b.  For this example, b = 2n = 2*7 = 14.
• Then, calculate n2.  This will be the value of c.  For this example, c = n2 = 72 = 49.
• Finally, your quadratic with one solution has a = 1, b = 2n, and c = n2.  So, the quadratic would look like x2 + 2n + n2.  The solution will be –n (the negative of the whole number n you chose at the beginning).

For this example, our quadratic with one solution is x2 + 14x + 49.  The solution is x = -7.

We can also change the sign of all the b values to get an entire new set of quadratic equations with only one solution:

• x2 – 2x + 1 = 0 (solution: x = 1)
• x2 – 4x + 4 = 0 (solution: x = 2)
• x2 – 6x + 9 = 0 (solution: x = 3)
• x2 – 8x + 16 = 0 (solution: x = 4)
• x2 – 10x + 25 = 0 (solution: x = 5)

Finally, we can take any given quadratic equation with one solution and multiply by any number (except zero) to get a new quadratic equation with the same solution.

## How Do You Write A Quadratic Equation With One Solution?

Now it’s time to think about working backwards.  That means taking one number and writing a quadratic equation whose only solution is that number.

This is easy to do.  Even better, there are infinitely many such equations for a given solution.

Let’s say we want to write a quadratic equation with only one solution.  Let’s also assume that we want the solution to be the number n.

To do this, we would simply write the equation:

• (x – n)(x – n) = 0

After using FOIL on the parentheses, we would get the equation

• x2 – 2nx + n2 = 0

This equation has exactly one solution, which is x = n (it is a double real root).

### Example: Writing A Quadratic Equation With One Real Solution

For example, let’s say I want a quadratic equation with a solution of n = 3.

Then the equation would be

• (x – 3)(x – 3) = 0

After using FOIL, we would get the equation

• x2 – 6x + 9 = 0

This equation has exactly one solution, which is the value x = 3.  We can also generate as many more equations as we like by simply multiplying both sides by 2, 3, 4, 5, and so forth:

• 2x2 – 12x + 18 = 0 (multiplied by 2)
• 3x2 – 18x + 27 = 0 (multiplied by 3)
• 4x2 – 24x + 36 = 0 (multiplied by 4)
• 5x2 – 30x + 45 = 0 (multiplied by 5)

## Does A Quadratic Equation Always Have Two Solutions?

In terms of real solutions, there are always either 0, 1, or 2 real solutions to a quadratic equation, depending on the sign of the discriminant.

However, there is always at least 1 solution if you count both real and complex numbers.

• If the discriminant is positive, there are exactly 2 real solutions (they are distinct, meaning we get two different real numbers as solutions).
• If the discriminant is zero, there is exactly one real solution (a repeated root: the same real number appears as a solution twice).
• If the discriminant is negative, there are exactly 2 complex solutions (they are distinct, and they are complex conjugates).

Remember that two complex conjugates have the form a + bi and a – bi.  In other words, they have the same real part, but opposite imaginary parts.

For example, 2 + 3i and 2 – 3i are complex conjugates.

## Conclusion

Now you know when a quadratic equation has exactly one solution.  You also know what to look out for in terms of the discriminant, the graph, and the coefficients.

You might also want to read my article on when to use the quadratic equation.

This article goes into detail on how to use a quadratic to find the nature of the solutions (real or complex) of a cubic function.

Don’t forget to subscribe to my YouTube channel & get updates on new math videos!

When working with quadratic equations, we often see one or two real solutions.  However, it is also possible that a quadratic will have no real solution.

So, when does a quadratic have no solution?  A quadratic equation has no solution when the discriminant is negative.  From an algebra standpoint, this means b2 < 4ac.  Visually, this means the graph of the quadratic (a parabola) will never touch the x axis.

Of course, a quadratic that has no real solution will still have complex solutions.

In this article, we’ll talk about how you can tell that a quadratic has no solution.  We’ll also look at some examples, along with how to write a quadratic equation given its complex solutions.

Having math trouble?

Looking for a tutor?

There are a few ways to tell when a quadratic equation has no solution:

• Look at the discriminant – if it is negative, there is no real solution to the quadratic.
• Look at the graph – if the parabola never touches the x-axis, there is no real solution to the quadratic.
• Look at the coefficients – there are some special cases that will tell you when there is no real solution to the quadratic (more on this later in the article!)

### Look At The Discriminant

The first way to tell if a quadratic has no real solution is to look at the discriminant. If the discriminant is negative, then the quadratic equation has no real solution.

Remember that for the quadratic equation given by:

• ax2 + bx + c = 0

the discriminant is the expression given by:

• b2 – 4ac

To get a negative discriminant, we need:

• b2 – 4ac < 0
• b2 < 4ac

Here is one example of a quadratic equation with no real solution:

•  x2 + 2x + 5 = 0

In this case, a = 1, b = 2, and c = 5.  This gives us:

• = (2)2
• = 4
• 4ac
• = 4(1)(5)
• = 20

So, b2 < 4ac (since 4 < 20), and thus the discriminant is negative.  This means that the quadratic has no real solution.

However, the equation does have two complex solutions, -1 + 2i and -1 – 2i.  (You can verify these solutions by using the quadratic formula with a = 1, b = 2, and c = 5).

You can also verify these two solutions by using FOIL on (x – (-1 + 2i))(x – (-1 – 2i)) to get back the original quadratic,  x2 + 2x  + 5.  (Remember that i2 = -1).

You can see the graph of this function below.

If you look closely, you will notice something interesting about the graph above: it never touches the x-axis (that is, the parabola never crosses the line y = 0).

Let’s take a closer look at why this is the case.

### Look At The Graph

Another way to tell if a quadratic has no real solution is to look at its graph.  For any quadratic equation, the graph will be a parabola.

Remember that one of the key features of a parabola is its vertex.  The vertex of a parabola is sort of like the “mountain top” (for negative values of a) or “valley bottom” (for positive values of a).

Note that a quadratic with no real solution could have a graph that is completely below the x axis (in that case, one of the conditions is that a < 0). An example is below:

As you can see in the first graph pictured above, the parabola does not touch the x axis. This means the quadratic equation x2 – 2x + 2 = 0 has no real solution.

Instead, it has two complex solutions: 1 + i and 1 – i.

Be careful: for a quadratic equation to have no real solution, its graph must never touch the x axis.  If the graph touches the x axis at all, then it has either one (repeated) real solution or two distinct real solutions.

Having math trouble?

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### Look At The Coefficients

You can also look at the coefficients of a quadratic equation in standard form to tell if it has no real solution.

Remember that the standard form of a quadratic equation has zero on one side, and terms in descending order on the other:

• ax2 + bx + c = 0

The quadratic formula becomes much simpler when b = 0.  After simplifying, we find that the solutions are the positive and negative square roots of –c / a.

If c and a are both positive, then c / a is positive, and –c / a is negative.

Likewise, if c and a are both negative, then c / a is positive, and –c / a is negative.

In either of those cases, we are taking the square root of a negative, which gives us two complex solutions to the quadratic (that is, no real solution).

For example, the quadratic equation x2 + 4 = 0 has no real solution.  In this case, a = 1, b = 0, and c = 4.

Since b = 0 when a and c have the same sign (both are positive), we know there is no real solution.  In fact, the solutions are 2i and -2i (you can verify this with the quadratic formula).

We can also verify this if we FOIL (x – 2i)(x + 2i), cancel like terms, and use i2 = -1 to simplify to get back to x2 + 4.

Similarly, the quadratic equation -2x2 – 18 has no real solution.  In this case, a = -2, b = 0, and c = -18.

Since b = 0 when a and c have the same sign (both are negative), we know there is no real solution.  In fact, the solutions are 3i and -3i.

Remember that you can always use a calculator to help you verify the solutions to a quadratic equation.  You can also use a quadratic equation solver, such as this one from WolframAlpha.

For WolframAlpha’s calculator, remember that:

• The quadratic coefficient (x2 coefficient) means a
• The linear coefficient (x coefficient) means b
• The constant coefficient means c

## Examples Of Quadratic Equations With No Real Solution

Here are some examples of quadratic equations with no real solution.  Look at them to see if you notice a pattern before reading further.

• x2 + x + 1 = 0
• x2 + x + 2 = 0
• x2 + x + 3 = 0
• x2 + x + 4 = 0
• x2 + x + 5 = 0

One thing you might notice is that the x2 coefficients (a values) are all equal to 1.

Another thing you might notice is that the x coefficients (b values) are all equal to 1.

One last thing you might notice is that the constant terms (c values) are the sequence of whole numbers:

• 1, 2, 3, 4, 5

After we notice the pattern, we can easily create more of these quadratic equations with no real solution.  When we plug a = 1 and b = 1 into the quadratic formula, the discriminant simplifies to 1 – 4c.

As long as c is a positive number greater than ¼, we will get a quadratic with no real solution.  Any positive whole number is greater than ¼, so we can choose any of them for c.

We can also change the sign of all the b values to get an entire new set of quadratic equations with no real solution:

• x2 – x + 1 = 0
• x2 – x + 2 = 0
• x2 – x + 3 = 0
• x2 – x + 4 = 0
• x2 – x + 5 = 0

After plugging in a = 1 and b = -1 into the quadratic formula, the discriminant still simplifies to 1 – 4c.

Once again, as long as c is a positive number greater than ¼, we will get a quadratic with no real solution.

There are many other possibilities for quadratic equations with no real solutions.  All you need to do is choose a, b, and c so that the discriminant is negative (that is, b2 < 4ac).

## Does A Quadratic Equation Always Have A Solution?

A quadratic equation does not always have a real solution.  However, a quadratic equation always has a solution if we consider complex/imaginary numbers.

In terms of real solutions, there are always either 0, 1, or 2 real solutions to a quadratic equation, depending on the sign of the discriminant.

• If the discriminant is positive, there are exactly 2 real solutions.
• If the discriminant is zero, there is exactly one real solution (a repeated root).
• If the discriminant is negative, there are exactly 2 complex solutions (they are complex conjugates), with no real solution.

Remember that two complex conjugates have the form a + bi and a – bi.  In other words, they have the same real part, but opposite imaginary parts.

Дополнительно:  Вопрос-ответ

For example, 2 + 3i and 2 – 3i are complex conjugates.

## How Do You Write A Quadratic Equation With No Real Solution?

Now it’s time to work backwards.  Given a pair of complex conjugate solutions, we want to find a quadratic equation that has those solutions.

This is easy to do, once you know the method.  If the complex solutions are r and s, then all we need to do is FOIL (x – r)(x – s) to find the quadratic.

This would give us the quadratic equation x2 – (r + s)x + rs.  Let’s try an example.

### Example: Writing A Quadratic Equation With Given Complex Solutions (No Real Solution)

Assume we want to find a quadratic equation with roots 2i and -2i.  Then we have r = 2i and s = -2i.

The equation would be (x – r)(x – s) = 0 with r = 2i and s = -2i.  This gives us:

• (x – 2i)(x – (-2i)) = 0
• (x – 2i)(x + 2i) = 0

Using FOIL on the left side gives us x2 – 2ix + 2ix – 4i2 = 0  The like terms cancel to give us x2 – 4i2 = 0.

Since i2 = -1, we can simplify to x2 – 4(-1) = 0, or x2 + 4 = 0.

You can see the graph of the parabola y = x2 + 4 below (note that it never touches the x-axis).

## Conclusion

Now you know when a quadratic equation has no solution (in the real numbers).  You also know what to look out for in terms of the discriminant, the graph, and the coefficients.

You might also want to read my article on when to use the quadratic equation or my article about how to factor a quadratic binomial (you can use a special case of the quadratic formula).

This article goes into detail on how to use a quadratic to find the nature of the solutions (real or complex) of a cubic function.

Don’t forget to subscribe to my YouTube channel & get updates on new math videos!

Quadratic equations are the equations where polynomial has the degree two. Quadratic equations are the equations of type ax2 + bx + c = 0 where x is unknown and a, b, c are known real numbers and a should not be zero. If a=0 then the equation will not remain quadratic, it will be then linear as a=0 will eliminate x2 term. As the quadratic equation has the highest degree two, so this equation has two roots, or we can say that we will find two values of x for a quadratic equation.

### How to find roots?

Method 1: The roots of the quadratic equations can be found by the Shridharacharaya formula.

Example: The length of sides of a rectangle is given by x – 3 and x – 5 and the area of the rectangle is 3 unit2. Find the sides of the rectangle.

Area of rectangle = length*breadth = (x – 3)(x – 5) = 3

Area = x2 – 8x + 15 = 3

= x2 – 8x + 12 = 0

Discriminant = b2 – 4ac = 64 – (4(1)(12)) = 64 – 48 = 16

x = 12/2 or 4/2

x = 2 or 6

When x is 2, sides are x – 3 = 2 – 3 = -1 and x – 5 = 2 – 5 = -3.

Since length of sides cannot be equal therefore x = 2 is not a valid ans.

When x is 6, sides are x – 3 = 6 – 3 = 3 and x – 5= 6 – 5 =1.

Therefore, x = 6 is the valid answer and the sides are 3 and 1.

Method 2. The other way is the factorizing method. A quadratic equation can be considered a factor of two terms. Like ax2 + bx + c = 0 can be written as (x – x1)(x – x2) = 0 where x1 and x2 are roots of quadratic equation.

1. Find two numbers such that there product = ac and their sum = b.
2. Then write x coefficient as sum of these two numbers and split them such that you get two terms for x.
3. factor the first two as a group and last two terms as a group.
4. Take common factors from these and on equating the two expression with zero after taking common factors and rearranging the equation we get the roots .

Example: Let be the quadratic equation x2 + 3x = 18

x2 + 3x – 18 = 0

1. 6 and -3 are the numbers whose sum is equal to b and product is equal ac.

2. x2 + (6-3)x – 18 = x2 + 6x -3x – 18 =0

3. x(x + 6) – x(x + 6) = 0

4. taking (x + 6) as common.

(x + 6)(x – 3) = 0

x = -6 or x = 3

In a factorizing method it is not necessary that you will always find these two numbers easily(especially in the case when roots are imaginary or irrational) so it is better to use the quadratic formula.

### Nature of roots

The nature of roots depends on the discriminant of the quadratic equation. The discriminant of a quadratic equation is given by b2 – 4ac.  It is so because in quadratic formula square root of discriminant is there.

Root 1: If b2 – 4ac > 0 roots are real and different. As the discriminant is >0 then the square root of it will not be imaginary. It has two cases.

• If b2 – 4ac is a perfect square then roots are rational. As the discriminant is a perfect square, so we will have an integer as a square root of the discriminant. Hence, the roots are rational numbers.

Example: Let the quadratic equation be x2-5x+6=0.

Then the discriminant of the given equation is b2 – 4ac=(-5)2 – 4*1*6 = 25-24 = 1

According to Shridharacharaya formula

Therefore, the roots are 3,2. Both are rational and different.

• If b2 – 4ac is not a perfect square then the square root of discriminant is irrational hence roots are irrational and occurs in pair.

Example: Let the quadratic equation be x2-7x+8 = 0.

Then the discriminant of the given equation is

b2 – 4ac=(-7)2 – 4*1*8 = 49-32 = 17

According to Shridharacharaya formula

Root 2: If b2 – 4ac = 0 roots are real and equal.

Example: Let the quadratic equation be 3x2-6x+3=0.

Then the discriminant of the given equation is

b2 – 4ac=(-6)2 – 4*3*3 = 36 – 36 = 0

According to Shridharacharaya formula

Therefore, the roots are 1,1. Both are real and equal.

Root 3: If b2 – 4ac < 0 roots are imaginary, or you can say complex roots. It is imaginary because the term under the square root is negative. These complex roots will always occur in pairs i.e, both the roots are conjugate of each other.

Example: Let the quadratic equation be x2+6x+11=0.

Then the discriminant of the given equation is

b2 – 4ac=(6)2 – 4*1*11 = 36-44 = -8

According to Shridharacharaya formula

Therefore, the roots are 3,2. Both are imaginary and conjugate of each other(in pair).

### Graphs for Roots

The maximum/minimum value of quadratic function is found at x = -b/2a

We get maxima or minima when d(f(x))/dx = 0.

On differentiating quadratic function f(x) = ax2 + bx + c.

2ax + b = 0

x = -b/2a

This x is either maxima(a<0) or minima(a>0).

1. When b2– 4ac > 0

2. When b2– 4ac = 0

3. When b2– 4ac < 0

Question 1. The height of a triangle is less than 4 cm than the base. The area of triangle is 30 cm2. Find the height and base of triangle.

Let the base of triangle be x cm then height is x-4 cm

Area of triangle = 1/2*height*base = 1/2*(x)(x – 4)=30

Area = x2 – 4x  = 30*2

= x2 – 4x = 60

= x2 – 4x – 60 = 0

Discriminant = (-4)2 – 4(1)(-60) = 16+240 = 256

x = 20/2 or -12/2

x = 10 or -6

As side cannot be negative, therefore -6 is not correct.

So, when x is 10, base =10 cm and height =x – 4 = 10 – 4 = 6 cm

Therefore, x = 10 is the valid answer and the base and height of the triangle are 10 and 6 respectively.

Question 2. The volume of a box is 600 inch2 . The length of box is 2 inches less than the width. The height of box is 5 inches . Find the dimensions of box.

Let the width of box be x inches then length = x – 2 inches.

Volume of box =Length* Width *Height = (x-2)(x)5= 600

x2 – 2x = 120 => x2 – 2x -120 = 0

x2 + 10x – 12x – 120 = 0

x(x + 10) – 12(x + 10)=0

(x – 12)(x + 10)=0

x = 12 or x = -10

As width can not be negative, therefore, -10 is not correct.

When x = 12, width = 12 inches, length = x – 2 = 12 – 2 = 10 inches, height = 5 inches

Question 3. A ball is thrown from the top of a building . its height in meters above the ground as a function of time is given by h(t) = -4t2 + 24t + 3. a) How much time it take to reach the maximum height and what is the maximum height. b) Find also the time at which ball hits the ground.

a) Since a<0, therefore the time to reach maximum height is = -b/2a. (refer graph part)

t = -24/(2(-4)) = -24/-8

t = 3 sec.

Height = h(t) = -4(3)2 +24(3)+3 = 39 meters.

b) When the ball hits the ground h(t)=0.

-4t2 + 24t + 3 = 0

The discriminant of the given equation is

b2 – 4ac=(24)2 – 4*(-4)*3 = 576 + 48 = 624

According to Shridharacharaya formula

Quadratic Equations are the polynomial equation of the second degree. These equations are very common in mathematics and their family of curve represent the conic section. The word quadratic is derived from the Latin word “quadratus” which means square. Thus, the name quadratic equation implies that these are equations with square terms.

Quadratic Equations are used to define various things they define the path and motion of various objects. Suppose we through a stone in the sky then its trajectory is defined using the quadratic equations. As quadratic equations are the equation with two degrees they can have a maximum of two roots. The roots of the quadratic equation are the values that satisfy the quadratic equation.

An algebraic equation of the second degree is called the quadratic equation. The standard form of the quadratic equation is ax2 + bx + c = 0. The most important condition for a quadratic equation to exist is that the coefficient of the highest degree term i.e. the coefficient of ‘x2‘ can never be zero. In standard form, we can imply that a  ≠ 0

Some examples of quadratic equations are,

• 3x2 – 11x + 23 = 0
• 5x2 = 0
• 11x2 -13 = 0

### Standard Form of Quadratic Equation

The standard form of the Quadratic Equation is,

ax2 + bx + c = 0

• x is the variable of the equation,
• a, b, and c are real numbers and constants and a ≠ 0.

In general, any second-degree polynomial P(x), when put like P(x) = 0 represents a quadratic equation.

Example: Rahul and Ravi together have 45 candies. Both of them lost 5 candies each. The product of number of candies both of them have now is 124.  We are asked to find out the number of candies each one had in the beginning. Formulate a quadratic equation for this problem.

Let’s say Rahul had “x” candies.

Then Ravi must have “45 – x” candies because both of them had 45 candies.

Now after losing the candies, we are given that product of the number of candies they have is 124, i.e.

x(45 – x) = 124

⇒ 45x – x2 = 124

⇒ x2 – 45x + 124 = 0

This is the required quadratic equation.

## Roots of a Quadratic Equation

The values of the xo which satisfies the quadratic equation q(x) are called the roots of the quadratic equation. This implies that for any xo if q(x) = 0. Then xo is the root of the q(x).

For example, the roots of the quadratic equation q(x): 3x2 – 10x – 8 = 0 are x = -2/3 and x = 4 as

For x = -2/3,

q(-2/3) = 3(-2/3)2 – 10(-2/3) – 8
⇒ q(-2/3)  = 4/3 + 20/3 – 8
⇒ q(-2/3) = 0

For x = 4,

q(4) = 3(4)2 – 10(4) – 8
⇒ q(4)  = 32 – 40 -8
⇒ q(4)  = 0

We use various methods to find the solution to the quadratic equation. Quadratic Formula is the most common way to solve quadratic equations which is discussed below in the article.

Note: The quadratic equation is a two degrees polynomial i.e., it can have a maximum of 2 roots.

The easiest and most efficient way to calculate the roots of the quadratic equation is by using the quadratic formula. It provides the roots of the quadratic equation in the least steps. Also, not all the methods of solving quadratic equations work on all types of equations, but quadratic formulas can be used to solve any type of quadratic equation.

The quadratic formula to solve the general quadratic equation ax2 + bx + c = 0 is,

### Steps for using Quadratic Formula

Step 1: Arrange the given quadratic equation in standard form.

Step 2: Compare the given equation with the standard quadratic equation to find the values of a, b and c.

Step 4: Simplify the value obtained in the above step to get the roots of the quadratic equation.

Example: Find the roots of the quadratic equation x2 – 5x – 6 = 0

Given equation, x2 – 5x – 6 = 0

comparing with ax2 + bx + c = 0 we get,

a = 1, b = -5, c = -6

Taking Positive Sign,
x = (5 + 7) / 2 = 12 /2 = 6

Taking Negative Sign
x = (5 – 7) / 2 = -2 /2 = -1

Thus, the roots of the equation x2 – 5x – 6 = 0 are x = 6 and x = -1

Take the standard quadratic equation as, ax2 + bx + c = 0, where a ≠ 0 now,

ax2 + bx = -c

x2 + bx/a = -c/a

Completing the perfect square on the left-hand side we get,

x2+ bx/a + (b/2a)2 = -c/a + (b/2a)2

⇒ (x + b/2a)2 = -c/a + b2/4a2

⇒ (x + b/2a)2 = (b2 – 4ac)/4a2

Taking square root on both sides we get,

x + b/2a = ±√(b2 – 4ac)/2a

Thus, the quadratic formula is calculated.

## Nature of Roots of Quadratic Equation

We can easily find the nature of the roots of the quadratic equations without actually finding the roots. Generally, we represent the roots of quadratic equations with α and β symbols.

We use the concept of discriminant to find the nature of the roots. The discriminant of the quadratic equation ax2 + bx + c = 0 is calculated using the formula,

### (Discriminant) D = b2 – 4ac

Using discriminant we can find the nature of the root as,

• If D > 0, the roots of the quadratic equation are real and distinct.
• If D = 0, the roots of the quadratic equation are real and equal.
• if D < 0, the roots of the quadratic equation do not exist or the roots are imaginary.

## Sum and Product of Roots of Quadratic Equation

For any given quadratic equation we can easily find the sum and product of the roots of the quadratic equation without actually calculating the roots of the quadratic equation.

For the given quadratic equation ax2 + bx + c = 0 the sum and the product of the roots are found with the help of the coefficient of x2, coefficient of x, and the constant term. For example, if the given quadratic equation is ax2 + bx + c = 0, then the sum and product of the roots are given using the formula,

• Sum of Roots: α + β = -b/a = -Coefficient of x/ Coefficient of x2
• Product of Roots: αβ = c/a = Constant term/ Coefficient of x2

Example: Find the sum and the product of the roots of equation 2x2 + 5x + 3 = 0.

Given quadratic equation, 2x2 + 5x + 3  = 0

Comparing with, ax2 + bx + c = 0

We get, a = 2, b = 5, c =3

• Sum of Roots = α + β = -b/a = -5/2
• Product of Roots = αβ = c/a = 3/2

### Writing Quadratic Equations with Roots

If the roots of the quadratic equation α, and β are given then we can easily write the quadratic equation by using the formula,

(x – α)(x – β) = 0

Example: Find the quadratic equation whose roots are, 1 and 2.

Given, α = 1 and β = 2

Then the required quadratic equation is,

(x – α)(x – β) = 0

⇒ (x – 1)(x – 2) = 0

⇒ x2 – 2x -x + 2 = 0

⇒ x2 – 3x + 2 = 0

This is the required quadratic equation.

We can also find the quadratic equation if the sum and the product of the quadratic equation are given. Suppose the sum (S) and the product (P) of the quadratic equation are (α + β) and αβ respectively. Then the quadratic equation is given using the formula,

• x2 – (Sum)x + (Product) = 0
• x2 – (α + β)x + αβ = 0

Example: Find the quadratic equation whose sum of the roots is, 3 and the product of the root are 2.

Given, α + β = 3 and αβ = 2

Then the required quadratic equation is,

x2 – (α + β)x + αβ = 0

⇒ x2 – 3x + 2 = 0

This is the required quadratic equation.

Various formulas are used for solving quadratic equations and finding the values and nature of the roots. Some of the important formulas of the quadratic equation are,

• The standard form of the quadratic equation is ax2 + bx + c = 0
• Quadratic formula used for finding the roots of the quadratic equation is x = [-b ± √(b2 – 4ac)]/2a
• Discriminant of the quadratic equation is found using the formula D = b2 – 4ac
• If D > 0 the roots are real and distinct.
• If D = 0 the roots are real and equal.
• If D < 0 the real roots do not exist, or the roots are imaginary.
• Sum of the roots of a quadratic equation is given by α + β = -b/a.
• Product of the Root of the quadratic equation is given by αβ = c/a.
• If the sum (α + β) and the product (αβ) of the quadratic equation then the quadratic equation is given using x2 – (α + β)x + αβ = 0.
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Let’s assume a quadratic equation P(x) = 0. The points which satisfy this equation are called solutions or zeros of this quadratic equation. There are three types of methods to find the solution of a quadratic equation:

• Factorization Method
• Completing Squares Method
• Shree Dharacharya or Quadratic Formula
• Graph Method to Find the Roots

Let’s look at all these methods of finding the roots of the quadratic equations one by one through examples.

## Factorization Method of Solving Quadratic Equations

A quadratic equation can be considered a factor of two terms. Like ax2 + bx + c = 0 can be written as (x – x1)(x – x2) = 0 where x1 and x2 are roots of quadratic equation.

### Steps of Solving Quadratic Equations Using Factorization

Step 1: Find two numbers such that the product of the numbers is ‘ac’ and the sum is ‘b’.

Step 2: Then write x coefficient as the sum of these two numbers and split them such that you get two terms for x.

Step 3: Factor the first two as a group and the last two terms as another group.

Step 4: Take common factors from these and on equating the two expressions with zero after taking common factors and rearranging the equation we get the roots.

Let’s consider an example of this for better understanding.

Example: Find out the solutions of the given quadratic equation using the factorization method.

2x2 – 3x + 1 = 0

Given, 2x2 – 3x + 1 = 0

⇒ 2x2 – 2x – x + 1 = 0

⇒ 2x(x – 1) – 1(x -1) = 0

⇒ (2x – 1)(x-1) = 0

Now this equation is zero when either of these two terms or both of these terms are zero

Putting 2x – 1 = 0, we get x = 1/2

Similarly, x – 1 = 0, we get x = 1

Thus, we get two roots x = 1 and 1/2

## Completing Squares Method

Any equation ax2 + bx + c = 0 can be converted in the form (x + m)2 – n2 = 0. After this take the square roots and get the roots of the equation. Completing the square is just a way to readjust the given quadratic equation in such a way that they come in the form of squares.

Let’s see this through an example.

Example: Find the root of the given equation through complete the square method.

x2 + 4x – 5 = 0

Given, x2 + 4x – 5 = 0

Solving by Completing Square Method

x2 + 4x – 5 = 0

⇒ x2 + 4x + 4 – 9 = 0

⇒ (x + 2)2 – 32 = 0

⇒ (x + 2)2 = 32

Taking square root both sides,

x + 2 = 3 and x + 2 = -3

⇒ x = 3 -2 and x = -3 -2

⇒ x = 1 and x = -5

This formula says, for a quadratic equation in general form, ax2 + bx + c = 0 If b2 – 4ac > 0,

Then roots are given by

Example: Find the roots of equation 3x2 – 5x + 2 = 0.

For finding out the roots using Shree Dharacharya formula,

We need to check If b2 – 4ac > 0,

In this particular equation, a = 3, b = -5 and c = 2.

b2 – 4ac = (-5)2 – 4(3)(2) = 25 – 24 = 1 > 0

Thus, the roots of the quadratic equation exist, which are

## Graphical Solution of Quadratic Equation

Suppose the general form of the quadratic equation is ax2 + bx + c = 0, where a ≠ 0. The quadratic equation is a polynomial equation of 2 degrees so it comes under the conic section. Further simplifying the standard form of quadratic equation,

y = ax2 + bx + c

This resembles a parabola and we can draw its curve easily the points where this curve cut the x-axis are the solution of the quadratic equation.

## Quadratic Equations Having Common Roots

We can have two quadratic equations which have the same roots. Suppose two quadratic equations are a1x2 + b1x + c1 = 0, and a2x2 + b2x + c2 = 0, then these equation have common roots if (a1b2 – a2b1) (b1c2 – b2c1) = (a2c1 – a1c2)2

Let’s consider an example based on above mentioned condition.

Example: Check whether the equation 3x2 + 7x – 6 = 0  and the equation 6x2 + 14x – 12 = 0 have common roots or not.

Given equations, 3x2 + 7x – 6 = 0 and 6x2 + 14x – 12 = 0

Comparing with a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 we get,

a1 = 3, b1 = 7 and c1 = -6

a2 = 6, b2 = 14 and c2 = -12

Using the above condition we get,

(a1b2 – a2b1) (b1c2 – b2c1) = (a2c1 – a1c2)2

⇒ (42 – 42)(-84  + 84) = (-36+36)2

⇒ 0 = 0

Thus, the above equations have common roots.

## Maximum and Minimum Value of Quadratic Equation

For any quadratic equation q(x) = ax2 + bx + c the maximum and minimum values of the quadratic can easily be calculated. We know that we can easily plot a graph of the quadratic equations and it comes out to be a parabola.

Now, performing the necessary calculation we can state that,

• If a > 0, q(x) is minimum at x = -b/2a
• If a < 0, q(x) is maximum at x = -b/2a

Thus we can easily get the range of the quadratic equation.

• When a > 0, Range of q(x) is [q(-b/2a), ∞)
• When a < 0, Range of q(x) is (-∞, q(-b/2a)]

## Solved Examples on Quadratic Equations

We know that a quadratic equation must be of degree 2.

Let’s simplify and check the given equation.

(x – 2)(x + 1) = (x – 1)(x + 3)

⇒ x2 + x – 2x – 2 = x2 + 3x – x – 3

⇒ x2 – x – 2 = x2 + 2x – 3

⇒ -x – 2 = 2x – 3

⇒ -3x + 1 = 0

This equation is of degree 1. Thus, it cannot be a quadratic equation.

2x2 – x – 6 = 0

2x2 – x – 6 = 0
⇒ 2x2 – 4x. +3x – 6 = 0
⇒ 2x (x – 2) +3(x – 2) = 0
⇒ (2x + 3) (x – 2) = 0

2x + 3 = 0
x = -3/2

x – 2 = 0
x = 2

Thus, this equation has roots x = 2 and -3/2

Example 3: Find the root of the given equation through complete the square method.

x2 + 6x + 9 = 0

Given, x2 + 6x + 9 = 0

⇒ x2 + 6x + 9 = 0
⇒ x2 + 2(3x) + 32 = 0
⇒ (x + 3)2 = 0

Taking square root,

x + 3 = 0
x = – 3

x = -3,-3

Example 4: Find the roots of the equation x + 1/x = 3

x + 1/x = 3

Simplify the above equation and using quadratic formula,

(x2 + 1)/x = 3
x2 + 1 = 3x
x2 – 3x + 1 = 0

Comparing with ax2 + bx + c = 0

a = 1, b = -3 and c = 1

Now, Checking if b2 – 4ac > 0

b2 – 4ac  = 9 – 4(1)(1)  = 5 > 0

Example 5: Find the quadratic equation having the roots 4 and 9 respectively.

The quadratic equation having the roots α, β, is (x – α)(x – β) = 0

α = 4, and β = 9

Therefore the required quadratic equation is,

(x – 4)(x – 9) = 0

x2 – 9x – 4x + 36 = 0

x2 – 13x + 36 = 0

Thus, the required quadratic equation is x2 – 13x + 36 = 0

Example 6: The quad equation 3x2 + 5x + 9 = 0 has roots α, and β. Find the quadratic equation having the roots 1/α, and 1/β.

Given equation 3x2 + 5x + 9 = 0

Comparing with ax2 + bx + c = 0

a = 3, b = 5 and c = 9

α + β = -b/a = -5/3

αβ = c/a = 9/3 = 3

Roots of the new equation are 1/α and 1/β.

Sum of Roots = 1/α + 1/β = (α + β)/α β = (-5/3)×(1/3) = -5/9

Product of Roots = 1/α β = 1/3

Thus, the required quadratic equation is,

x2 – Sumx + Product = 0

x2 – (-5/9)x + 1/3 = 0

9x2 + 5x + 3 = 0

### Q1: What is a Quadratic Equation?

A polynomial equation of the second degree in one variable is called the quadratic equation. The general form of the quadratic equation is ax2 + bx + c = 0. Here, a and b are the coefficients of x2 and x terms respectively, and c is the constant term.

### Q2: What is the Quadratic Formula?

The general formula to solve a quadratic equation of the form ax2 + bx + c = 0 is

This is called the quadratic formula.

### Q3: What is Determinant of Quadratic Equation?

For any quadratic equation of the form ax2 + bx + c = 0, we calculate the value b2 – 4ac this is called the determinant of the quadratic equation. It is denoted by D. It is used to tell the nature of the roots of the quadratic equation.

### Q4: What are Some Real-Life Applications of Quadratic Equations?

Quadratic Equations are highly used in various scenarios some of their real-life application are,

• They find or equate the price of the commodity with their demand.
• Quadratic equations are used to determine the rate and efficiency of the various machines.
• They are used to solve speed and distance problems.
• They are used in navigational systems and the calculation of the position of objects in space.

### Q5: How to Find the Value of the Discriminant?

The general form of the quadratic equation is ax2 + bx + c = 0 and its determinant is calculated using the formula,

D = b2 – 4ac

where D is the discriminant of the equation.

### Q6: What are the 3 quadratic equations or formulas?

• Standard Form: y = ax2 + bx + c
• Factored Form: y = (ax + c)(bx + d)
• Vertex Form: y = a(x + b)2 + c

### Q7: What is another name for Quadratic Equation?

As the quadratic equation uses the variable x along with x2 it is also called as the polynomial equation of two degrees.

### Q8: What is Biqudratic Equation?

The equation with power or degree four is called a biquadratic equation. The general form of the biquadratic equation is,

ax4 + bx3 + cx2 + dx + e = 0

### Q9: What is Shri Dharacharya Formula?

Quadratic formula is also called as Shri Dharacharya Formula. For any quadratic equation ax2 + bx + c = 0 we can find its root using the Dharacharya formula as,

Given a quadratic equation in the form ax2 + bx + c, (Only the values of a, b and c are provided) the task is to find the roots of the equation.

Input:  a = 1, b = -2, c = 1
Output:  Roots are real and same 1

Input  :  a = 1, b = 7, c = 12
Output:  Roots are real and different
-3, -4

Input  :  a = 1, b = 1, c = 1
Output :  Roots are complex
-0.5 + i1.73205, -0.5 – i1.73205

## Roots of Quadratic Equation using Sridharacharya Formula

The roots could be found using the below formula (It is known as the formula of Sridharacharya)

The values of the roots depends on the term (b2 – 4ac) which is known as the discriminant (D)

If D = 0:
=> This occurs when b2 = 4ac.
=> The roots are real and equal.
=> The roots are (-b/2a).

Below is the implementation of the above formula.

## C++

using namespace std;

void findRoots(int a, int b, int c)

        cout << "Invalid";

    int d = b * b - 4 * a * c;

    double sqrt_val = sqrt(abs(d));

        cout << "Roots are real and different \n";

        cout << (double)(-b + sqrt_val) / (2 * a) << "\n"

             << (double)(-b - sqrt_val) / (2 * a);

        cout << "Roots are real and same \n";

        cout << -(double)b / (2 * a);

        cout << "Roots are complex \n";

        cout << -(double)b / (2 * a) << " + i"

             << sqrt_val / (2 * a) << "\n"

             << -(double)b / (2 * a) << " - i"

             << sqrt_val / (2 * a);

    int a = 1, b = -7, c = 12;

    findRoots(a, b, c);

void findRoots(int a, int b, int c)

    int d = b * b - 4 * a * c;

    double sqrt_val = sqrt(abs(d));

        printf("Roots are real and different \n");

        printf("%f\n%f", (double)(-b + sqrt_val) / (2 * a),

               (double)(-b - sqrt_val) / (2 * a));

        printf("Roots are real and same \n");

        printf("%f", -(double)b / (2 * a));

        printf("Roots are complex \n");

        printf("%f + i%f\n%f - i%f", -(double)b / (2 * a),

               sqrt_val / (2 * a), -(double)b / (2 * a),

               sqrt_val / (2 * a));

    int a = 1, b = -7, c = 12;

    findRoots(a, b, c);

## Java

import static java.lang.Math.*;

    static void findRoots(int a, int b, int c)

        int d = b * b - 4 * a * c;

        double sqrt_val = sqrt(abs(d));

                "Roots are real and different \n");

                (double)(-b + sqrt_val) / (2 * a) + "\n"

                + (double)(-b - sqrt_val) / (2 * a));

                "Roots are real and same \n");

            System.out.println(-(double)b / (2 * a) + "\n"

                               + -(double)b / (2 * a));

            System.out.println("Roots are complex \n");

            System.out.println(-(double)b / (2 * a) + " + i"

                               + sqrt_val / (2 * a) + "\n"

                               + -(double)b / (2 * a)

                               + " - i"

                               + sqrt_val / (2 * a));

        int a = 1, b = -7, c = 12;

        findRoots(a, b, c);

## Python3

def findRoots(a, b, c):

    if a == 0:

    d = b * b - 4 * a * c

    sqrt_val = math.sqrt(abs(d))

    if d > 0:

        print("Roots are real and different ")

        print((-b + sqrt_val)/(2 * a))

        print((-b - sqrt_val)/(2 * a))

    elif d == 0:

        print("Roots are real and same")

        print(-b / (2*a))

        print("Roots are complex")

        print(- b / (2*a), " + i", sqrt_val / (2 * a))

        print(- b / (2*a), " - i", sqrt_val / (2 * a))

if __name__ == '__main__':

    b = -7

    c = 12

    findRoots(a, b, c)

    void findRoots(int a, int b, int c)

        int d = b * b - 4 * a * c;

        double sqrt_val = Math.Abs(d);

                "Roots are real and different \n");

                (double)(-b + sqrt_val) / (2 * a) + "\n"

                + (double)(-b - sqrt_val) / (2 * a));

            Console.Write("Roots are complex \n");

            Console.Write(-(double)b / (2 * a) + " + i"

                          + sqrt_val / (2 * a) + "\n"

                          + -(double)b / (2 * a) + " - i"

                          + sqrt_val / (2 * a));

    public static void Main()

        Quadratic obj = new Quadratic();

        int a = 1, b = -7, c = 12;

        obj.findRoots(a, b, c);

## PHP

function findRoots($a, $b, $c)  if ($a == 0)

    $d = $b * $b - 4 * $a * $c;  $sqrt_val = sqrt(abs($d));  if ($d > 0)

             (-$b - $sqrt_val) / (2 * $a);  else if ($d == 0)

              $sqrt_val / (2 * $a) , "\n" , -$b / (2 * $a),

                             " - i", $sqrt_val / (2 * $a) ;

$a = 1; $b = -7 ;$c = 12; findRoots($a, $b, $c);

## Javascript

    function findRoots(a, b, c)

        let d = b * b - 4 * a * c;

        let sqrt_val = Math.sqrt(Math.abs(d));

                "Roots are real and different \n" + "<br/>");

                (-b + sqrt_val) / (2 * a) + "<br/>"

                + (-b - sqrt_val) / (2 * a));

                "Roots are real and same \n" + "<br/>");

            document.write(-b / (2 * a) + "<br/>"

                               + -b / (2 * a)) ;

            document.write("Roots are complex \n");

            document.write(-b / (2 * a) + " + i"

                               + sqrt_val / (2 * a)  + "<br/>"

                               + -b / (2 * a)

                               + " - i" + sqrt_val) / (2 * a) ;

        let a = 1, b = -7, c = 12;

        findRoots(a, b, c);

Roots are real and different
4.000000
3.000000

Time Complexity: O(log(D)), where D is the discriminant of the given quadratic equation.
Auxiliary Space: O(1)

Self Paced Course

### Using Formula in python:

• Import the math module for square root and other mathematical operations.
• Declare the coefficients a, b, and c of the quadratic equation.
• Calculate the discriminant discriminant using the formula b^2 – 4ac.
• Check if the discriminant is greater than zero, zero, or less than zero.
• If the discriminant is greater than zero, calculate two real and distinct roots using the formula (-b + sqrt(discriminant)) / (2*a) and (-b – sqrt(discriminant)) / (2*a), and print the roots with the message “Roots are real and distinct”.
• If the discriminant is equal to zero, calculate one real and same root using the formula -b / (2*a), and print the root with the message “Roots are real and same”.
• If the discriminant is less than zero, calculate two complex and different roots using the formula (-b / 2*a) +/- (sqrt(-discriminant) / (2*a)), and print the roots with the message “Roots are complex and different”.

## Python3

b = -2

discriminant = b ** 2 - 4 * a * c

if discriminant > 0:

    root1 = (-b + math.sqrt(discriminant)) / (2 * a)

    root2 = (-b - math.sqrt(discriminant)) / (2 * a)

    print("Roots are real and distinct")

    print("Root 1:", root1)

    print("Root 2:", root2)

elif discriminant == 0:

    root = -b / (2 * a)

    print("Roots are real and same")

    realPart = -b / (2 * a)

    imaginaryPart = math.sqrt(-discriminant) / (2 * a)

    print("Roots are complex and different")

    print("Root 1:", realPart, "+", imaginaryPart, "i")

    print("Root 2:", realPart, "-", imaginaryPart, "i")

Roots are real and same
Root: 1.0

Time Complexity: O(1)
Space Complexity: O(1)