Why do extraneous roots occur?

The reason extraneous solutions exist is because some operations produce ‘extra’ answers, and sometimes, these operations are a part of the path to solving the problem. When we get these ‘extra’ answers, they usually don’t work when we try to plug them back into the original problem.

What is an extraneous root in a logarithmic equation?

An extraneous solution is a number for which an original equation is either undefined or false, but for which a later equation is true. Extraneous solutions are caused by actions that can take a false (or undefined) equation to a true equation.

What does it mean to check for extraneous solutions?

Extraneous solutions are ones that appear when you work through the problem, but don’t work when plugged back into the original equation. Be careful to check your work!.

How do you find extraneous roots?

Example: you work on an equation and come up with two roots (where it equals zero): “a” and “b”. When you put “a” into the original equation it becomes zero, but when you put in “b” it doesn’t. So “b” is an extraneous root. This often happens when we square both sides during our solution.

Is an extraneous solution?

In mathematics, an extraneous solution (or spurious solution) is a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the problem.

Can extraneous solutions be negative?

Extraneous solutions are not necessarily outside the domain. But they can appear as extra solutions when we square both sides of an equation, because when we square an equation, we would get the same result whether the original equation was positive or negative.

Why is it important to check for extraneous solutions?

Incorrect values of the variable, such as those that are introduced as a result of the squaring process are called extraneous solutions. Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation.

How do you solve rational equations with extraneous solutions?

1. Solution:
2. Step 1: Factor all denominators and determine the LCD.
3. Step 2: Identify the restrictions. In this case, they are x≠−2 x ≠ − 2 and x≠−3 x ≠ − 3 .
4. Step 3: Multiply both sides of the equation by the LCD.
5. Step 4: Solve the resulting equation.
6. Step 5: Check for extraneous solutions.

What is extraneous solution in algebra?

Extraneous solutions are values that we get when solving equations that aren’t really solutions to the equation.

Do rational equations have extraneous solutions?

Whenever you solve a rational equation, always check your (interim) solution against the denominators (and their disallowed values) from the original equation. It is entirely possible that a problem will have an invalid (that is, an “extraneous”) solution. This is especially true on tests. So always check!

Why do extraneous solutions sometimes occur in equations with rational expressions?

b. Why do extraneous solutions sometimes occur in the process of solving rational or radical equations? Answer: Squaring both sides of a square-root radical equation introduces two solutions (the positive and negative) that might be the solution to the equation.

Can cube roots have extraneous solutions?

First, they must realize that an equation with a single cube-root radical will not generate any extraneous solutions; only an even-root radical will. In other words, squaring the desired equation must produce a cubic polynomial.

Which equation has irrational and unequal roots?

What are the 4 nature of roots?

The word ‘nature’ refers to the types of numbers the roots can be — namely real, rational, irrational or imaginary.

What equations have no real roots?

Case 1: No Real Roots Since the quadratic formula requires taking the square root of the discriminant, a negative discriminant creates a problem because the square root of a negative number is not defined over the real line. An example of a quadratic function with no real roots is given by, f(x) = x2 − 3x + 4.

How do you know if a polynomial has real roots?

The terms solutions/zeros/roots are synonymous because they all represent where the graph of a polynomial intersects the x-axis. The roots that are found when the graph meets with the x-axis are called real roots; you can see them and deal with them as real numbers in the real world.

How do you find all real roots?

You can find the roots, or solutions, of the polynomial equation P(x) = 0 by setting each factor equal to 0 and solving for x. Solve the polynomial equation by factoring. Set each factor equal to 0. 2×4 = 0 or (x – 6) = 0 or (x + 1) = 0 Solve for x.

How do you prove a polynomial has no real roots?

Unless x is between 0 and 1, the first two terms are positive, and so the polynomial is positive. Even if x is between 0 and 1, the first two terms are tiny in magnitude, certainly each individually greater than −1, so that when 15 is added to their sum, the result is positive. Thus the polynomial has no real roots.

How many real roots does the polynomial?

How many real roots does the polynomial 2x 3 8x 7 have?

The polynomial has 1 real roots and 2 imaginary root.

How many real roots does the polynomial 2×5 8x 7 0 have?

It is negative at x=0 and positive at x=1, thus it has exactly one real root, and it’s somewhere between those two points.

How do you find all roots of a polynomial?

How Many Roots? Examine the highest-degree term of the polynomial – that is, the term with the highest exponent. That exponent is how many roots the polynomial will have. So if the highest exponent in your polynomial is 2, it’ll have two roots; if the highest exponent is 3, it’ll have three roots; and so on.

What are real roots?

In algebra, a real root is a solution to a particular equation. The term real root means that this solution is a number that can be whole, positive, negative, rational, or irrational. However, the solution to an equation can be real roots, complex roots or imaginary roots.

What are real and complex roots?

The factors that are first-degree polynomials are real roots of the original polynomial. The factors that are second-degree polynomials can’t be reduced using real numbers. Therefore, whenever a complex number is a root of a polynomial with real coefficients, its complex conjugate is also a root of that polynomial.

What is real and distinct roots?

If the discriminant is greater than zero, this means that the quadratic equation has two real, distinct (different) roots. If the discriminant is equal to zero, this means that the quadratic equation has two real, identical roots.

extraneous root

English-Russian scientific dictionary.
.

Смотреть что такое «extraneous root» в других словарях:

• С русского на:
• Английский
• С английского на:
• Все языки
• Болгарский
• Испанский
• Русский
• Сербский
• Тамильский

См. также в других словарях:

extraneous root

1) посторонний корень

2) лишний корень, побочный корень

Универсальный англо-русский словарь.
.
.

Смотреть что такое «extraneous root» в других словарях:

Solution 1

Suppose you have two expressions $e_1$ and $e_2$ and you know
$$e_1 = e_2.$$
Then, if you apply a function to both sides, you have
$$f(e_1) = f(e_2).$$
However, this logic in general does not reverse, unless the function
$f$ is 1-1. This is the mechanism by which extraneous roots get introduced.

When you square both sides of an equation, you are destroying information about the signs of the two sides. Now, the equality will match if the two sides have the same absolute value. This process can, and often does, introduce spurious roots.

Solution 2

Extraneous solutions are often the result of omitting a constraint during the formulation or solution of a problem. For example, the correct rule for solving absolute value equations is

If we use this rule then extraneous solutions do not occur.

Extraneous solutions often arise from using a rule of the form
$$x = y \implies f(x) = f(y).$$
Squaring both sides of an equation is an example of such a rule.

If $f$ is one-to-one, then the rule
$$f(x) = f(y) \iff x = y$$
is valid, provided that $x$ and $y$ are both in the domain of $f$. Ignoring this condition can lead to extraneous solutions. The equation $\log(x-4) = \log(2x-6)$ provides an example.

Extraneous solutions can also result from ignoring physical constraints in applied problems (e.g. length and mass are positive quantities).

Solution 3

extraneous roots can occur when solving rational equations — this results when multiplying the equation to clear the fractions and cancelling a 0/0 term. you can also end up with extraneous roots with logarithmic equations because you will end up with x values that make the argument negative (not in the domain of log functions)

Solution 4

Here is an unusual extraneous root, based on a video by Pesh Talwalker, , but slightly different. Consider . This has the roots or and . is not a root. So divide both sides by :

Now $x^2 = x+1$ so that

$x-1 = x+1-2 = x^2-2$

Substitute in (*) to get

Multiply both sides by to get

This equation has and as roots all right, but it also has the extraneous root .

Solution 5

Another less obvious way that extraneous roots may be introduced is through, lacking a better word, «careless commingling».

The less obvious part is that the solutions of the system do not necessarily satisfy the original problem, in other words solving it this way potentially introduces extraneous solutions. Reason is that the problem assumes to be the roots of the equation, with emphasis on the plural of «roots«. But the system of equations loses that distinction, and allows for and to be one and the same root.

Below is an example of this happening, based on an actual question.

Find knowing that the roots of the equation $x^2 — \lambda a x + b=0$ are and .

FWIW not all solutions where are necessarily extraneous, for example is a valid solution with , and so is when . The only safe recourse in this case is to verify each solution individually (unless the case is handled completely separately).

Related videos on Youtube

05 : 19

WCLN — Math — Extraneous Roots

03 : 30

Solving a Radical Equation When You Have One Extraneous Solution

08 : 27

When and why extraneous solution happen

05 : 22

09 : 29

Comments

• I asked this question some time ago in MO, where I got ground in the dirt like a wet french fry (as Joe Bob would say). So, I’m transferring the question here to MSE. 🙂

  edit (1.Jan.2017):
  In general, in mathematics and the real world, you get extraneous roots any time you are initially presented with (via some mechanical / automated / canonical process) a superset of the set that you want, and the sifting out of that set from the superset is left to you, for example, when panning for gold, or when reading the owner’s manual for your vehicle: “This owner’s manual covers all models of your vehicle. You may find descriptions of equipment and features that are not on your particular model.”

  • Extraneous roots come up in log equations.

  • Extraneous roots happen whenever you apply functions to both sides of an equation that aren’t invertible.

• Excellent! I’ve up-voted your answer.

• Ncmathsadist, can you be more specific as to how destroying signs will affect the number of solution?

• No square root can be equal to negative number. And square of positive is positive

• Notice that, when you substitute for x-1, you’re applying a function to only 1 occurrence of x in equation (*), and in only one side of this equation. So, even though the function you’re applying is reversible, the equation derivation isn’t. You would get the same result if you substituted 1 + 1/x for x in the quadratic equation $x^2 — x — 1$.

Example: you work on an equation and come up with two roots (where it equals zero): “a” and “b”. When you put “a” into the original equation it becomes zero, but when you put in “b” it doesn’t. So “b” is an extraneous root.

How do you know when a radical equation has an extraneous solution?

When you square a radical equation you sometimes get a solution to the squared equation that is not a solution to the original equation. Such an equation is called an extraneous solution. Remember to always check your solutions in the original equation to discard the extraneous solutions.

How do you tell if there are any extraneous solutions?

It’s not that your process is wrong; it’s just that this solution does not fit back into the equation (math is very complicating sometimes). To find whether your solutions are extraneous or not, you need to plug each of them back in to your given equation and see if they work.

What is extraneous root solution?

An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x , 1x − 2+1x + 2=4(x − 2)(x + 2) .

What’s an extraneous root?

Why do we get extraneous solutions when solving radical equations?

The reason extraneous solutions exist is because some operations produce ‘extra’ answers, and sometimes, these operations are a part of the path to solving the problem. When we get these ‘extra’ answers, they usually don’t work when we try to plug them back into the original problem.

What is the extraneous?

1 : existing on or coming from the outside extraneous light. 2a : not forming an essential or vital part extraneous ornamentation. b : having no relevance an extraneous digression. 3 : being a number obtained in solving an equation that is not a solution of the equation extraneous roots.

What are extraneous roots?

An “extraneous root” may be defined as “a value obtained. for an unknown in the solution of an equation which is not a. root of the equation.” It may be noted first that authors of. textbooks usually fail to give a definition of extraneous roots ; second, that the definition given above is faulty, in that it leaves.

Why do radical equations have extraneous solutions?

In general, extraneous solutions arise when we perform non-invertible operations on both sides of an equation. (That is, they sometimes arise, but not always.) Solving equations involving square roots involves squaring both sides of an equation.

What is an extraneous solution of an equation?

Extraneous solutions are values that we get when solving equations that aren’t really solutions to the equation.

How do you solve a radical equation in Algebra?

Isolate the radical term. Square both sides to remove the term x + 2 x + 2 from the radical. Now simplify and solve the equation. Combine like terms, and then factor. Set each factor equal to zero and solve for x. Now check both solutions by substituting them into the original equation.

How to remove the radical symbol from an equation?

Key Steps: 1 Isolate the radical symbol on one side of the equation 2 Square both sides of the equation to eliminate the radical symbol 3 Solve the equation that comes out after the squaring process 4 Check your answers with the original equation to avoid extraneous values More

How to solve a radical equation in chilimath?

Key Steps: 1 Isolate the radical symbol on one side of the equation 2 Square both sides of the equation to eliminate the radical symbol 3 Solve the equation that comes out after the squaring process 4 Check your answers with the original equation to avoid extraneous values

Why is 2 an extraneous solution to X?

x = 2 But 2 is excluded from the domain of the original equation because it would make the denominator of one of the fractions zero–and division by zero is not allowed! . Therefore, it cannot be a root of the original equation. So, 2 is an extraneous solution. So, the equation has no solutions.

https://youtube.com/watch?v=dq0p04tQFUY%3Ffeature%3Doembed

Extraneous Roots

Because equations involving rational expressions have variables in denominators, a
root to the equation might cause a 0 to appear in a denominator. In this case the root
does not satisfy the original equation, and so it is called an extraneous root.

An equation with an extraneous root

Because x² — 2x = x(x — 2), the LCD for x, x — 2, and x²
— 2x is x(x — 2).

Neither 0 nor 2 could be a solution because replacing x by either 0 or 2 would cause
0 to appear in a denominator in the original equation. So 2 is an extraneous root and
the solution set is the empty set, .

An equation with an extraneous root

Because the LCD is x — 2, we multiply each side by x — 2: