I b is not Root of Auxiliary Equation

Here are a few activities for you to practice.

Consider the nonhomogeneous linear second order ODE with constant coefficients:

$(1): \quad y» + p y’ + q y = \map R x$

Let $\map R x$ be a linear combination of sine and cosine:

$\map R x = \alpha \sin b x + \beta \cos b x$

such that $i b$ is not a root of the auxiliary equation to $(1)$.

Consider the nonhomogeneous linear second order ODE with constant coefficients:

$(1): \quad y» + p y’ + q y = \map R x$

Let $\map R x$ be a linear combination of sine and cosine:

$\map R x = \alpha \sin b x + \beta \cos b x$

such that $i b$ is not a root of the auxiliary equation to $(1)$.

The method of undetermined coefficients is a technique for solving a nonhomogeneous linear second order ODE with constant coefficients:

$(1): \quad y» + p y’ + q y = \map R x$
an exponential
a sine or a cosine
a polynomial

or a combination of such real functions.

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators … (previous) … (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.5$ Particular solution: trigonometric $\map f x$
• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators … (previous) … (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.2$ The general equation
• 1972: George F. Simmons: Differential Equations … (previous) … (next): $\S 3.18$: The Method of Undetermined Coefficients
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) … (previous) … (next): linear differential equation with constant coefficients

In this mini-lesson, we will explore about the nature of roots of a quadratic equation.

You will learn about the nature of roots of quadratic equation using the discriminant formula, quadratic formula, roots of a cubic equation, real roots, unreal roots, irrational roots, imaginary roots and other interesting facts around the topic. You can also check out the playful calculators to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page.

The word «Quadratic» is derived from the word «Quad» which means square.

In other words, a quadratic equation is an “equation of degree 2.”

There are many scenarios where quadratic equations are used.

Did you know that when a rocket is launched, its path is described by a quadratic equation?

Elsie has a two-digit secret number.

She gives a few hints to her friend Mia to crack it.

She says, «It is the value of the discriminant of the quadratic equation $$x^2+9x+14=0$$.»

Can you guess the lucky number?

The quadratic equation is $$x^2+9x+14=0$$

On comparing it with $$ax^2+bx+c=0$$, we get $$a=1$$, $$b=9$$, and $$c=14$$

Let’s use the discriminant formula to find the discriminant.

$$\therefore$$ The lucky number is 25

Flora loves collecting flowers.

Each of the flowers has a number and the answer to this question is shown on the flowers.

Can you help her find the answer?

We get $$a=k$$, $$b=-2k$$ and $$c=6$$

As both the roots of the quadratic equation are real and equal, the value of the discriminant $$b^2-4ac$$ is zero.

Now the flowers do not contain the number 0

$$\therefore$$, The correct number is 6

Riley is trying to draw the graph of a quadratic equation. The conditions for the nature of the roots of a quadratic equation are given as:

• Both the roots are real
• One of the roots is negative and the other root is positive

Can you help her draw the possible sketches of the quadratic equation?

Method and Proof

$y» + p y’ + q y = 0$

From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

$\map {y_g} x + \map {y_p} x$

is the general solution to $(1)$.

Let $\map R x = \alpha \sin b x + \beta \cos b x$.

Consider the auxiliary equation to $(1)$:

$(2): \quad m^2 + p m + q = 0$

We are given that $i b$ is not a root of $(2)$.

Assume that there is a particular solution to $(1)$ of the form:

$y_p = A \sin b x + B \cos b x$

From Euler’s Formula:

$\cos b x + i \sin b x = e^{i b x}$
$A \sin b x + B \cos b x$ is the real part of $\paren {A — i B} \paren {\cos b x + i \sin b x} = \paren {A — i B} e^{i b x}$

It is assumed that $A$, $B$, $p$ and $q$ are all real numbers.

Suppose we have found a solution $y$ of $(1)$ where:

$\map f x = \map {f_1} x + i \, \map {f_2} x$

where $\map y x$ and $\map f x$ are complex-valued.

${y_1}» + p {y_1}’ + q y_1 + i \paren { {y_2}» + p {y_2}’ + q y_2} = \map {f_1} x + i \, \map {f_2} x$

Equating real parts:

${y_1}» + p {y_1}’ + q y_1 = \map {f_1} x$

Equating imaginary parts:

${y_2}» + p {y_2}’ + q y_2 = \map {f_2} x$

Thus if $y$ is a particular solution to $(1)$ when the right hand side is $\map f x$:

$\map \Re y$ is a particular solution to $(1)$ when the right hand side is $\map \Re {\map f x}$
$\map \Im y$ is a particular solution to $(1)$ when the right hand side is $\map \Im {\map f x}$

Hence, when we have $A \cos b x + B \sin b x$ on the right hand side:

replace it with $\paren {A — i B} e^{i b x}$
use the Method of Undetermined Coefficients for Exponential functions

and then take its real part.

Method and Proof

$y» + p y’ + q y = 0$

From General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

$\map {y_g} x + \map {y_p} x$

is the general solution to $(1)$.

Let $\map R x = \alpha \sin b x + \beta \cos b x$.

Consider the auxiliary equation to $(1)$:

$(2): \quad m^2 + p m + q = 0$

We have that $i b$ is not a root of $(2)$.

Assume that there is a particular solution to $(1)$ of the form:

$y_p = A \sin b x + B \cos b x$

Inserting into $(1)$:

Hence $A$ and $B$ can be expressed in terms of $\alpha$ and $\beta$:

$y_p = \dfrac {\alpha \paren {q — b^2} + \beta b p} {\paren {q — b^2}^2 + b^2 p^2} \sin b x + \dfrac {\beta \paren {q — b^2} — \alpha b p} {\paren {q — b^2}^2 + b^2 b^2} \cos b x$

Assume that there is a particular solution to $(1)$ of the form:

$y_p = A \sin b x + B \cos b x$

From Euler’s Formula:

$\cos b x + i \sin b x = e^{i b x}$
$A \sin b x + B \cos b x$ is the real part of $\paren {A — i B} \paren {\cos b x + i \sin b x} = \paren {A — i B} e^{i b x}$

It is assumed that $A$, $B$, $p$ and $q$ are all real numbers.

Suppose we have found a solution $y$ of $(1)$ where:

$\map f x = \map {f_1} x + i \, \map {f_2} x$

where $\map y x$ and $\map f x$ are complex-valued.

${y_1}» + p {y_1}’ + q y_1 + i \paren { {y_2}» + p {y_2}’ + q y_2} = \map {f_1} x + i \, \map {f_2} x$

Equating real parts:

${y_1}» + p {y_1}’ + q y_1 = \map {f_1} x$

Equating imaginary parts:

${y_2}» + p {y_2}’ + q y_2 = \map {f_2} x$

Thus if $y$ is a particular solution to $(1)$ when the right hand side is $\map f x$:

$\map \Re y$ is a particular solution to $(1)$ when the right hand side is $\map \Re {\map f x}$
$\map \Im y$ is a particular solution to $(1)$ when the right hand side is $\map \Im {\map f x}$

Hence, when we have $A \cos b x + B \sin b x$ on the right hand side:

replace it with $\paren {A — i B} e^{i b x}$
use the Method of Undetermined Coefficients for Exponential functions

and then take its real part.

One proves in the curriculum that the number $$e$$, base of the Napierian logarithms, is not a rational number. We should, it seems to me, add that the same method also proves that $$e$$ cannot be the root of a second-degree equation with rational coefficients, so that we cannot have $$ae + b / e = c$$, where $$a$$ is a positive integer and $$b, c$$ are positive or negative integers. Indeed, if we replace $$e$$ and $$1 /e$$ or $$e^{-1}$$ in this equation by their expansions derived from that of $$e^{x}$$, then multiply each side of the equation by $$1 \cdot 2 \cdot 3 \ldots n$$, we easily find that: ${a \over {n+1}}\left(1 + {1 \over {n+2}} + \cdots \right) \pm {b \over {n+1}}\left(1 — {1 \over {n+2}} + \cdots \right) = \mu,$ $$\mu$$ being an integer. We can always make sure that the factor $\pm {b \over {n+1}}$ is positive; it will suffice to suppose $$n$$ is even if $$b \lt 0$$ and $$n$$ is odd if $$b \gt 0$$; then taking $$n$$ very large, the equation we just wrote leads to a contradiction; because the left side is positive and very small, being between $$0$$ and $$1$$, and so cannot be equal to an integer $$\mu$$. So, etc.

Delving into Liouville’s Proof

his is pretty straightforward down to the equation; let’s flesh it out past there, starting by adding one more term to those sums to make the pattern clear:

making the key equation:

Which sign obtains depends on the parity of $$n$$. To see this, work through the example of $$n = 2$$. Multiplying the original equation by $$2!$$ results in:

The minus sign appears similarly whenever $$n$$ is even, and in like way a positive sign is in order when $$n$$ is odd. The upshot is that (1) can be written more specifically as:

The sum in $$\tau_n$$ is an alternating series with initial term equal to one and subsequent terms decreasing monotonically in absolute value, so that sum converges and $$\tau_n \rightarrow 0$$ like $$\sigma_n$$ does. In fact, $$\tau_n \lt \sigma_n$$. The $$\tau_n$$ are positive too. To see this, note that the alternating nature of the sum in $$\tau_n$$ implies that:

Putting this together:

and the chart here shows that the four values are close (note that the ratio of the lower and upper limits is one). Recall that $$a$$ and $$b$$ are fixed integers, coefficients of the original quadratic equation. Like Liouville says, we can assume $$a \gt 0.$$ We want to show that the left side of (2) can be constrained to be strictly between $$0$$ and $$1$$, which would be a contradiction, considering that the right side of (2), $$\mu$$, is an integer. If $$b$$ is positive, taking increasingly large odd values of $$n$$ results in everything on the left side of (2) being positive and as small as desired — less than one, in particular. Contrarily if $$b$$ is negative, take large positive values of $$n$$ to reverse the sign of $$\tau_n$$, again driving the value of the left side of (2) down to a small positive value. Either way there is a contradiction — there can be not quadratic equation with integer coefficients satisfied by $$e$$. QED.

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February 11, 2018

2. Joseph Liouville 1809–1882: Master of Pure and Applied Mathematics, by Jesper Lützen (Springer-Verlag, 1990), ISBN 978-1-4612-6973-1. Liouville regarded it as such: p. 79.

3. Sur la fonction exponentielle, by C. Hermite, Comptes rendus hebdomadaires des séances de l’Académie 77, 1873, pp. 18-24, 74-79; 226-233, 285-293. The original work was spread across four articles in Comptes rendus, as indicated. The entire article is in Volume 3 of Hermite’s Oeuvres, pp. 150-181.

4. Ueber die Transcendenz der Zahlen e und π, by David Hilbert, Mathematische Annalen, Vol. 43 (2-3), June 1893, pp. 216-219.

Now, we shall have two important theorems on the nature of the roots an algebraic equation. These theorems have wide range of applications in the theory of equations.

In an algebraic equation with rational coefficients irrational roots are occur in conjugate pairs.

That is α – √β is another root of the equation f(x) = 0, if α + √β be a root.

In an algebraic equation with real coefficients, imaginary roots occur in conjugate pairs.

That is, if α + iβ (β ≠ 0) is a root of the equation f(x) = 0, α – iβ is another root.

The occurrence of irrational or imaginary roots of multiplicity m is an algebraic equation with rational or real coefficients implies the occurrence of conjugate irrational or imaginary roots of the same multiplicity.

General Properties of Algebraic Equations

In this section a few general properties of algebraic equations will be discussed. The properties indicate the existence and location of real roots of an equation.

If, for two distinct real numbers α and β, f(α) and f(β) are of opposite signs, then the polynomial equation f(x) = 0 must have one real root lying between α and β.

(i) If f(α) and f(β) are of opposite signs, then the equation f(x) = 0 has an odd number of real roots between α and β.

(ii) If f(α) and f(β) are of same sign, then the equation f(x) = 0 has either an even number of real roots or no real root between α and β.

(i) An equation of an odd degree must have at least one real root, opposite in sign to that of the last term, the leading term being positive.

(ii) An equation of an odd degree, whose last term is negative, has at least two real roots, one positive and the other negative.

Multiple Roots

If α1 is a root of the equation f(x) = 0 of multiplicity r, then α1 is a root of the equation f’(x) = 0 of multiplicity (r – 1), where f’(x) is the first derived function of f(x).

In order to obtain the multiple root of the equation f(x) = 0, the highest common factor (H.C.F) of f(x) and f’(x) are worked out. The HCF will give the common root of f(x) = 0 and f’(x) = 0.

If the HCF is of the form (x – α)m, then α is a root of the equation f(x) = 0 of multiplicity (m + 1).

If one root of the equation x4 – 3x3 – 5x2 + 9x – 2 = 0 is 2 – √3, find the other roots.

The coefficients of the given equation are rational numbers. Irrational roots in such equation must occur in conjugate pairs.

If 2 – √3 is a root of this equation, 2 + √3 must also be another root.

should be a factor of (x4 – 3x3 – 5x2 + 9x – 2).

On being divided by this factor, the given equation reduces to

Therefore, the other roots of the equation are: -2, 1, 2 + √3.

Solve x4 – x3 + 2x2 – 2x + 4 = 0, (1 + i) being one of its roots.

Since the coefficients of the equation are real and 1 + i is a root of the equation, so 1 – i is also a root of the equation.

is a factor of (x4 – x3 + 2x2 – 2x + 4).

Thus the other roots is given by,

Thus the roots of the given equation are

Has all its roots real.

If possible, let α + iβ be an imaginary root of the equation.

Then we have,

Again, since the coefficients of the equations are real, α – iβ is also a root of the equation. Thus we have

Subtracting (2) from (1), we get

Find the equation of the fourth degree with rational coefficients, one root of which is √2 + √3i.

Since one root is √2 + √3i, the other root is √2 – √3i.

Since the equation is of degree four with rational coefficients. Therefore other quadratic factor must be

Hence the required equation is

Show that the equation (x – a)3+ (x – b)3 + (x – c)3 + (x – d)3 = 0 where a, b, c, d are positive and not all equal has only one real root.

Let f(x) = (x – a)3+ (x – b)3 + (x – c)3 + (x – d)3  since the equation is of degree 3, it has either only one real root or three real roots.

Let us assume that the equation has more than one real root. Let α, β be such. Then f’(x) = 0 has a real root between α and β.

The discriminant of the equation f’(x) = 0 is

Since a, b, c, d are all positive and not all equal, so

Therefore, the discriminant of the equation f’(x) = 0 is negative and so f’(x) has no real root. This proves that f(x) = 0 has only one real root.

Prove that the equation xn – qxn-m + r = 0 has two equal roots if

Let f(x) = xn – qxn-m + r = 0

If f(x) = 0 has two equal roots then f(x) = 0 and f’(x) = 0 have a common root and the condition is obtained by eliminating x between these two equations.

From f’(x) = 0 we have,

Putting this value of x in f(x) = 0, we have

Show that the equation xn – nqx + (n – 1)r = 0 will have a pair of equal roots if qn = rn-1.

Let f(x) = xn – nqx + (n – 1)r

If f(x) = 0 has a pair of equal roots, then this root must be a root of f’(x) = 0.

Hence q1/(n-1) is a root of f(x) = 0

Determine the multiple roots of the equation x5 + 2x4 + 2x3 + 4x2 + x + 2 = 0

The HCF of f(x) and f’(x) is obtained as x2 + 1.

Therefore f(x) = 0 has two multiple roots i of order 2 and –i of order 2.

Solve the equation x4 – x3 + 2×2 – x + 1 = 0 which has four distinct roots of equal moduli.

Let r be the modulus. Since the roots are all distinct, two possibilities may occur.

I. Two roots are real and two complexes. The roots are r, -r, r(cosθ ± isinθ)

II. All are complex roots. The roots are r(cosθ ± isinθ) and r(cosφ ± isinφ).

The given equation is identical with

This cannot happen, since there is a term containing x2 in the given equation.

The given equation in identical with

Show that the equation x3 + 2x2 – 2x – 1 = 0 has one positive root and two negative roots – one lying between -3 and -1 and another lying between -1 and 0.

Therefore there is at least one positive root.

Thus there is at least one root lying between -1 and 0.

Hence there is at least one root lying between -3 and -1.

Since the equation has exactly three roots, only one is positive, only one lies between -1 and 0 and only one lies between -3 and -1.

Method and Proof

$y» + p y’ + q y = 0$

Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

$\map {y_g} x + \map {y_p} x$

is the general solution to $(1)$.

Consider the auxiliary equation to $(1)$:

$(2): \quad m^2 + p m + q = 0$

There are three cases to consider.

$a$ is not a root of $(2)$

Assume that there is a particular solution to $(1)$ of the form:

$y_p = A e^{a x}$

Inserting into $(1)$:

$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$

From Solution of Constant Coefficient Homogeneous LSOODE, $y_g$ depends on whether $(2)$ has equal or unequal roots.

Let $m_1$ and $m_2$ be the roots of $(2)$.

$y = \begin{cases} C_1 e^{m_1 x} + C_2 e^{m_2 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 \ne m_2: m_1, m_2 \in \R \\ C_1 e^{m_1 x} + C_2 x e^{m_1 x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = m_2 \\ e^{r x} \paren {C_1 \cos s x + C_2 \sin s x} + \dfrac {K e^{a x} } {a^2 + p a + q} & : m_1 = r + i s, m_2 = r — i s \end{cases}$

is the general solution to $(1)$.

$a$ is a root of $(2)$

Let the auxiliary equation to $(2)$ have two unequal real roots $a$ and $b$.

Assume that there is a particular solution to $(1)$ of the form:

$y_p = A x e^{a x}$

Inserting into $(1)$:

$y_p = \dfrac {K x e^{a x} } {2 a + p}$

and so from Solution of Constant Coefficient Homogeneous LSOODE:

$y = C_1 e^{a x} + C_2 e^{b x} + \dfrac {K x e^{a x} } {2 a + p}$

is the general solution to $(1)$.

$a$ is a repeated root of $(2)$

If the auxiliary equation to $(2)$ has two equal real roots $a$, then:

$a = — \dfrac p 2$

and so not only:

$a^2 + p a + q = 0$
$2 a + p = 0$

So, assume that there is a particular solution to $(1)$ of the form:

$y_p = A x^2 e^{a x}$

Inserting into $(1)$:

$y_p = \dfrac {K x^2 e^{a x} } 2$
$y = y_g + \dfrac {K x^2 e^{a x} } 2$

and so from Solution of Constant Coefficient Homogeneous LSOODE:

$y = C_1 e^{a x} + C_2 x e^{a x} + \dfrac {K x^2 e^{a x} } 2$

is the general solution to $(1)$.

Sine and Cosine Functions

Let $\map R x = \alpha \sin b x + \beta \cos b x$.

Consider the auxiliary equation to $(1)$:

$(2): \quad m^2 + p m + q = 0$

There are two cases which may apply.

$i b$ is not Root of Auxiliary Equation

First we investigate the case where $i b$ is not a root of the auxiliary equation to $(1)$.

Assume that there is a particular solution to $(1)$ of the form:

$y_p = A \sin b x + B \cos b x$

Inserting into $(1)$:

Hence $A$ and $B$ can be expressed in terms of $\alpha$ and $\beta$:

$y_p = \dfrac {\alpha \paren {q — b^2} + \beta b p} {\paren {q — b^2}^2 + b^2 p^2} \sin b x + \dfrac {\beta \paren {q — b^2} — \alpha b p} {\paren {q — b^2}^2 + b^2 b^2} \cos b x$

Assume that there is a particular solution to $(1)$ of the form:

$y_p = A \sin b x + B \cos b x$

From Euler’s Formula:

$\cos b x + i \sin b x = e^{i b x}$
$A \sin b x + B \cos b x$ is the real part of $\paren {A — i B} \paren {\cos b x + i \sin b x} = \paren {A — i B} e^{i b x}$

It is assumed that $A$, $B$, $p$ and $q$ are all real numbers.

Suppose we have found a solution $y$ of $(1)$ where:

$\map f x = \map {f_1} x + i \, \map {f_2} x$

where $\map y x$ and $\map f x$ are complex-valued.

${y_1}» + p {y_1}’ + q y_1 + i \paren { {y_2}» + p {y_2}’ + q y_2} = \map {f_1} x + i \, \map {f_2} x$

Equating real parts:

${y_1}» + p {y_1}’ + q y_1 = \map {f_1} x$

Equating imaginary parts:

${y_2}» + p {y_2}’ + q y_2 = \map {f_2} x$

Thus if $y$ is a particular solution to $(1)$ when the right hand side is $\map f x$:

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$\map \Re y$ is a particular solution to $(1)$ when the right hand side is $\map \Re {\map f x}$
$\map \Im y$ is a particular solution to $(1)$ when the right hand side is $\map \Im {\map f x}$

Hence, when we have $A \cos b x + B \sin b x$ on the right hand side:

replace it with $\paren {A — i B} e^{i b x}$
use the Method of Undetermined Coefficients for Exponential functions

and then take its real part.

$i b$ is Root of Auxiliary Equation

Now suppose that $(1)$ is of the form $y» + b^2 y = A \sin b x + B \cos b x$.

Thus one of the $i b$ is one of the roots of the auxiliary equation to $(1)$.

From Linear Second Order ODE: $y» + k^2 y = 0$ the general solution to $(2)$ is:

$y = C_1 \sin b x + C_2 \cos b x$

and it can be seen that an expression of the form $A \sin b x + B \cos b x$ is already a particular solution of $(2)$.

Thus we have:

$\paren {q — b^2}^2 + b^2 p^2 = 0$.

But using the Method of Undetermined Coefficients in the above manner, this would result in an attempt to calculate:

$\dfrac {\alpha \paren {q — b^2} + \beta b p} {\paren {q — b^2}^2 + b^2 p^2}$
$\dfrac {\beta \paren {q — b^2} — \alpha b p} {\paren {q — b^2}^2 + b^2 p^2}$

both of which are are undefined.

Assume that there is a particular solution to $(1)$ of the form:

$y_p = x \paren {A \sin b x + B \cos b x}$

Inserting into $(1)$:

Hence $A$ and $B$ can be expressed in terms of $\alpha$ and $\beta$:

$y_p = \dfrac {\beta x \sin b x} {2 b} — \dfrac {\alpha x \cos b x} {2 b}$

Assume that there is a particular solution to $(1)$ of the form:

$y_p = x \paren {A \sin b x + B \cos b x}$

From Euler’s Formula:

$\cos b x + i \sin b x = e^{i b x}$
$x \paren {A \sin b x + B \cos b x}$ is the real part of $x \paren {A — i B} \paren {\cos b x + i \sin b x} = x \paren {A — i B} e^{i b x}$

It is assumed that $A$, $B$, $p$ and $q$ are all real numbers.

Suppose we have found a solution $y$ of $(1)$ where:

$\map f x = \map {f_1} x + i \, \map {f_2} x$

where $\map y x$ and $\map f x$ are complex-valued.

${y_1}» + p {y_1}’ + q y_1 + i \paren { {y_2}» + p {y_2}’ + q y_2} = \map {f_1} x + i \, \map {f_2} x$

Equating real parts:

${y_1}» + p {y_1}’ + q y_1 = \map {f_1} x$

Equating imaginary parts:

${y_2}» + p {y_2}’ + q y_2 = \map {f_2} x$

Thus if $y$ is a particular solution to $(1)$ when the right hand side is $\map f x$:

$\map \Re y$ is a particular solution to $(1)$ when the right hand side is $\map \Re {\map f x}$
$\map \Im y$ is a particular solution to $(1)$ when the right hand side is $\map \Im {\map f x}$

Hence, when we have $A \cos b x + B \sin b x$ on the right hand side:

replace it with $x \paren {A — i B} e^{i b x}$
use the Method of Undetermined Coefficients for Exponential functions

and then take its real part.

Exponential of Sine and Cosine Functions

Substitute a trial solution of similar form, either:

$e^{a x} \paren {A \sin b x + B \cos b x}$

or replace the right hand side of $(1)$ by:

$\paren {\alpha — i \beta} e^{i \paren {a + i b} x}$

find a solution, and take the real part.

$x e^{a x} \paren {A \sin b x + B \cos b x}$
$x \paren {\alpha — i \beta} e^{i \paren {a + i b} x}$

Assume that there is a particular solution to $(1)$ of the form:

$\ds y_p = \sum_{j \mathop = 0}^n A_j x^j$

Inserting into $(1)$:

The coefficients $A_0$ to $A_n$ can hence be solved in terms of $a_0$ to $a_n$ using the techniques of simultaneous equations.

The general form is tedious and unenlightening to evaluate.

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.

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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.

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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.

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.

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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.

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What Do You Mean By Nature of Roots?

An equation of the form $$ax^2+bx+c=0$$, where $$a \neq 0$$ is called a quadratic equation.

The standard form of a quadratic equation looks like this:

This is a quadratic equation in variable $$x$$.

But always remember that $$a$$ is a non-zero value.

Now, to find the nature of the roots of the quadratic equation, we will first calculate the roots of the equation.

Nature of roots specifies that the equation has real roots, irrational roots, or imaginary roots.

Imaginary roots are also known as unreal roots.

Discriminant Formula

The discriminant (Δ or D) of any polynomial is in terms of its coefficients. Here are the discriminant formulas for a cubic equation and quadratic equation.

Let us see how to use these formulas to find the discriminant.

Discriminant

The discriminant is widely used in the case of quadratic equations and is used to find the nature of the roots. Though finding a discriminant for any polynomial is not so easy, there are formulas to find the discriminant of quadratic and cubic equations that make our work easier.

What is Discriminant in Math?

Discriminant of a polynomial in math is a function of the coefficients of the polynomial. It is helpful in determining what type of solutions a polynomial equation has without actually finding them. i.e., it discriminates the solutions of the equation (as equal and unequal; real and nonreal) and hence the name «discriminant». It is usually denoted by Δ or D. The value of the discriminant can be any real number (i.e., either positive, negative, or 0).

The mini-lesson targeted the fascinating concept of Nature of Roots. The math journey around Nature of Roots starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that is not only relatable and easy to grasp but will also stay with them forever. Here lies the magic with Cuemath.

How to Find Discriminant?

To find the discriminant of a cubic equation or a quadratic equation, we just have to compare the given equation with its standard form and determine the coefficients first. Then we substitute the coefficients in the relevant formula to find the discriminant.

The discriminant of a quadratic equation ax2 + bx + c = 0 is in terms of its coefficients a, b, and c. i.e.,

• Δ OR D = b2 − 4ac

Example: Find the discriminant of the quadratic equation 2x2 — 3x + 8 = 0.

Comparing the equation with ax2 + bx + c = 0, we get a = 2, b = -3, and c = 8. So the discriminant is,
Δ OR D = b2 − 4ac = (-3)2 — 4(2)(8) = 9 — 64 = -55.

Discriminant of Cubic Equation

The discriminant of a cubic equation ax3 + bx2 + cx + d = 0 is in terms of a, b, c, and d. i.e.,

• Δ or D = b2c2 − 4ac3 − 4b3d − 27a2d2 + 18abcd

Example: Find the discriminant of the cubic equation x3 — 3x + 2 = 0.

Comparing the equation with ax3 + bx2 + cx + d = 0, we have a = 1, b = 0, c = -3, and d = 2. So its discriminant is,

Δ or D = b2c2 − 4ac3 − 4b3d − 27a2d2 + 18abcd
= (0)2(-3)2 − 4(1)(-3)3 − 4(0)3(2) − 27(1)2(2)2 + 18(1)(0)(-3)(2)
= 0 + 108 — 0 — 108 + 0
= 0

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Discriminant and Nature of the Roots

If Discriminant is Equal to Zero

Important Notes on Discriminant:

• The discriminant of a quadratic equation ax2 + bx + c = 0 is Δ OR D = b2 − 4ac.

• A quadratic equation with discriminant D has:
(i) two unequal real roots when D > 0
(ii) only one real root when D = 0
(iii) no real roots or two complex roots when D < 0

FAQs on Discriminant

What is Discriminant Meaning?

The discriminant in math is defined for polynomials and it is a function of coefficients of polynomials. It tells the nature of roots or in other words, it discriminates the roots. For example, the discriminant of a quadratic equation is used to find:

• How many roots it has?
• Whether the roots are real or non-real?

What is Discriminant Formula?

There re different discriminant formulas for different polynomials:

• The discriminant of a quadratic equation ax2 + bx + c = 0 is Δ OR D = b2 − 4ac.

• The discriminant of a cubic equation ax3 + bx2 + cx + d = 0 is Δ or D = b2c2 − 4ac3 − 4b3d − 27a2d2 + 18abcd.

How to Calculate the Discriminant of a Quadratic Equation?

To calculate the discriminant of a quadratic equation:

• Identify a, b, and c by comparing the given equation with ax2 + bx + c = 0.
• Substitute the values in the discriminant formula D = b2 − 4ac.

What if Discriminant = 0?

If the discriminant of a quadratic equation ax2 + bx + c = 0 is 0 (i.e., if b2 — 4ac = 0), then the quadratic formula becomes x = -b/2a and hence the quadratic equation has only one real root.

What Does Positive Discriminant Tell Us?

If the discriminant of a quadratic equation ax2 + bx + c = 0 is positive (i.e., if b2 — 4ac > 0), then the quadratic formula becomes x = (-b ± √(positive number) ) / 2a and hence the quadratic equation has only two real and distinct roots.

What Does Negative Discriminant Tell Us?

If the discriminant of a quadratic equation ax2 + bx + c = 0 is negative (i.e., if b2 — 4ac < 0), then the quadratic formula becomes x = (-b ± √(negative number) ) / 2a and hence the quadratic equation has only two complex and distinct roots.

What is the Formula for Discriminant of Cubic Equation?

A cubic equation is of the form ax3 + bx2 + cx + d = 0 and its discriminant is in terms of its coefficients which is given by the formula D = b2c2 − 4ac3 − 4b3d − 27a2d2 + 18abcd.

How Do You Find the Nature of the Roots?

Here is the quadratic formula used to find the roots of a quadratic equation.

Do you observe the term $$b^2-4ac$$ in the quadratic formula?

This is called a «discriminant« because it discriminates the roots of the quadratic equation based on its sign.

The discriminant is used to find the nature of roots of a quadratic equation.

• $$b^2-4ac>0$$  —  In this case, the quadratic equation has two distinct real roots.
• $$b^2-4ac=0$$  —  In this case, the quadratic equation has one repeated real root.
• $$b^2-4ac<0$$  —  In this case, the quadratic equation has no real root.

Let us consider the quadratic equation $$2x^2+x-3=0$$

Let’s find the determinant to find the nature of the roots.

This means that the quadratic equation has two real and distinct roots.

Look at the graph of this quadratic equation shown below.

This is how the quadratic equation is represented on a graph.

This curve is called a parabola.

Observe that the parabola intersects the $$x$$-axis at two distinct points.

There are three types of roots of a quadratic equation $$ax^2+bx+c=0$$:

1. Real and distinct roots
2. Real and equal roots
3. Complex roots

Real and Distinct Roots

• The discriminant is positive, that is, $$b^2-4ac>0$$
• The curve intersects the $$x$$-axis at two distinct points.

Real and Equal Roots

• The discriminant is equal to zero, that is, $$b^2-4ac=0$$
• The curve intersects the $$x$$-axis at only one point.

• The discriminant is negative, that is, $$b^2-4ac<0$$
• The curve does not intersect the $$x$$-axis.

Nature of Roots Calculator

Use the below simulation to find the nature of roots of a quadratic equation graphically.

Use the sliders to adjust the values of $$a,b$$, and $$c$$, the simulation then gives the value of the discriminant, the number of real roots, and shows the graph of the quadratic equation.

1. The standard form of the quadratic equation is $$ax^2+bx+c=0$$ where $$a\neq0$$.

2. The curve of the quadratic equation is in the form of a parabola.

3. The discriminant is given by $$b^2-4ac$$. This is used to determine the nature of the roots of a quadratic equation.

4. A quadratic equation has at most two roots.

FAQs on Nature of Roots

1. How do you find the nature of roots without solving them?

To find the nature of roots of a quadratic equation, the discriminant can be used.

The value of the discriminant tells us the nature of the roots of the quadratic equation.

2. What is the nature of roots when the discriminant is 0?

When the value of the discriminant of a quadratic equation $$ax^2+bx+c=0$$ is zero, the equation will have real and equal roots.

3. What are the three types of nature of roots?

There are three types of roots for a quadratic equation $$ax^2+bx+c=0$$:

1. Real and distinct roots
2. Real and equal roots
3. Complex roots

4. What is a real root?

A real root is a root of a quadratic equation $$ax^2+bx+c=0$$ which belongs to the set of real numbers.

5. What are imaginary roots?

The roots that are not represented on a number line are imaginary roots. Imaginary roots occur when the quadratic equation does not cut the $$x$$-axis.

6. How will you approximate the nature of roots that are irrational?

We can approximate the nature of roots that are irrational by rounding the number to two digits after the decimal point.

8. What is the nature of the roots of the quadratic equation?

There are three types of roots of a quadratic equation $$ax^2+bx+c=0$$:

1.     Real and distinct roots
2.     Real and equal roots
3.     Complex roots

9. How do you find the nature of the roots of a cubic equation?

To determine the nature of the roots of a cubic equation, calculate the value of its discriminant.

If the value of the discriminant is zero and all the coefficients of the cubic equations are real, then the cubic equation has real roots.

Therefore, the value of discriminant and the coefficients of the cubic equation helps us find the nature of roots of a cubic equation.

10. How to identify types of roots in quadratic equations?

To find the nature of roots of a quadratic equation, the discriminant can be used. The value of the discriminant tells us the types of roots of quadratic equations.

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