How can there be 3 solutions?

Algebra Level 3

x 4 + x 2 ( 1 − 2 k ) + k 2 − 1 = 0 \large x^{4}+x^{2}(1-2k)+k^{2}-1=0

Find the value of constant k k for which the equation above has 3 distinct real solutions.


The answer is 1.

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Aditya Chauhan by Aditya Chauhan , 20, India

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3 solutions

Appan Rakaraddi
Jun 7, 2015

Clearly, the graph of the equation is symmetric about x=0; so, the no. of roots on left hand side and right side would be equal which implies that there will be even number of roots unless there is a root( which is actually a double root) at the origin. So, x=0 is a root of the equation; hence k=1,-1. But, since x is real, k=1

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Moderator note:

Nice interpretation of the problem.

x 4 + ( 1 − 2 k ) x 2 + k 2 − 1 = 0 x^4+(1-2k)x^2+k^2-1 = 0 is a quadratic equation for x 2 x^2 . In general there are either no real root or two real root of x 2 x^2 . When there are two positive real roots of x 2 x^2 , there can be four roots of x x . For the equation to have three real roots of x x , one of the real root of x 2 x^2 must be 0 0 . Therefore, the equation is of the form x 2 ( x 2 − a ) = 0 x^2(x^2-a) = 0 since there is no x 3 x^3 term. This is possible when k 2 − 1 = 0 k^2-1=0 ⟹ k = ± 1 \Longrightarrow k = \pm 1 and the equation is either:

{ k = 1 ⇒ x 2 ( x 2 − 1 ) ⇒ x = − 1 , 0 , 1 3 real roots k = − 1 ⇒ x 2 ( x 2 + 3 ) ⇒ x = 0 1 real root – rejected ⇒ k = 1 \begin{cases} k = 1 & \Rightarrow x^2(x^2-1) & \Rightarrow x = -1,0,1 & \text{3 real roots} \\ k = -1 & \Rightarrow x^2(x^2+3) & \Rightarrow x = 0 & \text{1 real root -- rejected} \end{cases} \\ \Rightarrow k = \boxed{1}

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Moderator note:

Why is this true?

For the equation to have three real roots of x x , one of the real root of x 2 x^2 must be 0 0 .

When there is two real roots of x 2 x^2 , then there are four real roots of x x ? x 4 − x 2 − 2 = 0 x^4 - x^2 - 2 = 0 have real roots of x 2 = 2 , − 1 x^2=2,-1 but only two real roots of x x .

Pi Han Goh - 6 years ago

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Thanks. My mistakes.

Chew-Seong Cheong - 6 years ago

I was trying to explain that the quadratic equation is of the form of ( x 2 − α ) ( x 2 − β ) (x^2-\alpha)(x^2-\beta) . For real roots of x x , α \alpha and β \beta must be non-negative. If α , β < 0 \alpha, \beta < 0 , then there are four real roots of x x . To have only three roots a pair of roots must be repeated. This is only possible only when either α \alpha or β \beta is 0 0 , because + 0 = − 0 +0=-0 .

Chew-Seong Cheong - 6 years ago
Shubhendra Singh
Jun 6, 2015

Let x 2 = t x^{2}=t then the given equation transforms into a quadratic.

t 2 + t ( 1 − 2 k ) + k 2 − 1 t^{2}+t(1-2k)+k^{2}-1

Now since the biquadratic equation has 3 solutions, the quadratic expression formed will satisfy the following conditions.

  • D > 0 ⟹ k < 5 4 D>0 \implies k<\dfrac{5}{4}

  • − b 2 a > 0 ⟹ k > 1 2 \dfrac{-b}{2a} >0 \implies k>\dfrac{1}{2}

  • t = 0 t=0 is a zero of the quadratic ⟹ k = ± 1 \implies k=±1

Intersection of all these conditions gives k = 1 k=1 .

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There is a need to modify the question for clarity.

First of all, an equation needs to have an equality relation.

The question should read

x 4 + x 2 ( 1 − 2 k ) + k 2 − 1 = 0 x^4+x^2(1-2k)+k^2-1=0

Secondly, the question does not specify if the three solutions need to be 'real'. Without the restriction k = − 1 k=-1 would also be a valid answer.

Janardhanan Sivaramakrishnan - 6 years ago

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Sorry for the inconvenience caused. It has been fixed now.

Aditya Chauhan - 6 years ago

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