Find a value to satisfy an equation

Algebra Level 4

Given that the equation x 4 + 2 m x 2 + 4 = 0 x^4 + 2mx^2 + 4 = 0 has four distinct real solutions x 1 x_1 , x 2 x_2 , x 3 x_3 , and x 4 x_4 for a certain real value of m m .

Find the value of m m such that x 1 4 + x 2 4 + x 3 4 + x 4 4 = 32 x_1^4 + x_2^4 + x_3^4 +x_4^4 = 32 . Type your answer as m 2 m^2 . If there is no such m m , type 1 -1 .


The answer is 6.

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3 solutions

Pi Han Goh
Nov 6, 2020

Plugging in x 1 , x 2 , x 3 , x 4 x_1, x_2, x_3, x_4 into the original equation gives { x 1 4 + 2 m x 1 2 + 4 = 0 x 2 4 + 2 m x 2 2 + 4 = 0 x 3 4 + 2 m x 3 2 + 4 = 0 x 4 4 + 2 m x 4 2 + 4 = 0 ( x 1 4 + x 2 4 + x 3 4 + x 4 4 ) = 32 + 2 m ( x 1 2 + x 2 2 + x 3 2 + x 4 2 ) + 16 = 0 x 1 2 + x 2 2 + x 3 2 + x 4 2 = 24 m \begin{cases} x_1^4 + 2mx_1 ^2 + 4 = 0 \\ x_2^4 + 2mx_2 ^2 + 4 = 0 \\ x_3^4 + 2mx_3 ^2 + 4 = 0 \\ x_4^4 + 2mx_4 ^2 + 4 = 0 \\ \end{cases} \quad \implies \quad \underbrace{(x_1 ^4 + x_2 ^4 + x_3^4 + x_4^4)}_{=\, 32} + 2m(x_1^2 + x_2^2 + x_3^2 + x_4^2) + 16 = 0 \quad \implies \quad x_1^2 + x_2^2 + x_3^2 + x_4^2 = - \frac{24}m And by Vieta's formula , x 1 2 + x 2 2 + x 3 2 + x 4 2 = ( x 1 + x 2 + x 3 + x 4 = 0 ) 2 2 ( x 1 x 2 + x 1 x 3 + x 1 x 4 + x 2 x 3 + x 2 x 4 + x 3 x 4 = 2 m ) = 4 m x_1^2 + x_2 ^2 + x_3^2 + x_4^2 = (\underbrace{x_1 + x_2 + x_3 + x_4}_{=\, 0})^2 - 2(\underbrace{x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4}_{=\, 2m}) = -4m Hence, 24 m = 4 m m 2 = 6 . -\dfrac{24}m = 4m \implies m^2 = \boxed{6} .

But is it true that m = ± 6 m = \pm \sqrt 6 satisfy condition? Turns out m = 6 m = \sqrt6 is not satisfied. Here's why:

Because f ( x ) : = x 4 + 2 m x 2 + 4 f(x):= x^4 + 2mx^2 + 4 has 4 distinct roots, then f ( x ) = 4 x 3 + 4 m x f'(x) = 4x^3 + 4mx has 3 distinct roots, namely 0 , ± m 0, \pm \sqrt{-m} , hence m < 0 m < 0 , so m = 6 m = -\sqrt6 only.

The 4 unordered values of ( x 1 , x 2 , x 3 , x 4 ) (x_1, x_2, x_3, x_4) are ± 6 ± 2 , 6 ± 2 . \pm \sqrt{ \sqrt6 \pm \sqrt2 } , \mp \sqrt{ \sqrt6 \pm \sqrt2 } .

Nice solution! A very minor shortcut (or just a non-Vieta alternative) is to just go ahead and solve the biquadratic; you get x 2 = m ± m 2 4 x^2=-m\pm \sqrt{m^2-4} so immediately x 1 2 + x 2 2 + x 3 2 + x 4 2 = 4 m x_1^2+x_2^2+x_3^2+x_4^2=-4m .

Chris Lewis - 6 months, 3 weeks ago

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Pi Han Goh - 6 months, 3 weeks ago

By Newton's sum or identities , we have

k = 1 4 x k 4 = 0 k = 1 4 x k 3 2 m k = 1 4 x k 2 + 0 k = 1 4 x k 4 4 32 = 2 m ( x 1 2 + x 2 2 + x 3 2 + x 4 2 ) 16 24 = m ( ( x 1 + x 2 + x 3 + x 4 ) 2 2 ( 2 m ) ) 24 = m ( 0 2 4 m ) 24 = 4 m 2 m 2 = 6 \begin{aligned} \sum_{k=1}^4 x_k^4 & = 0 \cdot \sum_{k=1}^4 x_k^3 - 2m \sum_{k=1}^4 x_k^2 + 0 \cdot \sum_{k=1}^4 x_k - 4 \cdot 4 \\ 32 & = -2m \left(x_1^2+x_2^2+x_3^2+x_4^2 \right) - 16 \\ 24 & = - m \left(\left(x_1+x_2+x_3+x_4 \right)^2 - 2(2m)\right) \\ 24 & = - m \left(0^2 - 4m\right) \\ 24 & = 4m^2 \\ \implies m^2 & = \boxed 6 \end{aligned}

Now we have to prove if m = ± 6 m=\pm \sqrt 6 are there four real roots. There are no real roots for m = 6 m=\sqrt 6 . For m = 6 m=-\sqrt 6 , we have:

x 4 2 6 x 2 + 4 = 0 x 4 2 6 x 2 + 6 = 2 ( x 2 6 ) 2 = 2 x 2 = 6 ± 2 x = ± 6 ± 2 \begin{aligned} x^4 - 2\sqrt 6 x^2 + 4 & = 0 \\ x^4 - 2\sqrt 6 x^2 + 6 & = 2 \\ (x^2 - \sqrt 6)^2 & = 2 \\ x^2 & = \sqrt 6 \pm \sqrt 2 \\ \implies x & = \pm \sqrt{\sqrt 6 \pm \sqrt 2} \end{aligned}

There are four real roots.

This is incomplete. If instead, you're given x 1 4 + x 2 4 + x 3 4 + x 4 4 = 0 x_1^4 + x_2^4 + x_3^4 +x_4^4 = 0 , then you will wrongly conclude that the answer to this problem is 1, when in fact, it should be -1.

For completeness, you got to prove that the four roots are real and distinct.

Pi Han Goh - 7 months, 1 week ago

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Yes, you are right.

Chew-Seong Cheong - 7 months, 1 week ago
Tin Le
Nov 6, 2020

Let t = x 2 t=x^2 ( t 0 t \geq 0 )

Therefore, we can rewrite the equation as t 2 + 2 m t + 4 = 0 t^2 + 2mt + 4 = 0 ( 1 ) (1)

In order for the original equation has 4 distinct real solutions, ( 1 ) (1) must have two distinct real solutions.

Therefore, according to Vieta's formula for quadratic equation we have:

{ Δ > 0 m 2 4 > 0 m < 2 OR m > 2 t 1 + t 2 = 2 m > 0 m < 0 t 1 × t 2 = 4 \left\{\begin{matrix} \Delta ' >0 \Leftrightarrow m^2 - 4 > 0 \Leftrightarrow m < -2 \text{ OR } m>2 \\ t_1 + t_2 = -2m > 0 \Leftrightarrow m<0 \\ t_1 \times t_2 = 4 \end{matrix}\right.

m < 2 \Rightarrow m<-2 .

With m < 2 m<-2 , let t 1 , t 2 t_1, t_2 be the solutions of ( 1 ) (1) .

Therefore, the original equation has 4 distinct solutions: ( x 1 , x 2 , x 3 , x 4 ) = ( t 1 , t 1 , t 2 , t 2 ) (x_1, x_2, x_3, x_4) = (\sqrt{t_1},-\sqrt{t_1},\sqrt{t_2},-\sqrt{t_2})

Substituting these results in S S

S = 2 ( t 1 2 + t 2 2 ) = 32 t 1 2 + t 2 2 = 16 S= 2(t_1^2+t_2^2)= 32 \Leftrightarrow t_1^2 + t_2^2 = 16

( t 1 + t 2 ) 2 2 t 1 t 2 = 16 ( 2 m ) 2 2 × 4 = 32 \Leftrightarrow (t_1+t_2)^2 - 2t_1t_2 = 16 \Leftrightarrow (-2m)^2 -2\times4 = 32

4 m 2 = 24 m 2 = 6 4m^2 = 24 \Leftrightarrow m^2 = \boxed{6}

As m < 2 m<-2 , there is only one solution, which is m = 6 m=-\sqrt{6}

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