Given that the equation x 4 + 2 m x 2 + 4 = 0 has four distinct real solutions x 1 , x 2 , x 3 , and x 4 for a certain real value of m .
Find the value of m such that x 1 4 + x 2 4 + x 3 4 + x 4 4 = 3 2 . Type your answer as m 2 . If there is no such m , type − 1 .
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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 so immediately x 1 2 + x 2 2 + x 3 2 + x 4 2 = − 4 m .
By Newton's sum or identities , we have
k = 1 ∑ 4 x k 4 3 2 2 4 2 4 2 4 ⟹ m 2 = 0 ⋅ k = 1 ∑ 4 x k 3 − 2 m k = 1 ∑ 4 x k 2 + 0 ⋅ k = 1 ∑ 4 x k − 4 ⋅ 4 = − 2 m ( x 1 2 + x 2 2 + x 3 2 + x 4 2 ) − 1 6 = − m ( ( x 1 + x 2 + x 3 + x 4 ) 2 − 2 ( 2 m ) ) = − m ( 0 2 − 4 m ) = 4 m 2 = 6
Now we have to prove if m = ± 6 are there four real roots. There are no real roots for m = 6 . For m = − 6 , we have:
x 4 − 2 6 x 2 + 4 x 4 − 2 6 x 2 + 6 ( x 2 − 6 ) 2 x 2 ⟹ x = 0 = 2 = 2 = 6 ± 2 = ± 6 ± 2
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 , 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.
Let t = x 2 ( t ≥ 0 )
Therefore, we can rewrite the equation as t 2 + 2 m t + 4 = 0 ( 1 )
In order for the original equation has 4 distinct real solutions, ( 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
⇒ m < − 2 .
With m < − 2 , let t 1 , t 2 be the solutions of ( 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 )
Substituting these results in S
S = 2 ( t 1 2 + t 2 2 ) = 3 2 ⇔ t 1 2 + t 2 2 = 1 6
⇔ ( t 1 + t 2 ) 2 − 2 t 1 t 2 = 1 6 ⇔ ( − 2 m ) 2 − 2 × 4 = 3 2
4 m 2 = 2 4 ⇔ m 2 = 6
As m < − 2 , there is only one solution, which is m = − 6
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Plugging in 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 ⟹ = 3 2 ( x 1 4 + x 2 4 + x 3 4 + x 4 4 ) + 2 m ( x 1 2 + x 2 2 + x 3 2 + x 4 2 ) + 1 6 = 0 ⟹ x 1 2 + x 2 2 + x 3 2 + x 4 2 = − m 2 4 And by Vieta's formula , x 1 2 + x 2 2 + x 3 2 + x 4 2 = ( = 0 x 1 + x 2 + x 3 + x 4 ) 2 − 2 ( = 2 m x 1 x 2 + x 1 x 3 + x 1 x 4 + x 2 x 3 + x 2 x 4 + x 3 x 4 ) = − 4 m Hence, − m 2 4 = 4 m ⟹ m 2 = 6 .
But is it true that m = ± 6 satisfy condition? Turns out m = 6 is not satisfied. Here's why:
Because f ( x ) : = x 4 + 2 m x 2 + 4 has 4 distinct roots, then f ′ ( x ) = 4 x 3 + 4 m x has 3 distinct roots, namely 0 , ± − m , hence m < 0 , so m = − 6 only.
The 4 unordered values of ( x 1 , x 2 , x 3 , x 4 ) are ± 6 ± 2 , ∓ 6 ± 2 .