# What is m if roots of equation x^4-(3m+4)x^2+(m+1)^2=0 are real and in arithmetic progresion?

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### 1 Answer

Given `x^4-(3m+4)x^2+(m+1)^2=0` ; also the roots are real and in arithmetic progression. Find m:

We know that the product of the roots is `(m+1)^2` . Also the sum of the roots is zero. (If `y=ax^4+bx^3+cx^2+dx+e` then the sum of the roots is `-b/a` ; here b=0.)

Since the roots are in arithmetic progression with a sum of zero,they must be symmetric about the origin. The roots will be of the form -3k,-k,k,3k. (Given a root k, by symmetry -k is a root. Then d= k-(-k)=2k so the other terms are 3k and -3k.)

So `(x-3k)(x-k)(x+k)(x+3k)=x^4-(3m+4)x^2+(m+1)^2` `(x^2-9k^2)(x^2-k^2)=x^4-(3m+4)x^2+(m+1)^2`

`x^4-10k^2x^2+9k^4=x^4-(3m+4)x^2+(m+1)^2`

Equating coefficients we get:

`10k^2=3m+4` and `9k^4=(m+1)^2` ==>`3k^2=+-(m+1)`

Then `(3m+4)/10=(m+1)/3` or `(3m+4)/10=(-m-1)/3`

so `m=2` or `m=-22/19`

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The possible values for m are 2 or `-22/19`

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When m=2 we get `x^4-10x^2+9=0` with roots -3,-1,1,3

When `m=-22/19` we get `x^4-10/19 x^2+9/361=0` with roots `(-3)/sqrt(19),(-1)/sqrt(19),1/sqrt(19),3/sqrt(19)`