Question

Find an m > 0 such that the the equation x^4−(3m+2)x^2+m^2=0 has four real solutions that form an arithmetic sequence.

Answer 1

The value of m is 6.

Explanation:

Here, the given equation,

`x^4-(3m+2)x^2+m^2=0`

`x^4+0x^3-(3m+2)x^2+0x+m^2=0`

Let the roots of the equation are a-3b, a-b, a+b and a + 3b, ( they must be form an AP )

Thus, we can write,

`a-3b+a-b+a+b+a+3b=\frac{\text{coefficient of }x^3}{\text{coefficient of }x^4}`

`=(0)/(1)=0`

`\implies a=0----(1)`

`(-3b)(-b)+(-b)(b)+(b)(3b)+(3b)(-3b)+(-b)(3b)+(-3b)(b)=\frac{\text{coefficient of }x^2}{\text{coefficient of }x^4}}`

`=(-3m-2)/(1)`

`3b^2-b^2+3b^2-9b^2-3b^2-3b^2=-3m-2`

`-10b^2=-3m-2`

`\implies b^2=(3m+2)/(10)-----(2)`

`(-3b)(-b)(b)(3b)=\frac{\text{Constant term}}{\text{coefficient of}x^4}= m^2`

`9b^4=m^2`

`9((3m+2)/(10))^2=m^2`

`9((9m^2+4+12m)/(100))=m^2`

`81m^2+36+108m=100m^2`

`-19m^2+108m+36=0`

`19m^2-108m-36=0`

`19m^2-114m+6m-36=0`

`19m(m-6)+6(m-6)=0`

`(19m+6)(m-6)=0`

`\implies m=-(6)/(19)\text{ or }m=6`

But m > 0,

Hence, the value of m is 6.