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Julli [10]
3 years ago
5

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.

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
4 0

Answer:

The value of m is 6.

Step-by-step 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}

=\frac{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}}

=\frac{-3m-2}{1}

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

-10b^2=-3m-2

\implies b^2=\frac{3m+2}{10}-----(2)

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

9b^4=m^2

9(\frac{3m+2}{10})^2=m^2

9(\frac{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=-\frac{6}{19}\text{ or }m=6

But m > 0,

Hence, the value of m is 6.

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