If one root of the equation x^3-6x^2+3x+10=0 | vedclass

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(A) Let the roots of the equation $x^3-6x^2+3x+10=0$ be $\alpha, \beta, \gamma$. Given that one root is the average of the other two,let $\beta = \fra

Vedclass · Accepted Answer

$642$

Vedclass · Answer

(A) Let the roots of the equation $x^3-6x^2+3x+10=0$ be $\alpha, \beta, \gamma$. Given that one root is the average of the other two,let $\beta = \frac{\alpha+\gamma}{2} \Rightarrow \alpha+\gamma = 2\beta$. From the sum of roots,$\alpha+\beta+\gamma = 6$. Substituting $\alpha+\gamma = 2\beta$,we get $2\beta+\beta = 6$ $\Rightarrow 3\beta = 6$ $\Rightarrow \beta = 2$. Since $\beta=2$ is a root,it must satisfy the equation: $(2)^3 - 6(2)^2 + 3(2) + 10 = 8 - 24 + 6 + 10 = 0$. Now,divide the polynomial by $(x-2)$: $(x-2)(x^2-4x-5) = 0$. Factoring the quadratic: $(x-2)(x-5)(x+1) = 0$. The roots are $2, 5, -1$. The sum of the fourth powers of the roots is $2^4 + 5^4 + (-1)^4 = 16 + 625 + 1 = 642$.

If one root of the equation $x^3-6x^2+3x+10=0$ is the average of the other two,then the sum of the fourth powers of the roots of the equation is

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