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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If $2 + i\sqrt{3}$ is a root of the equation $x^2 + px + q = 0$,where $p$ and $q$ are real,then $(p, q) = $

Let $\alpha, \beta, \gamma, \delta$ be the roots of the equation $x^4 + x^2 + 1 = 0$. Then the equation whose roots are $\alpha^2, \beta^2, \gamma^2, \delta^2$ is:

The least positive value of $a$ for which the equation $2x^{2} + (a-10)x + \frac{33}{2} = 2a$ has real roots is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+2x^2-x-2=0$,then $\alpha^6+\beta^6+\gamma^6=$

If $a$ and $b$ are natural numbers such that $2013 + a^2 = b^2$,then the minimum possible value of $ab$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module