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

The solutions of the quadratic equation $(3|x| - 3)^2 = |x| + 7$ which belong to the domain of definition of the function $y = \sqrt{x(x - 3)}$ are given by

Let $\theta$ be an acute angle such that the equation $x^3+4 x^2 \cos \theta+x \cot \theta=0$ has multiple roots. Then the value of $\theta$ (in radians) is

What is the product of the real roots of the equation ${t^2}{x^2} + |x| + 9 = 0$?

The algebraic equation of degree $4$ whose roots are the translates of the roots of the equation $x^4+5x^3+6x^2+7x+9=0$ by $-1$ is

If ${b_1}{b_2} = 2({c_1} + {c_2})$,then at least one of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module