The equation x^4-x^3-6x^2+4x+8=0 has two equa | vedclass

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

Question

(B) Given the equation $x^4-x^3-6x^2+4x+8=0$. By testing values,we find $x=2$ is a root. Dividing the polynomial by $(x-2)$,we get $(x-2)(x^3+x^2-4x-4

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given the equation $x^4-x^3-6x^2+4x+8=0$. By testing values,we find $x=2$ is a root. Dividing the polynomial by $(x-2)$,we get $(x-2)(x^3+x^2-4x-4)=0$. Testing $x=2$ again in the cubic factor $x^3+x^2-4x-4$,we get $8+4-8-4=0$,so $x=2$ is a root again. Dividing $(x^3+x^2-4x-4)$ by $(x-2)$,we get $(x-2)(x^2+3x+2)=0$. Factoring the quadratic part: $(x^2+3x+2) = (x+1)(x+2)$. Thus,the roots are $x=2, 2, -1, -2$. The two equal roots are $2, 2$. The other two roots are $\alpha = -1$ and $\beta = -2$. Therefore,$\alpha^2+\beta^2 = (-1)^2 + (-2)^2 = 1 + 4 = 5$.

The equation $x^4-x^3-6x^2+4x+8=0$ has two equal roots. If $\alpha$ and $\beta$ are the other two roots of this equation,then $\alpha^2+\beta^2=$

Similar Questions

The number which exceeds its positive square root by $12$ is

The polynomial equation of degree $4$ having real coefficients with three of its roots as $2 \pm \sqrt{3}$ and $1+2i$ is:

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$ for a given $b$. As $b$ varies,the range of $m(b)$ is

In $\triangle ABC$,the value of $\angle A$ is obtained from the equation $3 \cos A + 2 = 0$. The quadratic equation,whose roots are $\sin A$ and $\tan A$,is

If the roots of the equation $(a^2 + b^2)t^2 - 2(ac + bd)t + (c^2 + d^2) = 0$ are equal,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module