Question

One root of $x^3-5x^2+2x+8=0$ is double the other, then the roots are

• A. $-1,3,6$
• B. $-1,2,4$
• C. $1,2,4$
• D. $1,3,6$

Answer 1

Answer: (B)

$-1,2,4$

Let $\alpha ,2\alpha ,\beta$ be the roots of $x^3-5x^2+2x+8=0$ (As one root is double of other)

Therefore, $S_1=\alpha +2\alpha +\beta =5$$\Rightarrow 3\alpha +\beta =5$ ...(1)

$S_2=\alpha \left(2\alpha \right)+\alpha \left(\beta \right)+2\alpha \left(\beta \right)=2$$\Rightarrow 2{\alpha }^2+3\alpha \beta =2\Rightarrow \alpha (2\alpha +3\beta )=2$ ...(2)

$S_3=\alpha \left(2\alpha \right)\left(\beta \right)=-8\Rightarrow 2{\alpha }^2\beta =-8$$\Rightarrow {\alpha }^2\beta =-4$ ...(3)

Dividing (2) by (3), we get

$\frac{2\alpha +3\beta }{\alpha \beta }=\frac{2}{-4}$

$\Rightarrow -4\alpha -3\beta =2\alpha \beta$ ...(4)

Now solving (1) and (4), we get

$\alpha =2$ and $\beta =-1$

Hence roots are $-1,2,4$.