Question

The solution of the equation $2x^3-x^2-22x-24=0$ when two of the roots are in the ratio $3:4$, is

• A. $3,4,\frac{1}{2}$
• B. $-\frac{3}{2},-2,4$
• C. $-\frac{1}{2},\frac{3}{2},2$
• D. $\frac{3}{2},2,\frac{5}{2}$

Answer 1

The correct option is B

$-\frac{3}{2},-2,4$

Explanation for correct option

Given equation is $2x^3-x^2-22x-24=0$

The roots are in the ration $3:4$

Consider the roots of the equation to be $3a,4a,b.$

$$\begin{gathered} \sum x=3a+4a+b=-\left(-\frac{1}{2}\right)=\frac{1}{2} \\ \therefore b=\frac{1}{2}–7a\dots .\left(i\right) \\ \sum xy=3a\left(4a\right)+4a\left(b\right)+b\left(3a\right)=-\frac{22}{2} \\ \therefore 12a^2+7ab=–11....\left(ii\right) \end{gathered}$$

Substituting $b$ from $\left(i\right)$

$$\begin{gathered} \Rightarrow 12a^2+7a\left(\frac{1}{2}-7a\right)=-11 \\ \Rightarrow 12a^2+\frac{7a}{2}-49a^2=-11 \\ \Rightarrow 74a^2-7a-22=0 \\ \therefore a=\frac{-\left(-7\right)\pm \sqrt{{\left(-7\right)}^2-4\left(74\right)\left(-22\right)}}{2·74} \\ =\frac{+7\pm 81}{148} \\ \therefore a=-\frac{1}{2},\frac{22}{37} \\ \therefore b=\frac{1}{7}-\left(-\frac{1}{2}\right)=4 \end{gathered}$$

Therefore the roots are $-\frac{3}{2},-2$ and $4$

Hence,option (B) is correct