Question

If one root of the equation $x^3+ax+b=0$ is twice the difference of the other two then one root is

• A. $\frac{13b}{3a}$
• B. $\frac{13a}{3b}$
• C. $\frac{a}{b}$
• D. $\frac{2b}{5a}$
• E. None

Answer 1

Answer: (A)

$\frac{13b}{3a}$

Consider the problem

we have,

$x^3+ax+b=0....\left(1\right)$

Now, consider it roots be $\alpha ,\beta ,\gamma$

where $\alpha =2\left(\beta -\gamma \right)$

$\begin{array}{c}{\sigma }_1=\alpha +\beta +\gamma =0\\ \Rightarrow \alpha =-\beta -\gamma .....\left(2\right)\end{array}$

From $\left(1\right)$ and $\left(2\right)$

$\begin{array}{c}2\beta -2\gamma =-\beta -\gamma \\ \Rightarrow \gamma =3\beta \\ \therefore \alpha =-4\beta \end{array}$

Therefore roots are $-4\beta ,\beta ,3\beta$

$\begin{array}{c}{\sigma }_2=\left(-4\beta \right)\beta +\beta \left(3\beta \right)+3\beta \left(-4\beta \right)=a\\ -13{\beta }^2=a....\left(3\right)\end{array}$

$\begin{array}{c}{\sigma }_3=\left(-4\beta \right)\left(\beta \right)\left(3\beta \right)=-b\\ 12{\beta }^3=b.......\left(4\right)\end{array}$

Now, dividing $\left(4\right)$ by $\left(3\right)$

$\begin{array}{c}-\frac{13}{12}\beta =\frac{b}{a}\\ -4\beta =\frac{13b}{3a}\end{array}$

Therefore, One root is $\frac{13b}{3a}$

Hence, Option $A$ is the correct answer.