Question

If $\alpha ,\beta ,\gamma$ are the roots $x^3-6x-4=0$, then the equation whose roots are $\beta \gamma +\frac{1}{\alpha },\gamma \alpha +\frac{1}{\beta },\alpha \beta +\frac{1}{\gamma }$ is

• A. $4x^3-30x^2+125=0$
• B. $x^3+15x^2-120=0$
• C. $4x^3+30x^2-125=0$
• D. $4x^3-30x^2-125=0$

Answer 1

The correct option is A $4x^3-30x^2+125=0$

Given equation: $x^3-6x-4=0$ roots of the given equation is given by

$\alpha +\beta +\gamma =0$

$\alpha \beta +\beta \gamma +\gamma \alpha =-6$

$\alpha \beta \gamma =4$

now the value of roots given in the question are

$\alpha \beta +\frac{1}{\gamma }=\frac{\alpha \beta \gamma +1}{\gamma }=\frac{5}{\gamma }$

$\alpha \gamma +\frac{1}{\beta }=\frac{\alpha \beta \gamma +1}{\beta }=\frac{5}{\beta }$

$\beta \gamma +\frac{1}{\alpha }=\frac{\alpha \beta \gamma +1}{\alpha }=\frac{5}{\alpha }$

Now calculating the sum of the roots, we get

$\frac{5}{\gamma }+\frac{5}{\beta }+\frac{5}{\alpha }=\frac{-15}{2}$

Now calculating product of the roots, we get

$\frac{5}{\gamma }\ast \frac{5}{\beta }\ast \frac{5}{\alpha }=\frac{125}{4}$

Now calculating sum of the products of the roots, we get

$\frac{5}{\gamma }\ast \frac{5}{\beta }+\frac{5}{\beta }\ast \frac{5}{\alpha }+\frac{5}{\gamma }\ast \frac{5}{\alpha }=0$

Therefore the general cubic equation is

$x^3$+(Sum of the roots)$x^2+$(sum of the products of the roots)$x+$(products of the roots)$=0$

$x^3+\left(\frac{-15}{2}\right)x^2+\left(0\right)x+\left(\frac{125}{4}\right)=0$

$4x^3-30x^2+125=0$ which is the required equation.

Option(A) is correct.