Question

If $\alpha ,\beta$are roots of equation $x^2-7x+8=0$, then equation whose roots are ${\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}\&{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}$ is

• A. $x^2-8x+7=0$
• B. $2x^2-7x+4=0$
• C. $2x^2-7x+8=0$
• D. $x^2-7x+2=0$

Answer 1

Answer: (D)

$x^2-7x+2=0$

$x^2-7x+8=0\to \alpha ,\beta$ roots

Sum and product of roots$\Rightarrow \alpha +\beta =7;\alpha \beta =8$

$\gamma ={\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}};\delta ={\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}$

Product of roots, $\gamma \delta ={\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}={\left(\alpha \beta \right)}^{\frac{1}{3}}=8^{\frac{1}{3}}=2$

Cross-checking from options

$\gamma \delta =2$ satisfies only for

$x^2-7x+2=0$

$\therefore$ $x^2-7x+2=0$ is the required equation.