Question

If $p$ and $q$ are non-zero real numbers and ${\alpha }^3+{\beta }^3=-p,\alpha \beta =q$, then a quadratic equation whose roots are $\frac{{\alpha }^2}{\beta },\frac{{\beta }^2}{\alpha }$ is :

• A. $p{x}^2-qx+p^2=0$
• B. $q{x}^2+px+q^2=0$
• C. $p{x}^2+qx+p^2=0$
• D. $q{x}^2-px+q^2=0$

Answer 1

The correct option is B $q{x}^2+px+q^2=0$

${\alpha }^3+{\beta }^3=-p$ ----- ( 1 )

$\alpha \beta =q$ ----- ( 2 )

Now,

$\Rightarrow$ $\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{{\alpha }^3+{\beta }^3}{\alpha \beta }$

$=\frac{-p}{q}$ [ Using ( 1 ) and ( 2 ) ]

$\therefore$ $\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{-p}{q}$ ---- ( 3 )

$\Rightarrow$ $\frac{{\alpha }^2}{\beta }\times \frac{{\beta }^2}{\alpha }=\alpha \beta$

$=q$ [ From ( 2 ) ]

$\therefore$ $\frac{{\alpha }^2}{\beta }\times \frac{{\beta }^2}{\alpha }=q$ ------ ( 4 )

Now, required equation,

$x^2-(\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha })x+(\frac{{\alpha }^2}{\beta }\times \frac{{\beta }^2}{\alpha })=0$

$\Rightarrow$ $x^2-\left(\frac{-p}{q}\right)x+q=0$ [ Using ( 3 ) and ( 4 ) ]

$\therefore$ $q{x}^2+px+q^2=0$

Hence, this is the answer.