Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-px+q=0$, then the quadratic equation whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ is ___________.

• A. $q{x}^2-px+1=0$
• B. $q{x}^2+(p^2-2q)x+q=0$
• C. $q{x}^2+(2q-p^2)x+q=0$
• D. $p{x}^2+(2p^2-q^2)x+q=0$

Answer 1

Answer: (C)

$q{x}^2+(2q-p^2)x+q=0$

If $\alpha ,\beta$ are the roots of the equation $x^2-px+q=0$,

$\Rightarrow$ sum of the roots $=\alpha +\beta =p$ ...(1)

Similarly, product of the roots=$\alpha \beta =q$ ....(2)

If $\frac{\alpha }{\beta },\frac{\beta }{\alpha }$ are roots of the a quadratic equation, then the quadratic equation is given by,

$(x-\frac{\alpha }{\beta })(x-\frac{\beta }{\alpha })=0$

$\Rightarrow x^2-(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })x+1=0$

$\Rightarrow \alpha \beta x^2-({\alpha }^2+{\beta }^2)x+1=0$

$\Rightarrow \alpha \beta x^2-\left(\right(\alpha +\beta {)}^2-2\alpha \beta )x+1=0$ ....(3)

Substituting (1),(2) in (3), we get

$q{x}^2-(p^2-2q)x+q=0$

$\Rightarrow q{x}^2+(2q-p^2)x+q=0$