Question

If the roots of $a{x}^2+bx+c=0$ are in the ratio $1:k$, then $c{k}^2+bk+c=0$

• A. True
• B. False

Answer 1

The correct option is A True

$a{x}^2+bx+c=0$

$\frac{\alpha }{\beta }=\frac{1}{k}$

$\alpha +\beta =-\frac{b}{a}$

$\alpha \beta =\frac{c}{a}$

So, $c{k}^2+bk+c=0$

$⟹k(ck+b)=-c$

$⟹k=-c,k=-\frac{b}{c}$

$\therefore c{k}^2+bk+c=0$

When put value $-\frac{b}{c}$