Question

What is the condition for the roots of $a{x}^2+bx+c=0$ to be in the ratio of p : q ?

• A. $a^2{c}^2(p+q)=p^2{q}^{2}b$
• B. $ac(p+q{)}^2=pq{b}^2$
• C. $pq(a+c{)}^2=ac{b}^2$
• D. $p^2{q}^2(a+c)=a^2{c}^{2}b$

Answer 1

The correct option is B $ac(p+q{)}^2=pq{b}^2$

Let the roots of $a{x}^2+bx+c=0$ be $\alpha$ and $\beta$
$\alpha +\beta =\frac{-b}{a}$
$\alpha \beta =\frac{c}{a}$
Let $\frac{\alpha }{\beta }=\frac{p}{q}$
$\frac{\frac{-b+\sqrt{b^2-4ac}}{2a}}{\frac{-b-\sqrt{b^2-4ac}}{2a}}=\frac{p}{q}$
$\frac{-b+\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}=\frac{p}{q}$
$-bq+q\sqrt{b^2-4ac}=-pb-p\sqrt{b^2-4ac}$
$(p+q)\sqrt{b^2-4ac}=bq-pb$
$(p+q{)}^2(b^2-4ac)=b^2{q}^2+p^2{b}^2-2pq{b}^2$
$p^2{b}^2-4ac{p}^2+q^2{b}^2-4ac{q}^2-8acpq+2pq{b}^2=b^2{q}^2+p^2{b}^2-2pq{b}^2$
$-4ac(p+q{)}^2=-4pq{b}^2$
$ac(p+q{)}^2=pq{b}^2$