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 $ac(p+q{)}^2=pq{b}^2$

Consider given the equation,

$a{x}^2+bx+c=0$ and its roots are in ratio of $p:q$ .

Let, the roots are $px$ and $qx$

Now,

The sum of the roots,

$px+qx=-\frac{b}{a}$

$x(p+q)=-\frac{b}{a}$

$x=-\frac{b}{a(p+q)}$

Product of the roots are,

$px\times qx=\frac{c}{a}$

$x^2=\frac{c}{a\left(pq\right)}$

Now put the value of x,

${(-\frac{b}{a(p+q)})}^2=\frac{c}{a\left(pq\right)}$

$\frac{b^2}{a^2{(p+q)}^2}=\frac{c}{a\left(pq\right)}$

$\frac{b^2}{a{(p+q)}^2}=\frac{c}{\left(pq\right)}$

$ac{(p+q)}^2=pq.b^2$

Hence, this is the answer.