Question

If p and q are the roots of the quadratic equation
ax2+bx+c=0,$\mathrm{(c}\ne 0)$
then $\frac{1}{ap+b}+\frac{1}{aq+b}$=
.

• A. $\frac{c}{ab}$
• B. $\frac{b}{ac}$
• C. $\frac{a}{bc}$
• D. 1

Answer 1

The correct option is B $\frac{b}{ac}$

$pandqaretherootsofequationa{x}^2+bx+c=0\therefore a{p}^2+bp+c=0\Rightarrow a{p}^2+bp=-c\Rightarrow p(ap+b)=-c\Rightarrow ap+b=\frac{-c}{p}anda{q}^2+bq+c=0\Rightarrow a{q}^2+bq=-c\Rightarrow q(aq+b)=-c\Rightarrow aq+b=\frac{-c}{q}Now,\frac{1}{ap+b}+\frac{1}{aq+b}=\frac{1}{\left(\frac{-c}{p}\right)}+\frac{1}{\left(\frac{-c}{q}\right)}=\frac{-p}{c}+\frac{-q}{c}=\frac{-1}{c}(p+q)=\frac{-1}{c}\times \frac{-b}{a}=\frac{b}{ac}$