Question

If $p(q-r)x^2+q(r-p)x+r(p-q)=0$ has equal roots, then $\frac{2}{q}=$

• A. $\frac{p-r}{p+r}$
• B. $\frac{p-r}{pr}$
• C. $\frac{p+r}{pr}$
• D. $\frac{p+r}{p-r}$

Answer 1

Answer: (C)

$\frac{p+r}{pr}$

condition to have equal roots is $D=0\Rightarrow b^2=4ac$
$\Rightarrow \left(q\right(r-p){)}^2=4r(p-q\left)p\right(q-r)$

$q^2(r^2+p^2-2rp)=4rp(pq-q^2+qr-pr)$

$\Rightarrow q^2(r+p{)}^2=4rp(pr)$

$\Rightarrow q(r+p)=2pr$

$\frac{2}{q}=\frac{1}{r}+\frac{1}{p}$

$\frac{2}{q}=\frac{p+r}{pr}$