Question

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

• A. $\frac{1}{p}+\frac{1}{r}$
• B. $p+r$
• C. $\frac{1}{p}+r$
• D. $p+\frac{1}{r}$

Answer 1

The correct option is A $\frac{1}{p}+\frac{1}{r}$

Given that $p(q-r)x^2+q(r-p)x+r(p-q)=0$ has equal roots

$\therefore △=0$

$q^2{(r-p)}^2-4pr(q-r)(p-q)=0$

$q^2{r}^2-2q^{2}rp+q^2{p}^2-4p^{2}qr-4pq{r}^2+4q^{2}rp+4p^2{r}^2=0$

$p^2{q}^2+q^2{r}^2+4p^2{r}^2+2p{q}^{2}r-4pq{r}^2-4p^2{r}^{2}qr=0$

${(pq+qr-2pr)}^2=0$

$\therefore pq+qr=2pr$

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