If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$. If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $$\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$$


My Attempt:
$$p(q-r)x^2+q(r-p)x+r(p-q)=0$$
Comapring with $ax^2+bx+c=0$, we get:
$$a=p(q-r)$$
$$b=q(r-p)$$
$$c=r(p-q)$$.
Since the roots are equal:
$$b^2-4ac=0$$
$$(q(r-p))^2-4p(q-r)r(p-q)=0$$
Multipying and simplfying a bit;
$$q^2r^2-2pq^2r+p^2q^2-4p^2qr+4pq^2r+4p^2r^2-4pr^2q=0$$.
Dividing by $p^2q^2r^2$

$$=\frac{1}{p^2} +\frac{2}{pr}+\frac{1}{r^2}-\frac{4}{qr}+\frac{4}{q^2}-\frac{4}{pq}$$
$$=\Bigl(\frac{1}{p}\Bigl)^2+\Bigl(\frac{-2}{q}\Bigl)^2+\Bigl(\frac{1}{r}\Bigl)^2$$
$$2\Bigl( \frac{1}{p}\Bigl)\Bigl(\frac{1}{r}\Bigl)+2\Bigl(\frac{1}{p}\Bigl)\Bigl(\frac{-2}{q}\Bigl)$$
$$+2\Bigl(\frac{1}{r}\Bigl)\Bigl(\frac{-2}{q}\Bigl)$$


  Using
  $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$


$$=\Bigl(\frac{1}{p}+\frac{1}{r}-\frac{2}{q}\Bigl)^2$$
You can carry on
Let A=p(q-r), B=q(r-p) &amp; C=r(p-q). A+B+C=0 =&gt; B=-(A+C). Since roots are equal, B²-4AC=0 =&gt; (A+C)²-4AC=0 =&gt; (A-C)²=0 =&gt; A=C
=&gt; p(q-r)=r(p-q)
=&gt; (q-r)/qr=(p-q)/pq [Dividing throughout by pqr]
=&gt; 1/r-1/q=1/p-1/q
=&gt; 2/q=1/p+1/r
Hence QED