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;
$$q^2r^2-2pq^2r+p^2q^2-4p^2qr+4pq^2r+4p^2r^2-4pr^2q=0$$.
 A: 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
A: By the given $p(q-r)\neq0$, which gives $p\neq0$.
If $q=0$ we have $pr(x^2-1)=0$ and this equation has no equal roots. Contradiction.
Thus, $q\neq0$.
If $r=0$ we get $pq(x^2-1)=0$, which is contradiction again.
Thus, $pqr\neq0$.
Now, we need $$q^2(p-r)^2-4pr(q-r)(p-q)=0$$ or
$$q^2(p-r)^2+4pr(q^2-qp-qr+pr)=0$$ or
$$q^2(p+r)^2-4pqr(p+r)+4p^2r^2=0$$ or
$$(q(p+r)-2pr)^2=0$$ or
$$q(p+r)=2pr$$ or
$$\frac{1}{p}+\frac{1}{r}=\frac{2}{q}$$ and we are done!
A: 
If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal $\;\dots$

The equation $(pq-pr)x^2+(qr-pq)x+pr-qr=0$ has the root $x=1\,$, regardless of $\,p,q,r\,$.
Assuming a proper quadratic with leading coefficient $pq-pr \ne 0$ so that there are in fact two roots, the product of the roots is $\;\displaystyle \frac{pr-qr}{pq-pr}\,$, so the condition for the second root to also equal $\,1\,$ is:
$$
\frac{pr-qr}{pq-pr} = 1 \;\;\iff\;\; pr -qr=pq - pr \;\;\iff\;\; 2pr = pq+qr \;\;\iff\;\;\frac{2}{q}=\frac{1}{p}+\frac{1}{r}
$$
A: factorizing your term we get $$- \left( p-r \right)  \left( 4\,r{p}^{2}-p{q}^{2}-4\,pqr+{q}^{2}r
 \right) 
=0$$
A: Let A=p(q-r), B=q(r-p) & C=r(p-q). A+B+C=0 => B=-(A+C). Since roots are equal, B²-4AC=0 => (A+C)²-4AC=0 => (A-C)²=0 => A=C
=> p(q-r)=r(p-q)
=> (q-r)/qr=(p-q)/pq [Dividing throughout by pqr]
=> 1/r-1/q=1/p-1/q
=> 2/q=1/p+1/r
Hence QED
