If p, q are real and p ≠ q, then show that the roots of the equation

The quadric equation is $(p-q) x^{2}+5(p+q) x-2(p-q)=0$

Here,

$a=(p-q), b=5(p+q)$ and, $c=-2(p-q)$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=(p-q), b=5(p+q)$ and, $c=-2(p-q)$

$D=\{5(p+q)\}^{2}-4(p-q)(-2(p-q))$

$=25\left(p^{2}+2 p q+q^{2}\right)+8\left(p^{2}-2 p q+q^{2}\right)$

$=25 p^{2}+50 p q+25 q^{2}+8 p^{2}-16 p q+8 q^{2}$

$=33 p^{2}+34 p q+33 q^{2}$

Since, $P$ and $q$ are real and $p \neq q$, therefore, the value of $D \geq 0$.

Thus, the roots of the given equation are real and unequal.

Hence, proved