If a and b are real and a ≠ b then show that the roots of the equation

The given equation is $(a-b) x^{2}+5(a+b) x-2(a-b)=0$.

$\therefore D=[5(a+b)]^{2}-4 \times(a-b) \times[-2(a-b)]$

$=25(a+b)^{2}+8(a-b)^{2}$

Since $a$ and $b$ are real and $a \neq b$, so $(a-b)^{2}>0$ and $(a+b)^{2}>0$.

$\therefore 8(a-b)^{2}>0$     ..........(1)           (Product of two positive numbers is always positive)      

Also, $25(a+b)^{2}>0$           ..........(2)             (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25(a+b)^{2}+8(a-b)^{2}>0$                                (Sum of two positive numbers is always positive)

$\Rightarrow D>0$

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