Question

If $a$ and $b$ are real and $a\ne b$ then show that the roots of the equation $(a-b)x^2+5(a+b)x-2(a-b)=0$ are real and unequal.

Answer 1

Compare given equation with the general form of quadratic equation, which $a{x}^2+bx+c=0$

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

Find Discriminant:

$D=b^2-4ac$

$=\left(5\right(a+b){)}^2-4(a-b\left)\right(-2(a-b))$

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

$=17(a+b{)}^2+\left\{8\right(a+b{)}^2+8(a-b{)}^2\}$

$=17(a+b{)}^2+16(a^2+b^2)$

Which is always greater than zero.

Equation has real and unequal roots.