Question

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

Answer 1

The given equation is $\left(a-b\right)x^2+5\left(a+b\right)x-2\left(a-b\right)=0$.

$$\begin{gathered} \therefore D={\left[5\left(a+b\right)\right]}^2-4\times \left(a-b\right)\times \left[-2\left(a-b\right)\right] \\ =25{\left(a+b\right)}^2+8{\left(a-b\right)}^2 \end{gathered}$$
Since a and b are real and a ≠ b, so ${\left(a-b\right)}^2>0$ and ${\left(a+b\right)}^2>0$.

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

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

Adding (1) and (2), we get

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

$\Rightarrow D>0$

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