Question

If p, q are real and p ≠ q, then show that the roots of the equation (p − q) x² + 5(p + q) x− 2(p − q) = 0 are real and unequal.

Answer 1

The quadric equation is (p − q) x² + 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 = b² - 4ac

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

D = {5(p + q)}² - 4 x (p - q) x (-2(p - q))

= 25(p² + 2pq + q²) + 8(p² - 2pq + q²)

= 25p² + 50pq + 25q² + 8p² - 16pq + 8q²

= 33p² + 34pq + 33q²

Since, P and q are real and p ≠ q, therefore, the value of D ≥ 0.

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

Hence, proved