If a and b are re

Comparing with standard quadratic equation ax² + bx + c = 0

a = (a – b) b = 5(a + b) c = – 2(a – b)

Discriminant D = b² – 4ac

= [5(a + b)]² – 4(a – b)(– 2(a – b))

= 25(a + b)² + 8(a – b)²

Since a and b are real and ab then (a + b)² > 0 (a – b)² > 0

8(a – b)² > 0 – – – – (1) product of two positive numbers is always positive

25(a + b)² > 0 – – – – (2) product of two positive numbers is always positive

Adding (1) and (2) we get

8(a – b)² + 25(a + b)² > 0 (sum of two positive numbers is always positive)

D > 0

Hence the roots of given equation are real and unequal.