Question

If a and b are real and a ≠ b then show that the roots of the equation (a – b)x² + 5(a + b)x – 2(a – b) = 0 are real and unequal.

Answer 1

The given equation is (a – b)x² + 5(a + b)x – 2(a – b) = 0

∴ D = [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 a ≠ b, so (a – b)² > 0 and (a + b)² > 0 

∴ 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) 

⇒ D > 0 

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