Question

If a, b are real then show that the roots of the equation (a – b)x² – 6(a + b)x – 9(a – b) = 0 are real and unequal.

Answer 1

(a – b)x² – 6(a + b)x – 9(a – b) = 0 

Here a = a – b; b = – 6(a + b); c = – 9(a – b) 

∆ = b² – 4ac 

= [- 6(a + b)]² – 4(a – b)[-9(a – b)] 

= 36(a + b)² + 36(a – b)(a – b) 

= 36 (a + b)² + 36 (a – b)² 

= 36 [(a + b)² + (a – b)²]

The value is always greater than 0 

∆ = 36 [(a + b)² + (a – b)²] > 0 

∴ The roots are real and unequal.