If a, b are real then show that the roots of the equation (a – b)x2

1.	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» (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.

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»

(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.