Question

If a, b, c ∈ R, show that the roots of the equation (a – b)x² + (b + c – a)x – с = 0 are rational.

[Hint: D = (a + c – b)² ≥ 0]

Answer 1

Given: a, b, c ∈ R and the quadratic equation (a – b)x² + (b + c – a)x – c = 0.

Step-wise calculation:

1. Write A, B, C for the quadratic:

A = a – b, B = b + c – a, C = –c

2. Use the discriminant D = B² – 4AC (discriminant formula).

3. Compute D:

D = (b + c – a)² – 4(a – b)(–c) 

= (b + c – a)² + 4c(a – b) 

= a² + b² + c² – 2ab + 2ac – 2bc

= (a + c – b)²

Thus `sqrt(D) = ±(a + c - b)`.

4. Apply the quadratic formula:

`x = (-B ± sqrt(D))/(2A)` 

= `(-(b + c - a) ± (a + c - b))/(2(a - b))`

5. Evaluate the two signs:

With +: Numerator = –(b + c – a) + (a + c – b) 

= 2(a – b)

⇒ x = 1   ...(For a ≠ b)

With –: Numerator = –(b + c – a) – (a + c – b) 

= –2c 

⇒ `x = (-c)/(a - b)`   ...(For a ≠ b)

6. Special case a = b:

The equation becomes 0·x² + cx – c = 0, i.e. c(x – 1) = 0.

If c ≠ 0 then x = 1; if c = 0 the equation is 0 = 0 every x is a solution.