If a,b,c are real numbers and a not equal to b then prove that (a-b)x²-5(a+b)k-2(a-b)=0 has two distinct real roots?

　
I assume that "k" should actually be x??
(a−b)x² − 5(a+b)x − 2(a−b) = 0

Quadratic equation has 2 distinct real roots when discriminant > 0

Discriminant = B²−4AC, where: A = (a−b), B = −5(a+b), C = −2(a−b)

D = B²−4AC
= (−5(a+b))² − 4(a−b)(−2(a−b))
= 25(a+b)² + 8(a−b)²
= 25(a²+2ab+b²) + 8(a²−2ab+b²)
= 33a² + 34ab + 33b²

Since a ≠ b, then they cannot both be 0:

When a = 0, b ≠ 0 ---> D = 33b² > 0

When b = 0, a ≠ 0 ---> D = 33a² > 0

When a ≠ 0, b ≠ 0 ---> ab ≠ 0
If ab > 0
D = 33a² + 34ab + 33b² = 33(a−b)² + 100ab > 0
If ab < 0, −ab > 0
D = 33a² + 34ab + 33b² = 33(a+b)² − 32ab > 0

In all cases, D > 0, so quadratic equation has 2 distinct real roots

The equation should be (a - b)x²- 5(a + b)x - 2(a - b) = 0 instead.

The discriminant of the equation, Δ
= [-5(a + b)]² - 4(a - b)[-2(a - b)]
= 25(a + b)² + 8(a - b)²
> 0 for a ≠ b

As Δ > 0, the equation has two distinct real roots.