Question

If the equation (k-2)x2 – (k-4)x-2=0 has difference of
roots as 3 then the value of k is
(a)1,3
(b) 3,3/2
(c)2,3/
2 014
(d) 3/2, 1​

Answer 1

Answer:

b) 3, 3/2

Step-by-step explanation:

Let α and β be the roots of the equation.

α + β = k-4/k-2

αβ = -2/k-2

given α - β = 3

(α - β)² = 9

(α + β)² - 4αβ = 9

[k-4/k-2]² - 4[-2/k-2] = 9

(k-4)² + 8(k-2) = 9 (k-2)²

k² - 8k + 16 +8k - 16 = 9k² - 36k + 36

8k² - 36k + 36 = 0

2k² - 9k + 9 = 0

2k² - 6k - 3k + 9 = 0

2k(k-3) - 3(k-3) = 0

(2k-3)(k-3) = 0

k = 3, 3/2