The value of k for which the equation (k - 2)x² + 8x + k + 4 = 0 has both real, distinct and negative roots is

Concept:

If ax² + bx + c = 0 is a quadratic equation with D = b² - 4ac as its discriminant then,

• D > 0 then the quadratic equation has real and distinct roots
• D = 0 then the quadratic equation has real and repeated roots
• D < 0 then the quadratic equation has complex and conjugate roots

Calculation:

Given: The quadratic equation (k - 2)x2 + 8x + k + 4 = 0 has both real, distinct and negative roots

As we know that, for a quadratic equation ax2 + bx + c = 0 if the discriminant D > 0 then roots are real and distinct.

Here, a = k - 2, b = 8 and c = k + 4

⇒ 64 - 4 × (k - 2) × (k + 4) > 0

⇒ 16 - k² - 2k + 8 > 0

⇒ k² + 2k - 24 < 0

⇒ (k + 6) (k - 4) < 0

⇒ k ∈ (- 6, 4)----------(1)

For both the roots to be negative - b/2a < 0

⇒ \(-\frac{8}{2 \times (k -2)} < 0\)

⇒ \(\frac{4}{k-2}>0\)

⇒ k ∈ (2 , ∞) --------(2)

Now from (1) and (2), we get k ∈ (2, 4)

Hence, option C is the correct answer.