Question

Find the value of k for which the equation x⁴ – 14x² + 24x – k = 0 has four real and unequal roots.

Answer 1

Let f(x) = x⁴ – 14x² + 24x – k

Then f'(x) = 4x³ – 28x + 24

= 4(x³ – 7x + 6)

= 4(x – 1)(x – 2)(x + 3)

So, f(x) = 0 has four unequal roots.

Now, f(– 3) = –117 – k, f (1) = 11 – k and f(2) = 8 – k

By the sign scheme,

f(1) < 0, f(2) > 0 & f(– 3) < 0

⇒ (8 – k) < 0, 11 – k > 0 & – 117 – k < 0

⇒ k > 8, k < 11 and k > – 117

⇒ 8 < k < 11

Therfore, 8 < k < 11