If 8k⁶ + 15k³ - 2 = 0, ten te positive value of (k + 1/k ) is:

If 8k⁶ + 15k³ - 2 = 0, ten te positive value of (k + 1/k )…

If 8k⁶ + 15k³ - 2 = 0, ten te positive value of (k + $\frac{1}{k}$ ) is:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. 2 $\frac{1}{2}$

Let k³ = x, then

8k⁶ + 15k³ - 2 = 0 becomes 8x² + 15x - 2 = 0

⇒ 8x² + 15x - 2 = 0 ⇒ 8x² + 16x - x - 2 = 0

⇒ 8x(x + 2) - 1(x + 2) = 0

⇒ (8x - 1)(x + 2) = 0

Either 8x - 1 = 0 or x + 2 = 0

⇒ Either x = 1/8 or -2

We have to find the positive value of {k + 1/k} so x cannot be negative.

Thus, x = 1/8

⇒ k³ = 1/8 ⇒ k = 1/2

k + 1/k = 1/2 + 2 = 2 $\frac{1}{2}$