If 8k⁶ + 15k³ – 2 = 0, then the positive value of \( \left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right)\) is :

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Question

This question was previously asked in

SSC CGL 2021 Tier-I (Held On : 12 April 2022 Shift 3)

Answer 1

Given:

8k6 + 15k3 – 2 = 0

Calculation:

Let, k3 = x

So, 8x² + 15x - 2 = 0

⇒ 8x2 + 16x - x - 2 = 0

⇒ 8x (x + 2) - 1 (x + 2) = 0

⇒ (8x - 1) (x + 2) = 0

⇒ 8x - 1 = 0 ⇒ x = 1/8

⇒ x + 2 = 0 ⇒ x = - 2 [Not possible because of negative value]

Now, k3 = 1/8

⇒ k = 1/2 ⇒ 1/k = 2

Then, (k + 1/k) = (1/2 + 2) = 5/2 = \(2\frac{1}{2}\)

∴ The value of (k + 1/k) is \(2\frac{1}{2}\)

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