Question

If α and β are the roots of the equation 2x² – 3x – 6 = 0, then equation whose roots are α² + 2, β² + 2 is

• A. 4x² + 49x + 118 = 0
• B. 4x² – 49x + 118 = 0
• C. 4x² – 49x – 118 = 0
• D. x² – 49x + 118 = 0

Answer 1

Answer: (B)

4x² – 49x + 118 = 0

Answer 2

Here α + β = 3/2, αβ = –6/2 = –3 so that

S = α² + β² + 4 = (α + β)² – 2αβ + 4 = $\frac{49}{4}$,

P = α²β² + 2(α² + β²) + 4 = α²β² + 4 + 2[(α + β)² – 2αβ ]

=$\frac{118}{4}$.

Therefore, the equation is x² – $\left( \frac{49}{4} \right)x + \left( \frac{118}{4} \right) = 0$

⇒ 4x² – 49x + 118 = 0.

Hence (2) is the correct answer.

Alternate: Let y = x² +2, then 2x² – 3x – 6 = 0

⇒ (3x)² = (2x² – 6)² ⇒ [2(y – 2) – 6]² = 9(y – 2)

⇒ 4y² – 49y + 118 = 0