Question

Let α, β be the roots of the equation `x^2 - sqrt(2)x + sqrt(6)` = 0 and `1/α^2 + 1, 1/β^2 + 1` be the roots of the equation x² + ax + b = 0. Then the roots of the equation x² – (a + b – 2)x + (a + b + 2) = 0 are ______.

विकल्प

• non-real complex numbers

• real and both negative

• real and both positive

• real and exactly one of them is positive

Answer 1

Let α, β be the roots of the equation `x^2 - sqrt(2)x + sqrt(6)` = 0 and `1/α^2 + 1, 1/β^2 + 1` be the roots of the equation x² + ax + b = 0. Then the roots of the equation x² – (a + b – 2)x + (a + b + 2) = 0 are real and both negative.

Explanation:

For x² + ax + b = 0

Sum of roots  = a = `(-1)/α^2 - 1/β^2 - 2`

Product of roots = b = `(-1)/α^2 + 1/β^2 + 1 + 1/(α^2β^2)`

a + b = `1/(αβ)^2 - 1`

= `1/6 - 1`

= `-5/6`  ...[∵ αβ = `sqrt(6)`]

`x^2 - (-5/6 - 2)x + (2 - 5/6)` = 0

⇒ 6x² + 17x + 7 = 0

x = `-7/3`, x = `-1/2` are the roots

Both roots are real and negative.