The sum of the non-real root of `(x^2+x-2)(x^2+x-3)=12` is `-1` b. `1` c. `-6` d. `6`

Options

A. -1
B. 1
C. -6
D. 6

Answer

1

Explanation

Put `x^(2) + x = y`, so that Eq. (1) becomes
`(y -2)(y - 3) = 12`
or `y^(2) -5y -6 =0`
or `(y - b)(y + 1) = 0` or y = 6, -1
When y = 6, we get
`x^(2) + x - 6 = 0`
`rArr (x + 3)(x - 2) = 0` or x = -3, 2
When y = -1, we get
`x^(2) + x + 1 = 0`
which has nonreal roots and sum of roots is -1.