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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`A. -1B. 1C. -6D. 6 |
Answer» Correct Answer - 1 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. |
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