First slide

Theory of equations

Question

The sum of the non-real roots of $(x^{2} + x - 2) (x^{2} + x - 3) = 12$ is

Moderate

A

-1
B

1
C

-6
D

6

Solution

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

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