Question

1. Find the solutions of the equation x^4−x^2−12. Show your work.

Answer 1

Answer:$x=\pm 2$

Step-by-step explanation:

Given

$x^4-x^2-12=0$

Let $x^2=y$

thus

$y^2-y-12=0$

splitting $-y as -4 y+3 y$

$y^2-4y+3y-12=0$

$y(y-4)+3(y-4)=0$

$(y-4)(y+3)=0$

thus y=4 or y=-3 but y=-3 is not a solution because square of a number cannot be negative

thus $x^2=4$

$x=\pm 2$