Question

solve the equation.
$x^4 + 2x^2 - 3 = 0$
$x = \pm ?, \pm \sqrt{\square}i$
enter the real solutions.

Explanation:

Step1: Substitute $y=x^2$

Let $y = x^2$, so the equation becomes:
$$y^2 + 2y - 3 = 0$$

Step2: Factor the quadratic equation

Factor the quadratic:
$$(y+3)(y-1) = 0$$

Step3: Solve for $y$

Set each factor equal to 0:
$y+3=0 \implies y=-3$; $y-1=0 \implies y=1$

Step4: Substitute back $y=x^2$

For $y=1$: $x^2=1 \implies x=\pm1$
For $y=-3$: $x^2=-3 \implies x=\pm\sqrt{3}i$

Step5: Identify real solutions

Real solutions come from $x^2=1$.

Answer:

The real solutions are $x = \pm1$