Question

1. what are the solutions of this quadratic like equation $x^{4}+15x^{2}-16=0?$

a $pm1, pm4i$
b $pm4, pm i$
c $pm1, pm4$
d $pm i, pm4i$

Explanation:

Step1: Substitute $y=x^2$

Let $y = x^2$, then the equation becomes:
$$y^2 + 15y - 16 = 0$$

Step2: Factor the quadratic equation

Factor the equation to find $y$:
$$(y + 16)(y - 1) = 0$$

Step3: Solve for $y$

Set each factor equal to 0:
$y + 16 = 0 \implies y = -16$
$y - 1 = 0 \implies y = 1$

Step4: Substitute back $y=x^2$ and solve for $x$

For $y=1$:
$x^2 = 1 \implies x = \pm 1$
For $y=-16$:
$x^2 = -16 \implies x = \pm 4i$

Answer:

A. $\pm 1, \pm 4i$