Question

The roots of the equation $x^4-8x^2-9=0$ are

• A. $\pm 3,\pm 1$
• B. $\pm 3,\pm i$
• C. $\pm 2,\pm i$
• D. None of these

Answer 1

Answer: (B)

$\pm 3,\pm i$

Let $x^2=y$

Now we have

$y^2-8y-9=0$

We can factor into

$(y-9)(y+1)$

Now substitute back in

$x^2$ for $y$.

$(x^2-9)(x^2+1)$

Since

$(x^2-9)$is a difference of two squares,

$(x-3)(x+3)$

Now we have

$(x-3)(x+3)(x^2+1)$

x can be $3,-3$ for the first two parts

$x^2+1=0$

can become

$x^2=-1$

Taking the positive and negative root means

$x=\pm \surd -1$

Thus $x=\pm i$ in addition to $3$ and $-3.$