Question

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

• A. $\pm 1,\pm i$
• B. $\pm 3,\pm i$
• C. $\pm 2,\pm i$
• D. $noneofthese$

Answer 1

Answer: (A)

$\pm 1,\pm i$

Let $x^2=y$, then the given equation $x^4-8x^2-9=0$ becomes $y^2-8y-9=0$. Now, we factorize the equation $y^2-8y-9=0$ as follows:

$y^2-8y-9=0\Rightarrow y^2-9y+y-9=0\Rightarrow y(y-9)+1(y-9)=0\Rightarrow (y+1)(y-9)=0\Rightarrow (x^2+1)(x^2-9)=0(\because x^2=y)\Rightarrow (x^2+1)(x^2-3^2)=0\Rightarrow (x^2+1)(x-3)(x+3)=0(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)\Rightarrow (x^2+1)=0,(x-3)=0,(x+3)=0\Rightarrow x^2=-1,x=3,x=-3\Rightarrow x=\pm \sqrt{-1},x=3,x=-3\Rightarrow x=\pm i,x=\pm 3(\because i^2=-1)$

Hence, the roots of $x^4-8x^2-9=0$ are $\pm 3,\pm i$.