Question

The imaginary roots of the equation ${(x^2+2)}^2+8x^2=6x(x^2+2)$ are

• A. $1\pm i$
• B. $2\pm i$
• C. $-1\pm i$
• D. $-2\pm i$

Answer 1

The correct option is A $1\pm i$

The given equation is $(x^2+2{)}^2+8x^2=6x(x^2+2)$

Let $x^2+2=t,$ then the equation becomes

$t^2-6xt+8x^2=0$

$t^2-4xt-2xt+8x^2=0$

$t(t-4x)-2x(t-4x)=0$

$(t-2x)(t-4x)=0$

Now, putting the value of $t$ in above equation, we get

$(x^2-4x+2)(x^2-2x+2)=0$

$\Rightarrow$ $x^2-4x+2=0$

$\Rightarrow$ $x^2-2x-2x+2=0$

$\Rightarrow$ $(x-2)(x-2)=0$

$\therefore$ $x=2,2$, which is real roots.

Now,

$\Rightarrow$ $x^2-2x+2=0$

$\Rightarrow$ $x=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(1\left)\right(2)}}{2\left(1\right)}$

$\Rightarrow$ $x=\frac{2\pm \sqrt{-4}}{2}$

$\Rightarrow$ $x=1\pm i$

$\therefore$ Required imaginary roots are $1\pm i$