Question

If product of the imaginary roots of ${(x^2+2)}^2+{8x}^2=6x(x^2+2)$ is

$i\pm k$ find $k.$

Answer 1

$(x^2+2{)}^2+8x^2=6x(x^2+2)$

Let $(x^2+2)=t$

$\therefore$ $t^2+8x^2=6xt$

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

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

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

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

Substitute value of $t$

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

$\Rightarrow$ $(x^2-4x+2)$ has real roots as $D>0$

So, take $x^2-2x+2=0$

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

$=\frac{2\pm \sqrt{4-8}}{2}$

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

$=\frac{2\pm 2i}{2}$

$=1\pm i$

$\therefore$ $x=1\pm i$ can also be written as $i\pm 1$

Now, $i\pm 1$ comparing it with $i\pm k$

we get, $k=1$