Question

Find the number of real roots of the equation ${(x^2+2)}^2+8x^2=6x(x^2+2)$

Answer 1

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

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

then,

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

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

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

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

Putting value of t

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

$(x^2-2x+2)\to$It is has imaginary roots

So, taking $x^2-4x+2=0$

Using Sri dharacharya formula

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

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

$x=\frac{4\pm \sqrt{16-8}}{2}$

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

$x=\frac{2(2\pm \sqrt{2})}{2}$

$x=2\pm \sqrt{2}$

$x=2+\sqrt{2},2-\sqrt{2}$

Hence number of real roots are 2.