Solving following quartic equation

Question

Solve in $\mathbb{R}$ :

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

My try: I tried to make the graph by calculating values for $x=1, 2, 3, 4$ and I found that the function is positive at $x=0$ but negative at $x=1$, so the graph must have crossed the $x$ axis, thus there will be a root between $0$ and $1$ and similarly this was the case between $3$ and $4$, but I was unable to solve it precisely. I was also unable to find about the other two roots.

What is the way to solve it just by algebra or rough plotting with the help of pen and paper and not using any computational tools?

Answer 1

Hint: Rewrite it as:

$$(x^2+2)^2-6x(x^2+2)+8x^2=0$$

Now use the substitution $$y=\frac{x^2+2}{x}$$

This gives

$$x^2y^2-6x^2y+8x^2=0$$

Therefore $$x^2=0 \vee y^2-6y+8=0$$

Answer 2

Expanding the equation:

$$ x^4 + 4x^2 + 4+ 8x^2 = 6x^3 + 12x$$

$$x^4 -6x^3 +12x^2-12x+4=0$$

You can then factorise the above expression

Alternatively, you can replace $x^2 + 2$ with $y$ which gives you:

$$ y^2 + 8x^2 = 6xy$$

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

$$ y = 4x$$ or $$ y = 2x$$ Now replace $y$ with $x^2 + 2$

$x^2 + 2 = 4x $ or $x^2 + 2 = 2x$

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

Answer 3

Let, $$y=x^2+2$$

Now,

$y^2-6xy+8x^2=0$

$\implies \left(\frac{y}{x}\right)^2 -6\left(\frac{y}{x}\right)+8=0$

$\implies y = 4x$

or

$y=2x$

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

or

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

$\implies x = 2 \pm \sqrt{2}, 1 \pm i$

Answer 4

Given that $$(x^2+2)^2+8x^2=6x(x^2+2)$$ Put $x^2+2=y$. Then $$y^2+8x^2=6xy$$ $$8x^2-6xy+y^2=0$$ $$8x^2-4xy-2xy+y^2=0$$ $$4x(2x-y)-y(2x-y)=0$$ $$(4x-y)(2x-y)=0$$ $$(4x-x^2-2)(2x-x^2-2)=0$$ $$(x^2-4x+2)(x^2-2x+2)=0$$ Now we have $$x^2-4x+2=0$$ This is quadratic eq $$x=\frac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}$$ $$x=\frac{4 \pm 2\sqrt{2}}{2}$$ $$ x=2 \pm \sqrt{2}$$ Similarly we have second eq $$x^2-2x+2=0$$ Bu quadratic formula $$x=\frac{-(-2) \pm \sqrt{(-2)^2-4(1)(2)}}{2(1)}$$ $$x=\frac{2 \pm \sqrt{4(-1)}}{2}$$ $$x=\frac{2 \pm 2\sqrt{-1}}{2}$$ $$x=1 \pm i$$ As $\sqrt{-1}=i$.

So all roots of "$x$" we have $$x=2 \pm \sqrt{2} , 1 \pm i $$

Answer 5

$$(x^2+2)^2+8x^2=6x(x^2+2)\Longleftrightarrow$$ $$x^4+4x^2+4+8x^2=6x(x^2+2)\Longleftrightarrow$$ $$x^4+12x^2+4=6x(x^2+2)\Longleftrightarrow$$ $$x^4+12x^2+4=6x^3+12x\Longleftrightarrow$$ $$x^4+12x^2-6x^3-12x+4=0\Longleftrightarrow$$ $$(x^2-4x+2)(x^2-2x+2)=0\Longleftrightarrow$$ $$(x^2-4x+2)=0\vee (x^2-2x+2)=0\Longleftrightarrow$$ $$x^2-4x=-2\vee x^2-2x=-2\Longleftrightarrow$$ $$x^2-4x+4=-2+4\vee x^2-2x+1=-2+1\Longleftrightarrow$$ $$(x-2)^2=2\vee (x-1)^2=-1\Longleftrightarrow$$ $$x-2=\sqrt{2}\vee x-2=-\sqrt{2}\vee -\Longleftrightarrow$$ $$x=2+\sqrt{2}\vee x=2-\sqrt{2}\vee -\Longleftrightarrow$$

"$-$" $\rightarrow$ $(x-1)^2=-1$ has no real solutions

So:

$$(x^2+2)^2+8x^2=6x(x^2+2)\Longleftrightarrow$$ $$x=2+\sqrt{2} \vee x=2-\sqrt{2}$$