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?


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



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

$\implies y = 4x$



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



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

  • $\begingroup$ You should solve in $\mathbb{R}$, so $1\pm i$ is not a solution. $\endgroup$
    – Frank Vel
    Jul 22 '15 at 17:40

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

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


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

  • $\begingroup$ I'm not sure I could have seen that $(x^2-4x+2)(x^2-2x+2)$ is a factorization of $x^4+12x^2-6x^3-12x+4$ just be examining the latter equation. How did you do it? $\endgroup$
    – David K
    Jul 17 '15 at 12:35
  • $\begingroup$ First you've write out your equation and subtract everything, then you get something that ends with $=0$ than you look for the way to write it $()()=0$ $\endgroup$ Jul 17 '15 at 12:39
  • $\begingroup$ Yes, we got the "something that ends with $=0$" part; that was easy. It's the "look for the way to write it $()()=0$" that is a mystery. $\endgroup$
    – David K
    Jul 17 '15 at 12:48
  • 3
    $\begingroup$ @DavidK Posit a solution $$x^4 - 6x^3 + 12x^2 - 12x + 4 = (x^2 + ax + b)(x^2 + cx + d)$$ Matching coefficients yields \begin{align*} a + c & = -6\\ ac + b + d & = 12\\ ad + bc & = -12\\ bd & = 4\end{align*} The choice $b = 1, d = 4$ leads to the contradiction $a + c = -6$, $ac = 7$. The choice $b = 2, d = 2$ yields $a + c = -6$ and $ac = 8$, from which you can determine that $a = -4$ and $c = -2$. $\endgroup$ Jul 18 '15 at 10:02

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


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 $x^2 + 2 = 2x$

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

  • $\begingroup$ I tried to expand but was unable to factorise it. $\endgroup$ Jul 17 '15 at 10:19
  • $\begingroup$ @John the solutions are ugly radicals. Of course, one could factorise any polynomial, but it doesn't work this case. $\endgroup$
    – wythagoras
    Jul 17 '15 at 10:22
  • $\begingroup$ @wythagoras, you're right, it doesnt factorise nicely. $\endgroup$
    – John_dydx
    Jul 17 '15 at 10:25

Hint: Rewrite it as:


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

This gives


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

  • $\begingroup$ How did you thought about it?I tried to put $y=x^2+2 $ $\endgroup$ Jul 17 '15 at 10:16
  • $\begingroup$ @KartikWatwani because the term with $(x^2+2)^2$ has no $x$, the term with $(x^2+2)$ has one $x$, the term without it has $x^2$. This is asymmetric, thus this technique can be used. $\endgroup$
    – wythagoras
    Jul 17 '15 at 10:20
  • $\begingroup$ What do you mean when you say asymmetric? @wythagoras $\endgroup$ Jul 17 '15 at 10:25
  • 1
    $\begingroup$ @Kartik Watwani: in the equation, set $y=x^2+2$. The equation rewrites as: $\;y^2+8x^2=6xy$, which is a homogeneous function of $x$ and $y$. The standard method to solve it is to set $t=y/x$ (or $x/y$), whichbrings us back to an equation in 1 variable. $\endgroup$
    – Bernard
    Jul 17 '15 at 10:44
  • 1
    $\begingroup$ @wythagoras And due to such 'false results' one should check before the division if the zero is a solution (so it appears soon, if valid), then explicitly exclude it from further calculations (so it won't appear late as a false solution). It's much better (and sometimes easier) to not introduce false results, than eliminate them by additional checking at the end. $\endgroup$
    – CiaPan
    Jul 17 '15 at 10:47

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