Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$

Question:

Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$

My attempts:

1. Here's the expanded form:$$x^4-6x^3+12x^2-12x+4=0$$
2. I've plugged this into several online "math problem solving" websites, all claim that "solution could not be determined algebraically, hence numerical methods (i suppose the quartic formula?) were used"
3. I've substituted $y=x^2+2$ but then the $6x$ remains to prevent me from solving.
4. I've tried factorizing in other ways, I've tried finding simple first solutions but they are actually not simple so I couldn't find them.
5. Factoring into circles and hyperbola to arrive at a geometric solution, by estimating the roots from the graph

For reference, the roots are: (credits to wolframalpha)

$$2+\sqrt{2}, 2-\sqrt{2},1+i, 1-i$$

Setting $y=x^2+2$ gives $$y^2+8x^2=6xy,$$ i.e. $$y^2-6xy+8x^2=0$$ from which we have $$(y-2x)(y-4x)=0$$ I think that you can take it from here.

• Easiest, thanks! Oct 21, 2017 at 6:20

$$(x^2+2)^2+8x^2=6x(x^2+2)$$ $$(x^2+2)^2-6x(x^2+2)+8x^2=0$$ Let $x^2+2=U, x=V$. Then $$U^2-6UV+8V^2=0$$ Then $$\left(\frac UV\right)^2-6\left(\frac UV\right)+8=0$$ Then $\frac UV=2$ or $\frac UV=4$

$\frac {x^2+2}{x}=2$ or $\frac {x^2+2}{x}=4$

$x^2-2x+2=0$ or $x^2-4x+2=0$

$$x=2\pm \sqrt2$$

• +1 The essence of this is the same as the accepted answer, albeit expressed more verbosely. Oct 21, 2017 at 6:40

The trick of this is "complete square". Add $x^2$ to both sides first, and move the $6x(x^2+2)$ to the left side of the equation:

$(x^2+2)^2 - 2\cdot (x^2+2)\cdot 3x + (3x)^2 = x^2\implies ((x^2+2) - 3x)^2 = x^2$ , and you have the form $A^2 = B^2$ and you know how to take it from here....

• +1 That's the quickest but difficult to think of for the less inexperienced like me :) Oct 21, 2017 at 6:42
• Duh I meant the less experienced -_- Oct 21, 2017 at 9:14

By your hint we obtain that it's $$x^4-6x^3+12x^2-12x+4=0$$ or $$x^4-2x^3+2x^2-4x^3+8x^2-8x+2x^2-4x+4=0$$ or $$(x^2-2x+2)(x^2-4x+2)=0,$$ which gives the answer: $$\{1+i,1-i,2+\sqrt2,2-\sqrt2\}.$$

If we don't know the answer then we can get the following.

For all real $k$ we have: $$x^4-6x^3+12x^2-12x+4=$$ $$=(x^2-3x+k)^2-9x^2-k^2+6kx-2kx^2+12x^2-12x+4=$$ $$=(x^2-3x+k)-2kx^2+3x^2+6kx-12x-k^2+4=$$ $$=(x^2-3x+k)^2-((2k-3)x^2-(6k-12)x+k^2-4).$$ Now, we need to choose a value of $k$ such that $(2k-3)x^2-(6k-12)x+k^2-4=(ax+b)^2$ and we see that $k=2$ is valid.

Thus, $$x^4-6x^3+12x^2-12x+4=(x^2-3x+2)^2-x^2=(x^2-4x+2)(x^2-2x+2)$$ and the rest you saw.

The best way it's the substitution $x^2+2=t$.

• That's easy to see if one already knows the answer $\ddot\smile$ Oct 21, 2017 at 6:06
• That's better - you are starting to use the conventional method for solving the quartic; next step, get a cubic equation for $k$. Oct 21, 2017 at 6:12
• How did you even arrive at the strange expression on the RHS in the line following "For all real k we have:" o_O? Oct 21, 2017 at 6:41
• @Gaurang Tandon I added something. See now. Oct 21, 2017 at 7:53
• Thanks! But, why are you doing what you're currently doing? I meant, is this some standard technique? Oct 21, 2017 at 9:15