How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?

Question

How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?

I don't know how to start...

Answer 1

Let $ y=x^2+4x+8 $. Your equation becomes $ 0=y^2+3yx+2x^2=(y+x)(y+2x)$. So you only have to solve 2 quadratic equations. Solve them and you'll see the answer is 2

Answer 2

Let's start by applying the quadratic formula.

We have: $(x^{2}+4x+8)^{2}+3x\hspace{1 mm}(x^{2}+4x+8)+2x^{2}=0$

$\Rightarrow x^{2}+4x+8=\dfrac{-3x\pm\sqrt{(3x)^{2}-4(1)(2x^{2})}}{2(1)}$

$\hspace{27.75 mm} =\dfrac{-3x\pm{x}}{2}$

$\hspace{27.75 mm} =-2x,-x$

Then,

$\Rightarrow x^{2}+4x+8=-2x$

$\Rightarrow x^{2}+6x+8=0$

$\Rightarrow x^{2}+2x+4x+8=0$

$\Rightarrow x\hspace{1 mm}(x+2)+4\hspace{1 mm}(x+2)=0$

$\Rightarrow (x+2)(x+4)=0$

$\Rightarrow x=-2,-4$

or

$\Rightarrow x^{2}+4x+8=-x$

$\Rightarrow x^{2}+5x+8=0$

$\Rightarrow x=\dfrac{-5\pm\sqrt{5^{2}-4(1)(8)}}{2(1)}$

$\hspace{9 mm} =\dfrac{-5\pm\sqrt{-7}}{2}$

$\hspace{9 mm} =\dfrac{-5\pm\sqrt{7}i}{2}$

$\hspace{9 mm} =\dfrac{-5-\sqrt{7}i}{2},\dfrac{-5+\sqrt{7}i}{2}$

However, $x\in\mathbb{Q}$.

Therefore, the solutions to the equation are $x=-2$ and $x=-4$.

Answer 3

We can still use the classic "complete square" trick to do this without having to use the "double" variables as shown by the first answer. To this end, we have: $\left((x^2+4x+8) + \dfrac{3x}{2}\right)^2 - \dfrac{9x^2}{4}+2x^2=0\implies \left((x^2+4x+8)+\dfrac{3x}{2}\right)^2 = \left(\dfrac{x}{2}\right)^2$. Now using the familiar formula $a^2 = b^2 \implies a = \pm b$ to settle the answer.