Simultaneous solution(s) to $a^2+4b^2+4ab=0$ and $a^2+4b^2+32+16a-8b=0$?

Question

Could you tell me just how should I solve this system: $$ a^2+4b^2+4ab=0\\ a^2+4b^2+32+16a-8b=0 $$

I can't remember the proceeding and it's driving me crazy.

Thanks a lot

Answer 1

Hint: Start by taking the difference between the two equations, to get $$ 32+16a+8b-4ab=0 $$ Or, dividing by 4, $$ 8+4a+2b-ab=0 $$ Now, turn your attention to the first equation. It can be factored.

Answer 2

The first equation gives $(a+2b)^2=0\Rightarrow a=-2b$. Sub in the second equation gives $b$ value.

Answer 3

We have: $$ a^2+4b^2+4ab=0 \tag{1}$$ $$ a^2+4b^2+32+16a-8b=0\tag{2} $$

$\bf (I)$ Subtract equation $(2)$ from $(1)$:

$$\begin{align} & a^2+4b^2+4ab & =0\\ - & a^2+4b^2+32+16a-8b &=0 \\ & \hline \\ = & 4ab -16a + 8b - 32 & = 0 \\ 4 & (ab - 4a + 2b - 8) & = 0 \\ \\ = & ab - 4a + 2b -8 = 0 \tag{3}\\ \end{align} $$

$\bf (II)$ Factor equation $(1)$ $$ a^2+4b^2+4ab=0 \iff (a + 2b)^2 = 0 \iff a = -2b \tag{4} $$

$\bf (III)$

Substitute $a = -2b$ into equation $(3)$. Then solve for $a$ the solution for $b$ to obtain $a = -2b$.