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

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

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

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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.

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You saved my day, thanks a lot. – Surfer on the fall May 3 '13 at 12:15

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 $a = -2b$.

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Let me know if this helps, surfer! ;-) – amWhy May 3 '13 at 16:45
I prefer to have this be mine. + – Babak S. May 3 '13 at 17:52
@amWhy: So beautifully formatted! +1 – Amzoti May 4 '13 at 0:27