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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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3 Answers 3

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

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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. + –  B. S. May 3 '13 at 17:52
    
@amWhy: So beautifully formatted! +1 –  Amzoti May 4 '13 at 0:27

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

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