Kristen Stokes

2022-07-14

Simultaneous solution(s) to
${a}^{2}+4{b}^{2}+4ab=0\phantom{\rule{0ex}{0ex}}{a}^{2}+4{b}^{2}+32+16a-8b=0$

behk0

Expert

The first equation gives $\left(a+2b{\right)}^{2}=0⇒a=-2b$. Sub in the second equation gives $b$ value.

Palmosigx

Expert

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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