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\[\sqrt{12+4x \ge6}\]
haha, I got this :P
hold on...
o.o im not getting anything....
\(\bf \sqrt{12+4x}\ge 6\implies \begin{cases} +(\sqrt{12+4x})\ge 6\\ \quad \\ \bf -(\sqrt{12+4x})\ge 6 \end{cases}\)
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ohh wait... smokes, nevermind that
thank yall!
okayy lol
for a sec I thought it was hehe absolute value.. ok well hmm
plus, the ">=" is under the square root symbol...
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its not supossed to be lol sorry
\(\bf \sqrt{12+4x}\ge 6 \qquad \textit{square both sides}\\ \quad \\ 12+4x\ge 6^2\implies 4x\ge \cfrac{36}{12} \)
LOL THATS WHY I'M NOT GETTING ANYTHING. x = >=6
so you'd solve it more or less in the same way you'd any linear equation
thank you!
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