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sqrt{x+4}+2=x
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subtract 2 from both sides and square both sides. you will end up with a quadratic equation. solve that for x
can you show me how to set that up ?
sqrt(x+4) = x-2 square both sides x+4 = (x-2)^2
i got it down to x^2-5x=0
explain some more please
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\[\sqrt{x+4}=x-2\] \[x+4=(x-2)^{2}\] \[x+4=x ^{2}-4x+4\] \[x^2-4x-x=0\] \[x ^{2}-5x=0\]
x(x-5)=0 x=0 x=5
actually 0 wouldn't work - put x=0 in your original equation... but x=5 is good
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