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Solve the equation for x: x+4 = sqrt(8-x)
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Square both sides of the equation first.
(x+4)^2=8-x
x^2+8x+16=8-x
\[(x+4)^2 = (\sqrt{8-x})^2\]
x^2+9x+8=0
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do you know quadratic formula? @nlom
You don't need it.
hmm yeah
Fidn two numebrs that add to give you 9, and multiply to give you 8.
\[x+4=\sqrt{8-x}\] \[8-x \ge0,8\ge x ~or~x \le 8.\] squaring \[\left( x+4 \right)^2=8-x\] \[x^2+8x+16-8+x=0\] \[x^2+9x+8=0\] find x by quadratic formula or by making factors.
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@nlon , you need to participate on your post.
@nlom
(-9+-sqrt(81-32))/2
(-9+-7)/2
tell me what you get
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From squaring both sides I got x^2 + 16 = 8-x ?
no
(a+b)^2=a^2+2ab+b^2
Thanks!
np
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