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solve. check for extraneous solutions the square root of x + 8 equals x - 4
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Square both sides.
how do i do that?
\[(\sqrt{x+8})^2=(x-4)^2\]
\[(\sqrt{x+8})(\sqrt{x+8})=(x-4)(x-4)\]
which would be x + 64 = x + 16?
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No. On the left use this property:
\[(\sqrt{x})^2=x\]
On the right use this property:
\[(a-b)(a-b)=a^2-2ab+b^2\]
x^2 - 2x8 + 64?
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x^2-8x+16
i don't understand, =/
\[(\sqrt{x+8})(\sqrt{x+8})=x+8\]
\[(x-4)(x-4)=x^2-4x-4x+16=x^2-8x+16\]
So now we have: \[x^2-8x+16=x+8\]
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Subtract x from both sides. Subtract 8 from both sides.
Then factor.
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