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Solve for x: (x - 9)2 = 1 So far I got (x-9)2=1 (√x-92)=√1 x-9=1 x=1=9 x=10 was told to...take the square root of both sides. So x - 9 = +-1. Then what? I'm confused.
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Hint: \[(x-9)^2 = 1\]\[x - 9 = \pm \sqrt{1}\]
x=10,8
In general, when \[a^2 = b\], \[a = \pm \sqrt{b}\]
\[\sqrt{a^2} = \left| a \right|\]
Expand getting: \[x ^{2}-18x + 81 = 1\]\[x ^{2}-18x + 80=0\]
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\[(x-8)(x-10)=0\]\[x=10\]\[x=8\]
Thanks.
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