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(x – 9)^2 = (x + 1)^2
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Are you aloud to just take the square root of both sides?
\[ (x – 9)^2 = (x + 1)^2 \\ (x – 9)^2 - (x + 1)^2 = 0 \\ \text{Since }a^2 - b^2 = (a+b)(a-b): \\ (x – 9)^2 - (x + 1)^2 = (x-9+x+1)(x-9-x-1) = 0 \\ (2x-8)(-10) = 0 \\ 2x - 8 = 0 \\ 2x = 8 \\ x = 8/2 = 4 \]
Another way to do this problem: If a^2 = b^2 then either a = b OR a = -b (x – 9)^2 = (x + 1)^2 x - 9 = x + 1 OR x - 9 = -x - 1 (no solution) OR x = -x - 1 + 9; x = -x + 8; 2x = 8; x = 4
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