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solve √x+1 +5=x (the "+5" isnt under the radical)
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|dw:1339273409346:dw|
\[\sqrt{x+1}=x-5\]
√x+1 = x - 5 square both sides x + 1 = (x - 5)^2
\[x+1=x^2-10x+25\]
\[x^2-11x+24=0\]
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this rearranges to x^2 -11x + 24 = 0
\[(x-3)(x-8)=0\]
\[x=3, x=8\]
Must always check the answers after squaring.
so what your saying is +3, and +8 are both x?? @Mertsj
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|dw:1339273434621:dw| x + 1 = x^2 - 10x + 25 x^2 - 11x + 24 = 0 (x - 8)(x - 3) = 0 x = 8, 3 Check for extraneous solutions by plugging it back in. √(x + 1) + 5 = x √(9) + 5 = 8 3 = 3 √(x + 1) + 5 = x √(4) + 5 = 3 2 + 5 ≠ 3 THe solution is x = 8
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