√(x + 7) = x − 5
square both sides
x + 7 = (x − 5)²
expand
x + 7 = x² − 10x + 25
x² − 11x + 18 = 0
(x − 2)(x − 9) = 0
x = 2 ....... solution1
x = 9 ....... solution2
Remember that the x=2 solution, when inserted into the original equation
is valid because the √ has a positive and a negative root
√(2 + 7) = 2 − 5
√9 = -3
-3 = -3
.
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Verified answer
√(x + 7) = x − 5
square both sides
x + 7 = (x − 5)²
expand
x + 7 = x² − 10x + 25
x² − 11x + 18 = 0
(x − 2)(x − 9) = 0
x = 2 ....... solution1
x = 9 ....... solution2
Remember that the x=2 solution, when inserted into the original equation
√(x + 7) = x − 5
is valid because the √ has a positive and a negative root
√(2 + 7) = 2 − 5
√9 = -3
-3 = -3
.