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Solve for x by rewriting the exponent as a radical

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3+(x+4)^(1/2)=x

Guest Mar 1, 2017
#1
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Solve for x:
3 + sqrt(x + 4) = x

Subtract 3 from both sides:
sqrt(x + 4) = x - 3

Raise both sides to the power of two:
x + 4 = (x - 3)^2

Expand out terms of the right hand side:
x + 4 = x^2 - 6 x + 9

Subtract x^2 - 6 x + 9 from both sides:
-x^2 + 7 x - 5 = 0

Multiply both sides by -1:
x^2 - 7 x + 5 = 0

Subtract 5 from both sides:
x^2 - 7 x = -5

x^2 - 7 x + 49/4 = 29/4

Write the left hand side as a square:
(x - 7/2)^2 = 29/4

Take the square root of both sides:
x - 7/2 = sqrt(29)/2 or x - 7/2 = -sqrt(29)/2

x = 7/2 + sqrt(29)/2 or x - 7/2 = -sqrt(29)/2

x = 7/2 + sqrt(29)/2 or x = 7/2 - sqrt(29)/2

3 + sqrt(x + 4) ⇒ 3 + sqrt(4 + (7/2 - sqrt(29)/2)) = 3 + sqrt(15/2 - sqrt(29)/2) ≈ 5.19258
x ⇒ 7/2 - sqrt(29)/2 = 1/2 (7 - sqrt(29)) ≈ 0.807418:
So this solution is incorrect

3 + sqrt(x + 4) ⇒ 3 + sqrt(4 + (7/2 + sqrt(29)/2)) = 3 + sqrt(1/2 (15 + sqrt(29))) ≈ 6.19258
x ⇒ 7/2 + sqrt(29)/2 = 1/2 (7 + sqrt(29)) ≈ 6.19258:
So this solution is correct

The solution is:
Answer: |x = 7/2 + sqrt(29)/2

Guest Mar 1, 2017