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# help

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Find all real numbers x such that $$\sqrt{x} + \sqrt[4]{x} = 12$$

Jan 3, 2020

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Solve for x:
x^(1/4) + sqrt(x) = 12

Subtract 12 from both sides:
-12 + x^(1/4) + sqrt(x) = 0

Simplify and substitute y = x^(1/4).
-12 + x^(1/4) + sqrt(x) = -12 + x^(1/4) + (x^(1/4))^2
= y^2 + y - 12:
y^2 + y - 12 = 0

The left hand side factors into a product with two terms:
(y - 3) (y + 4) = 0

Split into two equations:
y - 3 = 0 or y + 4 = 0

y = 3 or y + 4 = 0

Substitute back for y = x^(1/4):
x^(1/4) = 3 or y + 4 = 0

Raise both sides to the power of four:
x = 81 or y + 4 = 0

Subtract 4 from both sides:
x = 81 or y = -4

Substitute back for y = x^(1/4):
x = 81 or x^(1/4) = -4

Raise both sides to the power of four:
x = 81 or x = 256

x^(1/4) + sqrt(x) ⇒ 81^(1/4) + sqrt(81) = 12:
So this solution is correct

x^(1/4) + sqrt(x) ⇒ 256^(1/4) + sqrt(256) = 20:
So this solution is incorrect

The solution is:

x = 81

Jan 3, 2020