How do you solve $2(sqrt x) + x=8$ ?

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Answer 1 · 2016-03-03T14:54:02

You must square both sides to get rid of the radical.

$2sqrt(x) = 8 - x$

$(2sqrt(x))^2 = (8 - x)^2$

$4x = 64 - 16x + x^2$

$0 = x^2 - 20x + 64$

$0 = (x - 16)(x - 4)$

$x = 16 and 4$

You must check the solutions inside the original equation to ensure they work. 4 does but 16 doesn't, so the solution is ${4}$

Practice exercises:

Solve for x in the following equations. Don't forget to check your solutions in the original equation.

a) $sqrt(5x + 6) = x$

b) $sqrt(4x + 1) - sqrt(x + 2) = 1$

c) $sqrt(2x + 1) + sqrt(3x + 4) = sqrt(6x + 1)$

Good luck!

How do you solve 2(sqrt x) + x=82(sqrt x) + x=8?