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!