How do you solve #sqrt (x) + sqrt(x-5) = 5#?

1 Answer
Oct 17, 2015

Answer:

#x = 9#

Explanation:

The first thing to do is figure out which values of #x# cannot be a solution to the equation.

For real numbers, you need to take the square root of positive numbers only, which means that you will need

#x - 5 >= 0 implies x >= 5#

Next, square both sides of the equation to reduce the number of radical terms from two to one.

#(sqrt(x) + sqrt(x-5))^2 = 5^2#

#(sqrt(x))^2 + 2sqrt(x(x-5)) + (sqrt(x-5))^2 = 25#

#x + 2sqrt(x(x-5)) + x - 5 = 25#

Rearrange to get the radical term alone on one side of the equation

#2sqrt(x(x-5)) = 30 - 2x#

#sqrt(x(x-5)) = 15 - x#

Square both sides of the equation to get rid of the square root

#(sqrt(x(x-5)))^2 = (15 - x)^2#

#x(x-5) = 225 - 30x + x^2#

#color(red)(cancel(color(black)(x^2))) - 5x = 225 - 30x + color(red)(cancel(color(black)(x^2)))#

#25x = 225 implies x = color(green)(9)#

Since #x = 9# satisfies the condition #x>= 5#, this will be a valid solution to the original equation.

Do a quick check to make sure that the calculations came out right

#sqrt(9) + sqrt(9-5) = 5#

#3 + 2 = 5color(white)(x)color(green)(sqrt())#