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

Question

2337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-16T23:23:12

#x = 4#

Before doing any calculations, make a note of the fact that any possible solution to this equation must satisfy

#x > = 0#

because, for real numbers, you can only take the square root of a positive number.

The first thing to do is 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 + 2 * sqrt(x * (x+5)) * (sqrt(x+5))^2 = 25#

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

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

#2sqrt(x(x+5)) = 20 - 2x#

#sqrt(x(x+5)) = 10 - x#

Once again, square both sides of the equation to get rid of the square root

#(sqrt(x(x+5)))^2 = (10-x)^2#

#x(x+5) = 100 - 20x + x^2#

This is equivalent to

#color(red)(cancel(color(black)(x^2))) + 5x = 100 - 20x + color(red)(cancel(color(black)(x^2)))#

#25x = 100 implies x = 100/25 = color(green)(4)#

Since #x = 4 >= 0#, this will be a valid solution to the original equation.

Do a quick check to make sure that the calculations are correct

#sqrt(4) + sqrt(4 + 5) = 5#

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