How do you solve #sqrt(x+4) = sqrt(x) + sqrt(2)#?

1 Answer
Stefan V.
Oct 14, 2015

#x = 1/2#

Explanation:

Right from the start, you know that, for real numbers, you can't take the square root of a negative value, which means that you need to have

#x >=0#

Next, square both sides of the equation to get

#(sqrt(x+4))^2 = (sqrt(x) + sqrt(2))^2#

#x+4 = (sqrt(x))^2 + 2 sqrt(2x) + (sqrt(2))^2#

#color(red)(cancel(color(black)(x))) + 4 = color(red)(cancel(color(black)(x))) + 2sqrt(2x) + 2#

This is equivalent to

#sqrt(2x) = 1#

Once again, you have

#(sqrt(2x))^2 = 1^2#

#2x = 1 implies x = 1/2#

Since this value of #x# satisfies the condition #x >= 0#, it will be a valid solution to the original equation.

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

#sqrt(1/2 + 4) = sqrt(1/2) + sqrt(2)#

#sqrt(9)/sqrt(2) = (1 + sqrt(2) * sqrt(2))/sqrt(2)#

#3/sqrt(2) = 3/sqrt(2)color(white)(x)color(green)(sqrt())#

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