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