How do you solve #sqrt(x+6)=x#?

1 Answer
Oct 12, 2015

Answer:

#x=3#

Explanation:

Start by squaring both sides of the equation to get rid of the square root

#(sqrt(x+6))^2 = x^2#

#x + 6 = x^2#

Next, move all the tems on one side of the equation

#x^2 - x - 6 = 0#

Use the quadratic formula to find the two roots of this quadratic equation

#x_(1, 2) = (-(-1) +- sqrt( (-1)^2 - 4 * 1 * (-6)))/(2 * 1)#

#x_(1,2) = (1 +- sqrt(25))/2 = (1 +- 5)/2 = { (x_1 = (1 + 5)/2 = 3), (x_2 = (1 - 5)/2 = -2) :}#

Notice that #x=-2# does not satisfy the initial equation, since you would get

#sqrt(-2 + 6) = -2#

#color(red)(cancel(color(black)(sqrt(4) = -2)))#

This means that #x=-2# will eb an extraneous solution. The only valid solution will thus be #x = 3#, for which

#sqrt(3 + 6) = 3#

#sqrt(9) = 3 " "color(green)(sqrt())#