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

1 Answer
Aug 5, 2015

Answer:

#x = 3 + sqrt(6)#

Explanation:

Right from the start, you know that any solution you find must satisfy the conditions

  • #2x+1 >=0 => x>= -1/2# #-># you need to take the square root of a positive number;

  • #x-2>=0 => x>=2# #-># the square root of a positive number is a positive number.

all in all, you need your valid solution(s) to satisfy the condition #x>=2#.

Since the radical term is already isolated on one side of the equation, proceed to square both sides to get

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

#2x+1 = x^2 - 4x + 4#

Next, move all your terms on one side of the equation and use the quadratic formula to determine the two solutions of this quadratic equation

#x^2 -6x +3 = 0#

#x_(1,2) = (-(-6) +- sqrt( (-6)^2 - 4 * 1 * 3))/(2 * 1)#

#x_(1,2) = (6 +- sqrt(24))/2 = (6 +- 2sqrt(6))/2#

This means that you have

#x_1 = color(green)(3 + sqrt(6))# #-># valid solution because #x_1>=2#;

#x_2 = cancel(color(red)(3-sqrt(6))# #-># extraneous solution because #x_2 cancel(>=)2#