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

1 Answer
Aug 5, 2015

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