How do you solve $\sqrt{2 x + 1} = x - 2$ $sqrt(2x+1)=x-2$ ?

Question

4100 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-05T18:39:40

$x = 3 + \sqrt{6}$ $x = 3 + sqrt(6)$

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

$2 x + 1 \geq 0 \Rightarrow x \geq - \frac{1}{2}$ $2x+1 >=0 => x>= -1/2$ $\to$ $->$ you need to take the square root of a positive number;
$x - 2 \geq 0 \Rightarrow x \geq 2$ $x-2>=0 => x>=2$ $\to$ $->$ the square root of a positive number is a positive number.

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

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

${(\sqrt{2 x + 1})}^{2} = {(x - 2)}^{2}$ $(sqrt(2x+1))^2 = (x-2)^2$

$2 x + 1 = x^{2} - 4 x + 4$ $2x+1 = x^2 - 4x + 4$

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

$x^{2} - 6 x + 3 = 0$ $x^2 -6x +3 = 0$

$x_{1, 2} = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$ $x_(1,2) = (-(-6) +- sqrt( (-6)^2 - 4 * 1 * 3))/(2 * 1)$

$x_{1, 2} = \frac{6 \pm \sqrt{24}}{2} = \frac{6 \pm 2 \sqrt{6}}{2}$ $x_(1,2) = (6 +- sqrt(24))/2 = (6 +- 2sqrt(6))/2$

This means that you have

$x_{1} = 3 + \sqrt{6}$ $x_1 = color(green)(3 + sqrt(6))$ $\to$ $->$ valid solution because $x_{1} \geq 2$ $x_1>=2$ ;

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

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