Solve sqrt{2x-5}+2=x

$$\sqrt { 2 x - 5 } + 2 = x$$

Solve for x

$$x=3$$

Solution Steps

Subtract 2 from both sides of the equation.

$$\sqrt{2x-5}=x-2$$

Square both sides of the equation.

$$\left(\sqrt{2x-5}\right)^{2}=\left(x-2\right)^{2}$$

Calculate \sqrt{2x-5} to the power of 2 and get 2x-5.

$$2x-5=\left(x-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$2x-5=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$2x-5-x^{2}=-4x+4$$

Add 4x to both sides.

$$2x-5-x^{2}+4x=4$$

Combine 2x and 4x to get 6x.

$$6x-5-x^{2}=4$$

Subtract 4 from both sides.

$$6x-5-x^{2}-4=0$$

Subtract 4 from -5 to get -9.

$$6x-9-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+6x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite -x^{2}+6x-9 as \left(-x^{2}+3x\right)+\left(3x-9\right).

$$\left(-x^{2}+3x\right)+\left(3x-9\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-3\right)+3\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+3\right)$$

To find equation solutions, solve x-3=0 and -x+3=0.

$$x=3 x=3$$

Substitute 3 for x in the equation \sqrt{2x-5}+2=x.

$$\sqrt{2\times 3-5}+2=3$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

Substitute 3 for x in the equation \sqrt{2x-5}+2=x.

$$\sqrt{2\times 3-5}+2=3$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{2x-5}=x-2.

$$x=3 x=3$$