Solve sqrt{2x-3}+1=x

$$\sqrt { 2 x - 3 } + 1 = x$$

Solve for x

$$x=2$$

Solution Steps

Subtract 1 from both sides of the equation.

$$\sqrt{2x-3}=x-1$$

Square both sides of the equation.

$$\left(\sqrt{2x-3}\right)^{2}=\left(x-1\right)^{2}$$

Calculate \sqrt{2x-3} to the power of 2 and get 2x-3.

$$2x-3=\left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$2x-3=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$2x-3-x^{2}=-2x+1$$

Add 2x to both sides.

$$2x-3-x^{2}+2x=1$$

Combine 2x and 2x to get 4x.

$$4x-3-x^{2}=1$$

Subtract 1 from both sides.

$$4x-3-x^{2}-1=0$$

Subtract 1 from -3 to get -4.

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

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

$$-x^{2}+4x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-4\right)=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite -x^{2}+4x-4 as \left(-x^{2}+2x\right)+\left(2x-4\right).

$$\left(-x^{2}+2x\right)+\left(2x-4\right)$$

Factor out -x in the first and 2 in the second group.

$$-x\left(x-2\right)+2\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(-x+2\right)$$

To find equation solutions, solve x-2=0 and -x+2=0.

$$x=2 x=2$$

Substitute 2 for x in the equation \sqrt{2x-3}+1=x.

$$\sqrt{2\times 2-3}+1=2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Substitute 2 for x in the equation \sqrt{2x-3}+1=x.

$$\sqrt{2\times 2-3}+1=2$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

List all solutions of \sqrt{2x-3}=x-1.

$$x=2 x=2$$