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

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

Solve for x

$$x=6$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$2x-3=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$2x-3-x^{2}=-6x+9$$

Add 6x to both sides.

$$2x-3-x^{2}+6x=9$$

Combine 2x and 6x to get 8x.

$$8x-3-x^{2}=9$$

Subtract 9 from both sides.

$$8x-3-x^{2}-9=0$$

Subtract 9 from -3 to get -12.

$$8x-12-x^{2}=0$$

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

$$-x^{2}+8x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-12\right)=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=6 b=2$$

Rewrite -x^{2}+8x-12 as \left(-x^{2}+6x\right)+\left(2x-12\right).

$$\left(-x^{2}+6x\right)+\left(2x-12\right)$$

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

$$-x\left(x-6\right)+2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+2\right)$$

To find equation solutions, solve x-6=0 and -x+2=0.

$$x=6 x=2$$

Substitute 6 for x in the equation \sqrt{2x-3}=x-3.

$$\sqrt{2\times 6-3}=6-3$$

Simplify. The value x=6 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Equation \sqrt{2x-3}=x-3 has a unique solution.

$$x=6$$