Solve sqrt{2x-1}=x-2?

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{2x-1} to the power of 2 and get 2x-1.

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

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

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

Subtract x^{2} from both sides.

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

Add 4x to both sides.

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

Combine 2x and 4x to get 6x.

$$6x-1-x^{2}=4$$

Subtract 4 from both sides.

$$6x-1-x^{2}-4=0$$

Subtract 4 from -1 to get -5.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-5\right)=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=5 b=1$$

Rewrite -x^{2}+6x-5 as \left(-x^{2}+5x\right)+\left(x-5\right).

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

Factor out -x in -x^{2}+5x.

$$-x\left(x-5\right)+x-5$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x+1\right)$$

To find equation solutions, solve x-5=0 and -x+1=0.

$$x=5 x=1$$

Substitute 5 for x in the equation \sqrt{2x-1}=x-2.

$$\sqrt{2\times 5-1}=5-2$$

Simplify. The value x=5 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Equation \sqrt{2x-1}=x-2 has a unique solution.

$$x=5$$