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

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

Solve for x

$$x=19$$

$$x=1$$

Solution Steps

Subtract 2-x from both sides of the equation.

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

To find the opposite of 2-x, find the opposite of each term.

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

The opposite of -x is x.

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

Subtract 2 from 1 to get -1.

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

Square both sides of the equation.

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

Expand \left(3\sqrt{2x-2}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{2x-2}\right)^{2}=\left(-1+x\right)^{2}$$

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

$$9\left(2x-2\right)=\left(-1+x\right)^{2}$$

Use the distributive property to multiply 9 by 2x-2.

$$18x-18=\left(-1+x\right)^{2}$$

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

$$18x-18=1-2x+x^{2}$$

Subtract 1 from both sides.

$$18x-18-1=-2x+x^{2}$$

Subtract 1 from -18 to get -19.

$$18x-19=-2x+x^{2}$$

Add 2x to both sides.

$$18x-19+2x=x^{2}$$

Combine 18x and 2x to get 20x.

$$20x-19=x^{2}$$

Subtract x^{2} from both sides.

$$20x-19-x^{2}=0$$

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

$$-x^{2}+20x-19=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-19. To find a and b, set up a system to be solved.

$$a+b=20 ab=-\left(-19\right)=19$$

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.

$$a=19 b=1$$

Rewrite -x^{2}+20x-19 as \left(-x^{2}+19x\right)+\left(x-19\right).

$$\left(-x^{2}+19x\right)+\left(x-19\right)$$

Factor out -x in -x^{2}+19x.

$$-x\left(x-19\right)+x-19$$

Factor out common term x-19 by using distributive property.

$$\left(x-19\right)\left(-x+1\right)$$

To find equation solutions, solve x-19=0 and -x+1=0.

$$x=19 x=1$$

Substitute 19 for x in the equation 3\sqrt{2x-2}+2-x=1.

$$3\sqrt{2\times 19-2}+2-19=1$$

Simplify. The value x=19 satisfies the equation.

$$1=1$$

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

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

$$x=19 x=1$$