Solve sqrt{-x^2+3x-2}-1+x=0

$$\sqrt{ - { x }^{ 2 } +3x-2 } -1+x=0$$

Solve for x

$$x=1$$

Solution Steps

Subtract -1+x from both sides of the equation.

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

To find the opposite of -1+x, find the opposite of each term.

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

Square both sides of the equation.

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

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

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from -2 to get -3.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

$$-x^{2}+5x-3=x^{2}$$

Subtract x^{2} from both sides.

$$-x^{2}+5x-3-x^{2}=0$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$-2x^{2}+5x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=5 ab=-2\left(-3\right)=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

Rewrite -2x^{2}+5x-3 as \left(-2x^{2}+3x\right)+\left(2x-3\right).

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

Factor out -x in -2x^{2}+3x.

$$-x\left(2x-3\right)+2x-3$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-x+1\right)$$

To find equation solutions, solve 2x-3=0 and -x+1=0.

$$x=\frac{3}{2} x=1$$

Substitute \frac{3}{2} for x in the equation \sqrt{-x^{2}+3x-2}-1+x=0.

$$\sqrt{-\left(\frac{3}{2}\right)^{2}+3\times \frac{3}{2}-2}-1+\frac{3}{2}=0$$

Simplify. The value x=\frac{3}{2} does not satisfy the equation.

$$1=0$$

Substitute 1 for x in the equation \sqrt{-x^{2}+3x-2}-1+x=0.

$$\sqrt{-1^{2}+3\times 1-2}-1+1=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Equation \sqrt{-x^{2}+3x-2}=1-x has a unique solution.

$$x=1$$