Solve sqrt{y^2-y+2}+y=1

$$\sqrt { y ^ { 2 } - y + 2 } + y = 1$$

Solve for y

$$y=-1$$

Solution Steps

Subtract y from both sides of the equation.

$$\sqrt{y^{2}-y+2}=1-y$$

Square both sides of the equation.

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

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

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

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

$$y^{2}-y+2=1-2y+y^{2}$$

Add 2y to both sides.

$$y^{2}-y+2+2y=1+y^{2}$$

Combine -y and 2y to get y.

$$y^{2}+y+2=1+y^{2}$$

Subtract y^{2} from both sides.

$$y^{2}+y+2-y^{2}=1$$

Combine y^{2} and -y^{2} to get 0.

$$y+2=1$$

Subtract 2 from both sides.

$$y=1-2$$

Subtract 2 from 1 to get -1.

$$y=-1$$

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

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

Simplify. The value y=-1 satisfies the equation.

$$1=1$$

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

$$y=-1$$