Solve sqrt{2y^2-3y+2}=y

$$\sqrt { 2 y ^ { 2 } - 3 y + 2 } = y$$

Solve for y

$$y=2$$

$$y=1$$

Solution Steps

Square both sides of the equation.

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

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

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

Subtract y^{2} from both sides.

$$2y^{2}-3y+2-y^{2}=0$$

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

$$y^{2}-3y+2=0$$

To solve the equation, factor y^{2}-3y+2 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-3 ab=2$$

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

$$a=-2 b=-1$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-2\right)\left(y-1\right)$$

To find equation solutions, solve y-2=0 and y-1=0.

$$y=2 y=1$$

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

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

Simplify. The value y=2 satisfies the equation.

$$2=2$$

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

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

Simplify. The value y=1 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{2y^{2}-3y+2}=y.

$$y=2 y=1$$