Solve sqrt{y^2-4y+5}=sqrt{y^2+1}

$$\sqrt { y ^ { 2 } - 4 y + 5 } = \sqrt { y ^ { 2 } + 1 }$$

Solve for y

$$y=1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{y^{2}-4y+5}\right)^{2}=\left(\sqrt{y^{2}+1}\right)^{2}$$

Calculate \sqrt{y^{2}-4y+5} to the power of 2 and get y^{2}-4y+5.

$$y^{2}-4y+5=\left(\sqrt{y^{2}+1}\right)^{2}$$

Calculate \sqrt{y^{2}+1} to the power of 2 and get y^{2}+1.

$$y^{2}-4y+5=y^{2}+1$$

Subtract y^{2} from both sides.

$$y^{2}-4y+5-y^{2}=1$$

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

$$-4y+5=1$$

Subtract 5 from both sides.

$$-4y=1-5$$

Subtract 5 from 1 to get -4.

$$-4y=-4$$

Divide both sides by -4.

$$y=\frac{-4}{-4}$$

Divide -4 by -4 to get 1.

$$y=1$$

Substitute 1 for y in the equation \sqrt{y^{2}-4y+5}=\sqrt{y^{2}+1}.

$$\sqrt{1^{2}-4+5}=\sqrt{1^{2}+1}$$

Simplify. The value y=1 satisfies the equation.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

Equation \sqrt{y^{2}-4y+5}=\sqrt{y^{2}+1} has a unique solution.

$$y=1$$