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

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

Solve for y

$$y=3$$

Solution Steps

Subtract -\sqrt{y+1} from both sides of the equation.

$$\sqrt{5y+1}=2+\sqrt{y+1}$$

Square both sides of the equation.

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

Calculate \sqrt{5y+1} to the power of 2 and get 5y+1.

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

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

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

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

$$5y+1=4+4\sqrt{y+1}+y+1$$

Add 4 and 1 to get 5.

$$5y+1=5+4\sqrt{y+1}+y$$

Subtract 5+y from both sides of the equation.

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

To find the opposite of 5+y, find the opposite of each term.

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

Subtract 5 from 1 to get -4.

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

Combine 5y and -y to get 4y.

$$4y-4=4\sqrt{y+1}$$

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4y-4\right)^{2}.

$$16y^{2}-32y+16=\left(4\sqrt{y+1}\right)^{2}$$

Expand \left(4\sqrt{y+1}\right)^{2}.

$$16y^{2}-32y+16=4^{2}\left(\sqrt{y+1}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16y^{2}-32y+16=16\left(\sqrt{y+1}\right)^{2}$$

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

$$16y^{2}-32y+16=16\left(y+1\right)$$

Use the distributive property to multiply 16 by y+1.

$$16y^{2}-32y+16=16y+16$$

Subtract 16y from both sides.

$$16y^{2}-32y+16-16y=16$$

Combine -32y and -16y to get -48y.

$$16y^{2}-48y+16=16$$

Subtract 16 from both sides.

$$16y^{2}-48y+16-16=0$$

Subtract 16 from 16 to get 0.

$$16y^{2}-48y=0$$

Factor out y.

$$y\left(16y-48\right)=0$$

To find equation solutions, solve y=0 and 16y-48=0.

$$y=0 y=3$$

Substitute 0 for y in the equation \sqrt{5y+1}-\sqrt{y+1}=2.

$$\sqrt{5\times 0+1}-\sqrt{0+1}=2$$

Simplify. The value y=0 does not satisfy the equation.

$$0=2$$

Substitute 3 for y in the equation \sqrt{5y+1}-\sqrt{y+1}=2.

$$\sqrt{5\times 3+1}-\sqrt{3+1}=2$$

Simplify. The value y=3 satisfies the equation.

$$2=2$$

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

$$y=3$$