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

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

Solve for y

$$y=11$$

$$y=3$$

Solution Steps

Square both sides of the equation.

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

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

$$\left(\sqrt{y-2}\right)^{2}+4\sqrt{y-2}+4=\left(\sqrt{2y+3}\right)^{2}$$

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

$$y-2+4\sqrt{y-2}+4=\left(\sqrt{2y+3}\right)^{2}$$

Add -2 and 4 to get 2.

$$y+2+4\sqrt{y-2}=\left(\sqrt{2y+3}\right)^{2}$$

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

$$y+2+4\sqrt{y-2}=2y+3$$

Subtract y+2 from both sides of the equation.

$$4\sqrt{y-2}=2y+3-\left(y+2\right)$$

To find the opposite of y+2, find the opposite of each term.

$$4\sqrt{y-2}=2y+3-y-2$$

Combine 2y and -y to get y.

$$4\sqrt{y-2}=y+3-2$$

Subtract 2 from 3 to get 1.

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

Square both sides of the equation.

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

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

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

Calculate 4 to the power of 2 and get 16.

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

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

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

Use the distributive property to multiply 16 by y-2.

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

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

$$16y-32=y^{2}+2y+1$$

Subtract y^{2} from both sides.

$$16y-32-y^{2}=2y+1$$

Subtract 2y from both sides.

$$16y-32-y^{2}-2y=1$$

Combine 16y and -2y to get 14y.

$$14y-32-y^{2}=1$$

Subtract 1 from both sides.

$$14y-32-y^{2}-1=0$$

Subtract 1 from -32 to get -33.

$$14y-33-y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-y^{2}+14y-33=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-33. To find a and b, set up a system to be solved.

$$a+b=14 ab=-\left(-33\right)=33$$

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 33.

$$1,33 3,11$$

Calculate the sum for each pair.

$$1+33=34 3+11=14$$

The solution is the pair that gives sum 14.

$$a=11 b=3$$

Rewrite -y^{2}+14y-33 as \left(-y^{2}+11y\right)+\left(3y-33\right).

$$\left(-y^{2}+11y\right)+\left(3y-33\right)$$

Factor out -y in the first and 3 in the second group.

$$-y\left(y-11\right)+3\left(y-11\right)$$

Factor out common term y-11 by using distributive property.

$$\left(y-11\right)\left(-y+3\right)$$

To find equation solutions, solve y-11=0 and -y+3=0.

$$y=11 y=3$$

Substitute 11 for y in the equation \sqrt{y-2}+2=\sqrt{2y+3}.

$$\sqrt{11-2}+2=\sqrt{2\times 11+3}$$

Simplify. The value y=11 satisfies the equation.

$$5=5$$

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

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

Simplify. The value y=3 satisfies the equation.

$$3=3$$

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

$$y=11 y=3$$