Solve sqrt{4y+1}-sqrt{y-2}=3

$$\sqrt{ 4y+1 } - \sqrt{ y-2 } =3$$

Solve for y

$$y=6$$

$$y=2$$

Solution Steps

Subtract -\sqrt{y-2} from both sides of the equation.

$$\sqrt{4y+1}=3+\sqrt{y-2}$$

Square both sides of the equation.

$$\left(\sqrt{4y+1}\right)^{2}=\left(3+\sqrt{y-2}\right)^{2}$$

Calculate \sqrt{4y+1} to the power of 2 and get 4y+1.

$$4y+1=\left(3+\sqrt{y-2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{y-2}\right)^{2}.

$$4y+1=9+6\sqrt{y-2}+\left(\sqrt{y-2}\right)^{2}$$

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

$$4y+1=9+6\sqrt{y-2}+y-2$$

Subtract 2 from 9 to get 7.

$$4y+1=7+6\sqrt{y-2}+y$$

Subtract 7+y from both sides of the equation.

$$4y+1-\left(7+y\right)=6\sqrt{y-2}$$

To find the opposite of 7+y, find the opposite of each term.

$$4y+1-7-y=6\sqrt{y-2}$$

Subtract 7 from 1 to get -6.

$$4y-6-y=6\sqrt{y-2}$$

Combine 4y and -y to get 3y.

$$3y-6=6\sqrt{y-2}$$

Square both sides of the equation.

$$\left(3y-6\right)^{2}=\left(6\sqrt{y-2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-6\right)^{2}.

$$9y^{2}-36y+36=\left(6\sqrt{y-2}\right)^{2}$$

Expand \left(6\sqrt{y-2}\right)^{2}.

$$9y^{2}-36y+36=6^{2}\left(\sqrt{y-2}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$9y^{2}-36y+36=36\left(\sqrt{y-2}\right)^{2}$$

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

$$9y^{2}-36y+36=36\left(y-2\right)$$

Use the distributive property to multiply 36 by y-2.

$$9y^{2}-36y+36=36y-72$$

Subtract 36y from both sides.

$$9y^{2}-36y+36-36y=-72$$

Combine -36y and -36y to get -72y.

$$9y^{2}-72y+36=-72$$

Add 72 to both sides.

$$9y^{2}-72y+36+72=0$$

Add 36 and 72 to get 108.

$$9y^{2}-72y+108=0$$

Divide both sides by 9.

$$y^{2}-8y+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+12. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite y^{2}-8y+12 as \left(y^{2}-6y\right)+\left(-2y+12\right).

$$\left(y^{2}-6y\right)+\left(-2y+12\right)$$

Factor out y in the first and -2 in the second group.

$$y\left(y-6\right)-2\left(y-6\right)$$

Factor out common term y-6 by using distributive property.

$$\left(y-6\right)\left(y-2\right)$$

To find equation solutions, solve y-6=0 and y-2=0.

$$y=6 y=2$$

Substitute 6 for y in the equation \sqrt{4y+1}-\sqrt{y-2}=3.

$$\sqrt{4\times 6+1}-\sqrt{6-2}=3$$

Simplify. The value y=6 satisfies the equation.

$$3=3$$

Substitute 2 for y in the equation \sqrt{4y+1}-\sqrt{y-2}=3.

$$\sqrt{4\times 2+1}-\sqrt{2-2}=3$$

Simplify. The value y=2 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{4y+1}=\sqrt{y-2}+3.

$$y=6 y=2$$