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

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

Solve for y

$$y=\frac{1}{3}\approx 0.333333333$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3y+1\right)^{2}.

$$9y^{2}+6y+1=\left(2\sqrt{3y}\right)^{2}$$

Expand \left(2\sqrt{3y}\right)^{2}.

$$9y^{2}+6y+1=2^{2}\left(\sqrt{3y}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9y^{2}+6y+1=4\left(\sqrt{3y}\right)^{2}$$

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

$$9y^{2}+6y+1=4\times 3y$$

Multiply 4 and 3 to get 12.

$$9y^{2}+6y+1=12y$$

Subtract 12y from both sides.

$$9y^{2}+6y+1-12y=0$$

Combine 6y and -12y to get -6y.

$$9y^{2}-6y+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9y^{2}+ay+by+1. To find a and b, set up a system to be solved.

$$a+b=-6 ab=9\times 1=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

$$a=-3 b=-3$$

Rewrite 9y^{2}-6y+1 as \left(9y^{2}-3y\right)+\left(-3y+1\right).

$$\left(9y^{2}-3y\right)+\left(-3y+1\right)$$

Factor out 3y in the first and -1 in the second group.

$$3y\left(3y-1\right)-\left(3y-1\right)$$

Factor out common term 3y-1 by using distributive property.

$$\left(3y-1\right)\left(3y-1\right)$$

Rewrite as a binomial square.

$$\left(3y-1\right)^{2}$$

To find equation solution, solve 3y-1=0.

$$y=\frac{1}{3}$$

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

$$3\times \frac{1}{3}+1=2\sqrt{3\times \frac{1}{3}}$$

Simplify. The value y=\frac{1}{3} satisfies the equation.

$$2=2$$

Equation 3y+1=2\sqrt{3y} has a unique solution.

$$y=\frac{1}{3}$$