Solve sqrt{3y+4}=y

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

Solve for y

$$y=4$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{3y+4} to the power of 2 and get 3y+4.

$$3y+4=y^{2}$$

Subtract y^{2} from both sides.

$$3y+4-y^{2}=0$$

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

$$-y^{2}+3y+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=3 ab=-4=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -y^{2}+3y+4 as \left(-y^{2}+4y\right)+\left(-y+4\right).

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

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

$$-y\left(y-4\right)-\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(-y-1\right)$$

To find equation solutions, solve y-4=0 and -y-1=0.

$$y=4 y=-1$$

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

$$\sqrt{3\times 4+4}=4$$

Simplify. The value y=4 satisfies the equation.

$$4=4$$

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

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

Simplify. The value y=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Equation \sqrt{3y+4}=y has a unique solution.

$$y=4$$