Solve y+sqrt{y}=6

$$y + \sqrt { y } = 6$$

Solve for y

$$y=4$$

Solution Steps

Subtract y from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$y=36-12y+y^{2}$$

Subtract 36 from both sides.

$$y-36=-12y+y^{2}$$

Add 12y to both sides.

$$y-36+12y=y^{2}$$

Combine y and 12y to get 13y.

$$13y-36=y^{2}$$

Subtract y^{2} from both sides.

$$13y-36-y^{2}=0$$

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

$$-y^{2}+13y-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-36. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-36\right)=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=9 b=4$$

Rewrite -y^{2}+13y-36 as \left(-y^{2}+9y\right)+\left(4y-36\right).

$$\left(-y^{2}+9y\right)+\left(4y-36\right)$$

Factor out -y in the first and 4 in the second group.

$$-y\left(y-9\right)+4\left(y-9\right)$$

Factor out common term y-9 by using distributive property.

$$\left(y-9\right)\left(-y+4\right)$$

To find equation solutions, solve y-9=0 and -y+4=0.

$$y=9 y=4$$

Substitute 9 for y in the equation y+\sqrt{y}=6.

$$9+\sqrt{9}=6$$

Simplify. The value y=9 does not satisfy the equation.

$$12=6$$

Substitute 4 for y in the equation y+\sqrt{y}=6.

$$4+\sqrt{4}=6$$

Simplify. The value y=4 satisfies the equation.

$$6=6$$

Equation \sqrt{y}=6-y has a unique solution.

$$y=4$$