Solve x+sqrt{x}=6

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

Solve for x

$$x=4$$

Solution Steps

Subtract x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract 36 from both sides.

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

Add 12x to both sides.

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

Combine x and 12x to get 13x.

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

Subtract x^{2} from both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-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 -x^{2}+13x-36 as \left(-x^{2}+9x\right)+\left(4x-36\right).

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

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

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

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

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

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

$$x=9 x=4$$

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

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

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

$$12=6$$

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

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

Simplify. The value x=4 satisfies the equation.

$$6=6$$

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

$$x=4$$