Solve x+4sqrt{x}-12=0

$$x + 4 \sqrt { x } - 12 = 0$$

Solve for x

$$x=4$$

Solution Steps

Subtract x-12 from both sides of the equation.

$$4\sqrt{x}=-\left(x-12\right)$$

To find the opposite of x-12, find the opposite of each term.

$$4\sqrt{x}=-x-\left(-12\right)$$

The opposite of -12 is 12.

$$4\sqrt{x}=-x+12$$

Square both sides of the equation.

$$\left(4\sqrt{x}\right)^{2}=\left(-x+12\right)^{2}$$

Expand \left(4\sqrt{x}\right)^{2}.

$$4^{2}\left(\sqrt{x}\right)^{2}=\left(-x+12\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16\left(\sqrt{x}\right)^{2}=\left(-x+12\right)^{2}$$

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

$$16x=\left(-x+12\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+12\right)^{2}.

$$16x=x^{2}-24x+144$$

Subtract x^{2} from both sides.

$$16x-x^{2}=-24x+144$$

Add 24x to both sides.

$$16x-x^{2}+24x=144$$

Combine 16x and 24x to get 40x.

$$40x-x^{2}=144$$

Subtract 144 from both sides.

$$40x-x^{2}-144=0$$

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

$$-x^{2}+40x-144=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-144. To find a and b, set up a system to be solved.

$$a+b=40 ab=-\left(-144\right)=144$$

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 144.

$$1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12$$

Calculate the sum for each pair.

$$1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24$$

The solution is the pair that gives sum 40.

$$a=36 b=4$$

Rewrite -x^{2}+40x-144 as \left(-x^{2}+36x\right)+\left(4x-144\right).

$$\left(-x^{2}+36x\right)+\left(4x-144\right)$$

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

$$-x\left(x-36\right)+4\left(x-36\right)$$

Factor out common term x-36 by using distributive property.

$$\left(x-36\right)\left(-x+4\right)$$

To find equation solutions, solve x-36=0 and -x+4=0.

$$x=36 x=4$$

Substitute 36 for x in the equation x+4\sqrt{x}-12=0.

$$36+4\sqrt{36}-12=0$$

Simplify. The value x=36 does not satisfy the equation.

$$48=0$$

Substitute 4 for x in the equation x+4\sqrt{x}-12=0.

$$4+4\sqrt{4}-12=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Equation 4\sqrt{x}=12-x has a unique solution.

$$x=4$$