Solve sqrt{3x}+8=x+2

$$\sqrt { 3 x } + 8 = x + 2$$

Solve for x

$$x=12$$

Solution Steps

Subtract 8 from both sides of the equation.

$$\sqrt{3x}=x+2-8$$

Subtract 8 from 2 to get -6.

$$\sqrt{3x}=x-6$$

Square both sides of the equation.

$$\left(\sqrt{3x}\right)^{2}=\left(x-6\right)^{2}$$

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

$$3x=\left(x-6\right)^{2}$$

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

$$3x=x^{2}-12x+36$$

Subtract x^{2} from both sides.

$$3x-x^{2}=-12x+36$$

Add 12x to both sides.

$$3x-x^{2}+12x=36$$

Combine 3x and 12x to get 15x.

$$15x-x^{2}=36$$

Subtract 36 from both sides.

$$15x-x^{2}-36=0$$

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

$$-x^{2}+15x-36=0$$

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.

$$a+b=15 ab=-\left(-36\right)=36$$

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 15.

$$a=12 b=3$$

Rewrite -x^{2}+15x-36 as \left(-x^{2}+12x\right)+\left(3x-36\right).

$$\left(-x^{2}+12x\right)+\left(3x-36\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-12\right)+3\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x+3\right)$$

To find equation solutions, solve x-12=0 and -x+3=0.

$$x=12 x=3$$

Substitute 12 for x in the equation \sqrt{3x}+8=x+2.

$$\sqrt{3\times 12}+8=12+2$$

Simplify. The value x=12 satisfies the equation.

$$14=14$$

Substitute 3 for x in the equation \sqrt{3x}+8=x+2.

$$\sqrt{3\times 3}+8=3+2$$

Simplify. The value x=3 does not satisfy the equation.

$$11=5$$

Equation \sqrt{3x}=x-6 has a unique solution.

$$x=12$$