Solve sqrt{3x+6}=x-4

$$\sqrt { 3 x + 6 } = x - 4$$

Solve for x

$$x=10$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x+6}\right)^{2}=\left(x-4\right)^{2}$$

Calculate \sqrt{3x+6} to the power of 2 and get 3x+6.

$$3x+6=\left(x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$3x+6=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$3x+6-x^{2}=-8x+16$$

Add 8x to both sides.

$$3x+6-x^{2}+8x=16$$

Combine 3x and 8x to get 11x.

$$11x+6-x^{2}=16$$

Subtract 16 from both sides.

$$11x+6-x^{2}-16=0$$

Subtract 16 from 6 to get -10.

$$11x-10-x^{2}=0$$

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

$$-x^{2}+11x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=11 ab=-\left(-10\right)=10$$

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 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 11.

$$a=10 b=1$$

Rewrite -x^{2}+11x-10 as \left(-x^{2}+10x\right)+\left(x-10\right).

$$\left(-x^{2}+10x\right)+\left(x-10\right)$$

Factor out -x in -x^{2}+10x.

$$-x\left(x-10\right)+x-10$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(-x+1\right)$$

To find equation solutions, solve x-10=0 and -x+1=0.

$$x=10 x=1$$

Substitute 10 for x in the equation \sqrt{3x+6}=x-4.

$$\sqrt{3\times 10+6}=10-4$$

Simplify. The value x=10 satisfies the equation.

$$6=6$$

Substitute 1 for x in the equation \sqrt{3x+6}=x-4.

$$\sqrt{3\times 1+6}=1-4$$

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

$$3=-3$$

Equation \sqrt{3x+6}=x-4 has a unique solution.

$$x=10$$