Solve sqrt{x-4}+x=6

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

Solve for x

$$x=5$$

Solution Steps

Subtract x from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{x-4} to the power of 2 and get x-4.

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

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

$$x-4=36-12x+x^{2}$$

Subtract 36 from both sides.

$$x-4-36=-12x+x^{2}$$

Subtract 36 from -4 to get -40.

$$x-40=-12x+x^{2}$$

Add 12x to both sides.

$$x-40+12x=x^{2}$$

Combine x and 12x to get 13x.

$$13x-40=x^{2}$$

Subtract x^{2} from both sides.

$$13x-40-x^{2}=0$$

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

$$-x^{2}+13x-40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-40\right)=40$$

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 40.

$$1,40 2,20 4,10 5,8$$

Calculate the sum for each pair.

$$1+40=41 2+20=22 4+10=14 5+8=13$$

The solution is the pair that gives sum 13.

$$a=8 b=5$$

Rewrite -x^{2}+13x-40 as \left(-x^{2}+8x\right)+\left(5x-40\right).

$$\left(-x^{2}+8x\right)+\left(5x-40\right)$$

Factor out -x in the first and 5 in the second group.

$$-x\left(x-8\right)+5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x+5\right)$$

To find equation solutions, solve x-8=0 and -x+5=0.

$$x=8 x=5$$

Substitute 8 for x in the equation \sqrt{x-4}+x=6.

$$\sqrt{8-4}+8=6$$

Simplify. The value x=8 does not satisfy the equation.

$$10=6$$

Substitute 5 for x in the equation \sqrt{x-4}+x=6.

$$\sqrt{5-4}+5=6$$

Simplify. The value x=5 satisfies the equation.

$$6=6$$

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

$$x=5$$