Solve sqrt{x+3}+3=x

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

Solve for x

$$x=6$$

Solution Steps

Subtract 3 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x+3=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$x+3-x^{2}=-6x+9$$

Add 6x to both sides.

$$x+3-x^{2}+6x=9$$

Combine x and 6x to get 7x.

$$7x+3-x^{2}=9$$

Subtract 9 from both sides.

$$7x+3-x^{2}-9=0$$

Subtract 9 from 3 to get -6.

$$7x-6-x^{2}=0$$

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

$$-x^{2}+7x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=7 ab=-\left(-6\right)=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=6 b=1$$

Rewrite -x^{2}+7x-6 as \left(-x^{2}+6x\right)+\left(x-6\right).

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

Factor out -x in -x^{2}+6x.

$$-x\left(x-6\right)+x-6$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+1\right)$$

To find equation solutions, solve x-6=0 and -x+1=0.

$$x=6 x=1$$

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

$$\sqrt{6+3}+3=6$$

Simplify. The value x=6 satisfies the equation.

$$6=6$$

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

$$\sqrt{1+3}+3=1$$

Simplify. The value x=1 does not satisfy the equation.

$$5=1$$

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

$$x=6$$