Solve sqrt{x+3}-x=1

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

Solve for x

$$x=1$$

Solution Steps

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

$$\left(\sqrt{x+3}\right)^{2}=\left(1+x\right)^{2}$$

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

$$x+3=\left(1+x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$x+3=1+2x+x^{2}$$

Subtract 1 from both sides.

$$x+3-1=2x+x^{2}$$

Subtract 1 from 3 to get 2.

$$x+2=2x+x^{2}$$

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract x^{2} from both sides.

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

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

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

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

$$a+b=-1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-2$$

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

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

Factor out x in the first and 2 in the second group.

$$x\left(-x+1\right)+2\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+2\right)$$

To find equation solutions, solve -x+1=0 and x+2=0.

$$x=1 x=-2$$

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

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

Substitute -2 for x in the equation \sqrt{x+3}-x=1.

$$\sqrt{-2+3}-\left(-2\right)=1$$

Simplify. The value x=-2 does not satisfy the equation.

$$3=1$$

Equation \sqrt{x+3}=x+1 has a unique solution.

$$x=1$$