Solve sqrt{x+7}-x=1

$$\sqrt { x + 7 } - x = 1$$

Solve for x

$$x=2$$

Solution Steps

Subtract -x from both sides of the equation.

$$\sqrt{x+7}=1+x$$

Square both sides of the equation.

$$\left(\sqrt{x+7}\right)^{2}=\left(1+x\right)^{2}$$

Calculate \sqrt{x+7} to the power of 2 and get x+7.

$$x+7=\left(1+x\right)^{2}$$

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

$$x+7=1+2x+x^{2}$$

Subtract 1 from both sides.

$$x+7-1=2x+x^{2}$$

Subtract 1 from 7 to get 6.

$$x+6=2x+x^{2}$$

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract x^{2} from both sides.

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

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

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

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.

$$a+b=-1 ab=-6=-6$$

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. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

Rewrite -x^{2}-x+6 as \left(-x^{2}+2x\right)+\left(-3x+6\right).

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

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

$$x\left(-x+2\right)+3\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(x+3\right)$$

To find equation solutions, solve -x+2=0 and x+3=0.

$$x=2 x=-3$$

Substitute 2 for x in the equation \sqrt{x+7}-x=1.

$$\sqrt{2+7}-2=1$$

Simplify. The value x=2 satisfies the equation.

$$1=1$$

Substitute -3 for x in the equation \sqrt{x+7}-x=1.

$$\sqrt{-3+7}-\left(-3\right)=1$$

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

$$5=1$$

Equation \sqrt{x+7}=x+1 has a unique solution.

$$x=2$$