Solve sqrt{x+7}+5=x

$$\sqrt { x + 7 } + 5 = x$$

Solve for x

$$x=9$$

Solution Steps

Subtract 5 from both sides of the equation.

$$\sqrt{x+7}=x-5$$

Square both sides of the equation.

$$\left(\sqrt{x+7}\right)^{2}=\left(x-5\right)^{2}$$

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

$$x+7=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x+7=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$x+7-x^{2}=-10x+25$$

Add 10x to both sides.

$$x+7-x^{2}+10x=25$$

Combine x and 10x to get 11x.

$$11x+7-x^{2}=25$$

Subtract 25 from both sides.

$$11x+7-x^{2}-25=0$$

Subtract 25 from 7 to get -18.

$$11x-18-x^{2}=0$$

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

$$-x^{2}+11x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=11 ab=-\left(-18\right)=18$$

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 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 11.

$$a=9 b=2$$

Rewrite -x^{2}+11x-18 as \left(-x^{2}+9x\right)+\left(2x-18\right).

$$\left(-x^{2}+9x\right)+\left(2x-18\right)$$

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

$$-x\left(x-9\right)+2\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x+2\right)$$

To find equation solutions, solve x-9=0 and -x+2=0.

$$x=9 x=2$$

Substitute 9 for x in the equation \sqrt{x+7}+5=x.

$$\sqrt{9+7}+5=9$$

Simplify. The value x=9 satisfies the equation.

$$9=9$$

Substitute 2 for x in the equation \sqrt{x+7}+5=x.

$$\sqrt{2+7}+5=2$$

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

$$8=2$$

Equation \sqrt{x+7}=x-5 has a unique solution.

$$x=9$$