Solve sqrt{9-x}-x+7=0

$$\sqrt{ 9-x } -x+7 = 0$$

Solve for x

$$x=8$$

Solution Steps

Subtract -x+7 from both sides of the equation.

$$\sqrt{9-x}=-\left(-x+7\right)$$

To find the opposite of -x+7, find the opposite of each term.

$$\sqrt{9-x}=-\left(-x\right)-7$$

The opposite of -x is x.

$$\sqrt{9-x}=x-7$$

Square both sides of the equation.

$$\left(\sqrt{9-x}\right)^{2}=\left(x-7\right)^{2}$$

Calculate \sqrt{9-x} to the power of 2 and get 9-x.

$$9-x=\left(x-7\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$9-x=x^{2}-14x+49$$

Subtract x^{2} from both sides.

$$9-x-x^{2}=-14x+49$$

Add 14x to both sides.

$$9-x-x^{2}+14x=49$$

Combine -x and 14x to get 13x.

$$9+13x-x^{2}=49$$

Subtract 49 from both sides.

$$9+13x-x^{2}-49=0$$

Subtract 49 from 9 to get -40.

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

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{9-x}-x+7=0.

$$\sqrt{9-8}-8+7=0$$

Simplify. The value x=8 satisfies the equation.

$$0=0$$

Substitute 5 for x in the equation \sqrt{9-x}-x+7=0.

$$\sqrt{9-5}-5+7=0$$

Simplify. The value x=5 does not satisfy the equation.

$$4=0$$

Equation \sqrt{9-x}=x-7 has a unique solution.

$$x=8$$