Solve sqrt{2x+9}=x+5

$$\sqrt { 2 x + 9 } = x + 5$$

Solve for x

$$x=-4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x+9}\right)^{2}=\left(x+5\right)^{2}$$

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

$$2x+9=\left(x+5\right)^{2}$$

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

$$2x+9=x^{2}+10x+25$$

Subtract x^{2} from both sides.

$$2x+9-x^{2}=10x+25$$

Subtract 10x from both sides.

$$2x+9-x^{2}-10x=25$$

Combine 2x and -10x to get -8x.

$$-8x+9-x^{2}=25$$

Subtract 25 from both sides.

$$-8x+9-x^{2}-25=0$$

Subtract 25 from 9 to get -16.

$$-8x-16-x^{2}=0$$

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

$$-x^{2}-8x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-\left(-16\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite -x^{2}-8x-16 as \left(-x^{2}-4x\right)+\left(-4x-16\right).

$$\left(-x^{2}-4x\right)+\left(-4x-16\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(-x-4\right)+4\left(-x-4\right)$$

Factor out common term -x-4 by using distributive property.

$$\left(-x-4\right)\left(x+4\right)$$

To find equation solutions, solve -x-4=0 and x+4=0.

$$x=-4 x=-4$$

Substitute -4 for x in the equation \sqrt{2x+9}=x+5.

$$\sqrt{2\left(-4\right)+9}=-4+5$$

Simplify. The value x=-4 satisfies the equation.

$$1=1$$

Substitute -4 for x in the equation \sqrt{2x+9}=x+5.

$$\sqrt{2\left(-4\right)+9}=-4+5$$

Simplify. The value x=-4 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{2x+9}=x+5.

$$x=-4 x=-4$$