Solve sqrt{16x+17}=x+5

$$\sqrt{ 16x+17 } =x+5$$

Solve for x

$$x=4$$

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{16x+17}\right)^{2}=\left(x+5\right)^{2}$$

Calculate \sqrt{16x+17} to the power of 2 and get 16x+17.

$$16x+17=\left(x+5\right)^{2}$$

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

$$16x+17=x^{2}+10x+25$$

Subtract x^{2} from both sides.

$$16x+17-x^{2}=10x+25$$

Subtract 10x from both sides.

$$16x+17-x^{2}-10x=25$$

Combine 16x and -10x to get 6x.

$$6x+17-x^{2}=25$$

Subtract 25 from both sides.

$$6x+17-x^{2}-25=0$$

Subtract 25 from 17 to get -8.

$$6x-8-x^{2}=0$$

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

$$-x^{2}+6x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-8\right)=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=4 b=2$$

Rewrite -x^{2}+6x-8 as \left(-x^{2}+4x\right)+\left(2x-8\right).

$$\left(-x^{2}+4x\right)+\left(2x-8\right)$$

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

$$-x\left(x-4\right)+2\left(x-4\right)$$

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

$$\left(x-4\right)\left(-x+2\right)$$

To find equation solutions, solve x-4=0 and -x+2=0.

$$x=4 x=2$$

Substitute 4 for x in the equation \sqrt{16x+17}=x+5.

$$\sqrt{16\times 4+17}=4+5$$

Simplify. The value x=4 satisfies the equation.

$$9=9$$

Substitute 2 for x in the equation \sqrt{16x+17}=x+5.

$$\sqrt{16\times 2+17}=2+5$$

Simplify. The value x=2 satisfies the equation.

$$7=7$$

List all solutions of \sqrt{16x+17}=x+5.

$$x=4 x=2$$