Solve sqrt{4n+8}=n+3

$$\sqrt { 4 n + 8 } = n + 3$$

Solve for n

$$n=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4n+8}\right)^{2}=\left(n+3\right)^{2}$$

Calculate \sqrt{4n+8} to the power of 2 and get 4n+8.

$$4n+8=\left(n+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+3\right)^{2}.

$$4n+8=n^{2}+6n+9$$

Subtract n^{2} from both sides.

$$4n+8-n^{2}=6n+9$$

Subtract 6n from both sides.

$$4n+8-n^{2}-6n=9$$

Combine 4n and -6n to get -2n.

$$-2n+8-n^{2}=9$$

Subtract 9 from both sides.

$$-2n+8-n^{2}-9=0$$

Subtract 9 from 8 to get -1.

$$-2n-1-n^{2}=0$$

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

$$-n^{2}-2n-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn-1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-1$$

Rewrite -n^{2}-2n-1 as \left(-n^{2}-n\right)+\left(-n-1\right).

$$\left(-n^{2}-n\right)+\left(-n-1\right)$$

Factor out n in -n^{2}-n.

$$n\left(-n-1\right)-n-1$$

Factor out common term -n-1 by using distributive property.

$$\left(-n-1\right)\left(n+1\right)$$

To find equation solutions, solve -n-1=0 and n+1=0.

$$n=-1 n=-1$$

Substitute -1 for n in the equation \sqrt{4n+8}=n+3.

$$\sqrt{4\left(-1\right)+8}=-1+3$$

Simplify. The value n=-1 satisfies the equation.

$$2=2$$

Substitute -1 for n in the equation \sqrt{4n+8}=n+3.

$$\sqrt{4\left(-1\right)+8}=-1+3$$

Simplify. The value n=-1 satisfies the equation.

$$2=2$$

List all solutions of \sqrt{4n+8}=n+3.

$$n=-1 n=-1$$