Solve (2+x)^2=9

$$( 2 + x ) ^ { 2 } = 9$$

Solve for x

$$x=1$$

$$x=-5$$

Steps Using Factoring

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

$$4+4x+x^{2}=9$$

Subtract 9 from both sides.

$$4+4x+x^{2}-9=0$$

Subtract 9 from 4 to get -5.

$$-5+4x+x^{2}=0$$

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

$$x^{2}+4x-5=0$$

To solve the equation, factor x^{2}+4x-5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve x-1=0 and x+5=0.

$$x=1 x=-5$$

Steps Using Factoring By Grouping

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

$$4+4x+x^{2}=9$$

Subtract 9 from both sides.

$$4+4x+x^{2}-9=0$$

Subtract 9 from 4 to get -5.

$$-5+4x+x^{2}=0$$

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

$$x^{2}+4x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-5\right)=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite x^{2}+4x-5 as \left(x^{2}-x\right)+\left(5x-5\right).

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

Factor out x in the first and 5 in the second group.

$$x\left(x-1\right)+5\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve x-1=0 and x+5=0.

$$x=1 x=-5$$

Steps Using the Quadratic Formula

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

$$4+4x+x^{2}=9$$

Subtract 9 from both sides.

$$4+4x+x^{2}-9=0$$

Subtract 9 from 4 to get -5.

$$-5+4x+x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+4x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-5\right)}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$x=\frac{-4±\sqrt{16+20}}{2}$$

Add 16 to 20.

$$x=\frac{-4±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-4±6}{2}$$

Now solve the equation x=\frac{-4±6}{2} when ± is plus. Add -4 to 6.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-4±6}{2} when ± is minus. Subtract 6 from -4.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=1 x=-5$$

Steps for Completing the Square

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

$$4+4x+x^{2}=9$$

Subtract 4 from both sides.

$$4x+x^{2}=9-4$$

Subtract 4 from 9 to get 5.

$$4x+x^{2}=5$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+4x=5$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=5+2^{2}$$

Square 2.

$$x^{2}+4x+4=5+4$$

Add 5 to 4.

$$x^{2}+4x+4=9$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=3 x+2=-3$$

Subtract 2 from both sides of the equation.

$$x=1 x=-5$$