Solve {l}{text{If}49=(x+2)^2,text{then}pm7=}{text{So}x=-}

$$\left. \begin{array} { l } { \text { If } 49 = ( x + 2 ) ^ { 2 } , \text { then } \pm 7 = } \\ { \text { So } x = - } \end{array} \right.$$

Solve for x

$$x=5$$

$$x=-9$$

Steps Using Factoring

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

$$49=x^{2}+4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x+4=49$$

Subtract 49 from both sides.

$$x^{2}+4x+4-49=0$$

Subtract 49 from 4 to get -45.

$$x^{2}+4x-45=0$$

To solve the equation, factor x^{2}+4x-45 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.

$$a+b=4 ab=-45$$

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. List all such integer pairs that give product -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

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

$$\left(x-5\right)\left(x+9\right)$$

To find equation solutions, solve x-5=0 and x+9=0.

$$x=5 x=-9$$

Steps Using Factoring By Grouping

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

$$49=x^{2}+4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x+4=49$$

Subtract 49 from both sides.

$$x^{2}+4x+4-49=0$$

Subtract 49 from 4 to get -45.

$$x^{2}+4x-45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-45. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-45\right)=-45$$

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. List all such integer pairs that give product -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

Rewrite x^{2}+4x-45 as \left(x^{2}-5x\right)+\left(9x-45\right).

$$\left(x^{2}-5x\right)+\left(9x-45\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x-5\right)+9\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+9\right)$$

To find equation solutions, solve x-5=0 and x+9=0.

$$x=5 x=-9$$

Steps Using the Quadratic Formula

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

$$49=x^{2}+4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x+4=49$$

Subtract 49 from both sides.

$$x^{2}+4x+4-49=0$$

Subtract 49 from 4 to get -45.

$$x^{2}+4x-45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-45\right)}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-45\right)}}{2}$$

Multiply -4 times -45.

$$x=\frac{-4±\sqrt{16+180}}{2}$$

Add 16 to 180.

$$x=\frac{-4±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{-4±14}{2}$$

Now solve the equation x=\frac{-4±14}{2} when ± is plus. Add -4 to 14.

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

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{-4±14}{2} when ± is minus. Subtract 14 from -4.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=5 x=-9$$

Steps for Completing the Square

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

$$49=x^{2}+4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+4x+4=49$$

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}.

$$\left(x+2\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+2=7 x+2=-7$$

Subtract 2 from both sides of the equation.

$$x=5 x=-9$$