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# Completing the Square

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# Completing the Square

How can you use square roots to solve the quadratic equation $(x-3)^2=15?$

The equation $x^2-6x-6=0$ is equivalent to $(x-3)^2=15$ . How can you algebraically change $x^2-6x-6=0$ into $(x-3)^2=15$ in order to solve it?

### Guidance

Recall that some quadratics are known as perfect square trinomials because they can be represented as a binomial squared. For example:

• $x^2+10x+25$ can be written as $(x+5)^2$
• $x^2+6x+9$ can be written as $(x+3)^2$
• $x^2+2x+1$ can be written as $(x+1)^2$

You can solve a quadratic such as $(x+5)^2=0$ by taking the square root of both sides as shown:

$(x+5)^2&=0\\\sqrt{(x+5)^2}&=\pm\sqrt{0}\\x+5=+0 \ & \text{OR} \ x+5=-0\\x=-5$

You can also solve such a quadratic even if it is not set equal to zero. For example, you can solve $(x+5)^2=9$ in a similar way:

$(x+5)^2&=9\\\sqrt{(x+5)^2}&=\pm\sqrt{9}\\x+5=+3 \ & \text{OR} \ x+5=-3\\x=-2 \ & \ x=-8$

Completing the square is a technique for solving quadratic equations that turns a given equation into a perfect square trinomial set equal to a number so that it can be solved using the method above. Consider the equation:

$x^2-12x+20=0$

Step 1: Move 20 to the right side of the equation.

$x^2-12x=-20$

Step 2: Complete the square on the left side of the equation. This means, figure out what could be added to the left side to turn that equation into a perfect square trinomial. One method for figuring this out is to take the value of ‘ $b$ ’ (which is –12), divide it by 2, and square the result:

$& {\color{red}b}=-12 \\& {\color{red}\frac{b}{2}} = \frac{-12}{2} = -6 \\& {\color{red}\left(\frac{b}{2} \right)^2} = (-6)^2 = 36 \\& x^2-12x+{\color{red}36}=-20+36$

Remember that what is added to one side of the equation must be added to the other side of the equation. Once you simplify you get:

$x^2-12x+36=16$

Step 3: Rewrite the left side of the equation as a binomial squared.

$(x-6)^2 = 16$

Step 4: Take the square root of both sides of the equation.

$\sqrt{(x-6)^2} & = \sqrt{16} \\x-6 & = \pm 4$

Step 5: Set the left side of the equation equal to each of the roots on the right side and solve each linear equation.

$x-6=4$ and $x-6=-4$

$x-6 {\color{red}+6} = 4 {\color{red}+6}$ and $x-6 {\color{red}+6} = -4 {\color{red}+6}$

$x={\color{red}10}$ or $x={\color{red}2}$

The solutions to the equation are:

$\boxed{x=10 \ or \ x=2}$

#### Example A

What would you need to add to the following expression to turn it into a perfect square trinomial?

$x^2+16x$

Solution: To determine the value to add, you can divide $16$ by 2 and square the result.

$& {\color{red}b}=16 \\& {\color{red}\frac{b}{2}} = \frac{16}{2} = 8 \\& {\color{red}\left(\frac{b}{2} \right)^2} = (8)^2 = 64$

The expression is now $x^2+16x \ {\color{red} + 64}$ , which can be rewritten as $(x+8)^2$ .

#### Example B

Solve the following quadratic equation by completing the square: $x^2+2x-35=0$

Solution:

Step 1: Move –35 to the right side of the equation.

$x^2+2x=35$

Step 2: Complete the square on the left side of the equation. The value of $b$ is 2.

$& {\color{red}b}=2 \\& {\color{red}\frac{b}{2}} = \frac{2}{2} = 1 \\& {\color{red}\left(\frac{b}{2} \right)^2} = (1)^2 = 1 \\& x^2+2x{\color{red}+1}=35+1$

$x^2+2x+1=36$

Step 3: Rewrite the left side of the equation as a binomial squared.

$(x+1)^2 = 36$

Step 4: Take the square root of both sides of the equation.

$\sqrt{(x+1)^2} & = \sqrt{36} \\x+1 & = \pm 6$

Step 5: Set the left side of the equation equal to each of the roots on the right side and solve each linear equation.

$x+1=6$ and $x+1=-6$

$x+1 {\color{red}-1} = 6{\color{red}-1}$ and $x+1{\color{red}-1} = -6{\color{red}-1}$

$x={\color{red}5}$ or $x={\color{red}-7}$

The solutions to the equation are:

$\boxed{x=5 \ or \ x=-7}$

#### Example C

Solve the following quadratic equation by completing the square:

$\boxed{2x^2-5x+2=0}$

Solution: In this quadratic equation, the value of ‘ $a$ ’ is not one. The first step will be to factor out the value of $a$ .

Step 1: Factor out the ‘2’. Then, because the equation is set equal to zero, divide both sides by 2 in order to have a simpler equation.

$2(x^2 - \frac{5}{2}x +1)&=0 \\x^2 - \frac{5}{2}x +1=0$

Step 2: Move 1 to the right side of the equation.

$x^2 - \frac{5}{2}x =-1$

Step 3: Complete the square on the left side of the equation.

$& {\color{red}b}=-\frac{5}{2} \\& {\color{red}\frac{b}{2}} = -\frac{5}{2} \div 2 = -\frac{5}{2} \left(\frac{1}{2}\right)=-\frac{5}{4} \\& {\color{red}\left(\frac{b}{2}\right)^2} = \left(-\frac{5}{4}\right)^2 = \frac{25}{16} \\& x^2-\frac{5}{2}x \ {\color{red} + \frac{25}{16}} = -1 {\color{red} + \frac{25}{16}}$

$x^2 - \frac{5}{2}x \ + \frac{25}{16} = \frac{9}{16}$

Step 4: Rewrite the perfect square trinomial on the left side of the equation as a binomial squared.

$\left(x-\frac{5}{4}\right)^2 = \frac{9}{16}$

Step 4: Take the square root of both sides of the equation.

$\sqrt{\left(x -\frac{5}{4}\right)^2} & = \sqrt{\frac{9}{16}} \\x-\frac{5}{4} &= \pm \frac{3}{4}$

Step 5: Set the left side of the equation equal to each of the roots on the right side and solve each linear equation.

$x-\frac{5}{4} = \frac{3}{4}$ and $x-\frac{5}{4} = -\frac{3}{4}$

$x-\frac{5}{4} \ {\color{red}+ \frac{5}{4}} = \frac{3}{4} \ {\color{red}+ \frac{5}{4}}$ and $x- \frac{5}{4} \ {\color{red}+ \frac{5}{4}} = -\frac{3}{4} \ {\color{red}+ \frac{5}{4}}$

$x=\frac{8}{4}$ or $x=\frac{2}{4}$

$x=2$ and $x=\frac{1}{2}$

The solutions to the equation are:

$\boxed{x=2 \ or \ x=\frac{1}{2}}$

#### Concept Problem Revisited

To solve $(x-3)^2=15$ , take the square root of both sides. You will get that $x-3=\sqrt {15}$ or $x-3=-\sqrt{15}$ . The two solutions are therefore $x=3+\sqrt{15},3-\sqrt{15}$ .

You can turn $x^2-6x-6=0$ into $(x-3)^2=15$ by completing the square.

### Vocabulary

Completing the Square
Completing the square is a method used for solving quadratic equations. This method uses the principle of completing the square of an algebraic expression to make it a perfect square trinomial.
Perfect Square Trinomial
A perfect square trinomial is one of the form $(ax)^2 + 2abx + b^2$ where the first and last terms are perfect squares and the middle term is twice the product of the square root of the first term and the square root of the third term.
The roots of a quadratic function are the $x$ -intercepts of the function. These are the values for the variable ‘ $x$ ’ that will result in $y = 0$ .
The zeros of a quadratic function are also the $x$ -intercepts of the function. These are the values for the variable ‘ $x$ ’ that will result in $y = 0$ .

### Guided Practice

1. What would you need to add to the following expression to turn it into a perfect square trinomial?

$x^2-7x$

2. Solve the following quadratic equation by completing the square.

$x^2-4x+1=0$

3. Solve the following quadratic equation by completing the square.

$7x^2-2x-2=0$

1. Divide b by $\frac{1}{2}$ and square the result.

$(\frac{-7}{2})^2 = \frac{49}{4}$

2. $x^2-4x+1=0$

$x^2-4x&=-1 \\x^2-4x{\color{red}+4}&=-1{\color{red}+4} \\x^2-4x+4&=3 \\(x-2)^2&=3 \\\sqrt{(x-2)^2} &= \sqrt{3} \\x-2&= \pm \sqrt{3}$

The solutions to the equation are:

$\boxed{x=2+\sqrt{3} \ or \ x=2-\sqrt{3}}$

$\boxed{x=2 \pm \sqrt{3}}$

3. $7x^2-2x-2=0$

$7(x^2-\frac{2}{7}x-\frac{2}{7})&=0 \\x^2-\frac{2}{7}x-\frac{2}{7}&=0 \\x^2-\frac{2}{7}x&=\frac{2}{7} \\x^2-\frac{2}{7}x+\frac{1}{49}&=\frac{2}{7}+\frac{1}{49} \\\left(x-\frac{1}{7}\right)^2&=\frac{15}{49}$

The exact solutions to the equation are:

$\boxed{x=\frac{\sqrt{15}+1}{7} \ or \ x=\frac{-\sqrt{15}+1}{7}}$

The approximate solutions (which have been rounded) are:

$\boxed{x=0.70 \ or \ x=-0.41}$

### Practice

State the value of $m$ that makes each trinomial a perfect square:

1. $x^2-10x+m$
2. $x^2-22x+m$
3. $x^2+\frac{1}{2}x+m$
4. $x^2+9x+m$
5. $x^2+x+m$

Solve each of the following quadratic equations by completing the square:

1. $x^2+18x=85$
2. $x^2-\frac{2}{3}x=1$
3. $x^2-7x=3$
4. $x^2+\frac{1}{5}x=2$
5. $x^2-\frac{2}{3}x=5$

Solve each of the following quadratic equations by completing the square:

1. $x^2-2x-8=0$
2. $2x^2+8x+5=0$
3. $3x^2-6x-2=0$
4. $2x^2-3x-5=0$
5. $3x^2+4x-2=0$