Algebra 2 by Richard Wright

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3-04 Solve Quadratic Equations by Completing the Square (3.3)

Mr. Wright teaches the lesson.

Objectives:

SDA NAD Content Standards (2018): AII.4.1, AII.4.2, AII.5.1, AII.6.3

garden
Figure 1: Garden. (Pixabay/Tama66)

Freddie wants to put a gravel path around his rectangular garden. He wants the path to be the same width all around the garden, but he only can afford enough gravel to cover a certain area. To find out how wide the path should be requires solving a quadratic equation.

All the ways to solve quadratic equations covered so far only work in specific cases. Factoring only works when it is factorable. Graphing only works with the solutions are real. Taking square roots only works when the x appears only once. Completing the square is a method that solves all quadratic equations.

Solving by Completing the Square

A perfect square is something like x2. It could be in the form (x + k)2. Multiply this out.

(x + k)2

(x + k)(x + k)

x2 + kx + kx + k2

x2 + 2kx + k2

Compare this to general form ax2 + bx + c.

b = 2k → \(k=\frac{b}{2}\)

$$ c = k^2 = \left(\frac{b}{2}\right)^2 $$

This means that if the middle term is known, then third term is known for a perfect square.

Find c to Complete the Square

If a quadratic is in the form x2 + bx + c, then to make a perfect square

$$ c = \left(\frac{b}{2}\right)^2 $$

Example 1: Complete the Square

What is the value of c for the expression to be a perfect square?

x2 − 10x + c

Solution

For a perfect square, \(c=\left(\frac{b}{2}\right)^2\) and b = −10 for this example.

$$ c=\left(-\frac{10}{2}\right)^2 $$

$$ c=\left(-5\right)^2 $$

c = 25

Analysis: This is a perfect square because it can be factored into a square of a binomial.

x2 − 10x + 25

(x − 5)(x − 5)

(x − 5)2

Solve a Quadratic Equation by Completing the Square

To solve a quadratic equation by completing the square,

  1. Rewrite the quadratic so the x terms are on one side of the = and the constant on the other.
  2. If the leading coefficient is not 1, divide everything by the leading coefficient.
  3. Complete the square by adding \(\left(\frac{b}{2}\right)^2\) to both sides.
  4. Rewrite the quadratic side as a square by factoring. The form is always \(\left(x+\frac{b}{2}\right)^2\).
  5. Square root both sides. Remember the ±.
  6. Solve for x.

Example 2: Solve by Completing the Square

Solve 0 = x2 + 7x + 6 by completing the square.

Solution

Rewrite the quadratic so the x terms are on one side of the = and the constant is on the other.

0 = x2 + 7x + 6

−6 = x2 + 7x

The leading coefficient is already 1 because it is just x2.

Complete the square by adding \(\left(\frac{b}{2}\right)^2\) to both sides.

$$ -6+\left(\frac{7}{2}\right)^2=x^2+7x+\left(\frac{7}{2}\right)^2 $$

$$ -6+\frac{49}{4}=x^2+7x+\frac{49}{4} $$

$$ -\frac{24}{4}+\frac{49}{4}=x^2+7x+\frac{49}{4} $$

$$ \frac{25}{4}=x^2+7x+\frac{49}{4} $$

Rewrite the quadratic side as a square by factoring. The form is \(\left(x+\frac{b}{2}\right)^2\).

$$ \frac{25}{4}=\left(x+\frac{7}{2}\right)^2 $$

Square root both sides. Remember the ±.

$$ \pm\sqrt{\frac{25}{4}}=\sqrt{\left(x+\frac{7}{2}\right)^2} $$

$$ \pm\frac{5}{2}=x+\frac{7}{2} $$

Solve for x.

$$ -\frac{7}{2}\pm\frac{5}{2}=x $$

x = −1, −6

Example 3: Solve by Completing the Square

Solve 0 = 2x2 − 7x − 4 by completing the square.

Solution

Rewrite the quadratic so the x terms are on one side of the = and the constant is on the other.

4 = 2x2 − 7x

The leading coefficient is 2 because it is 2x2. Divide both sides by 2.

$$ 2 = x^2-\frac{7}{2}x $$

Complete the square by adding \(\left(\frac{b}{2}\right)^2\) to both sides.

$$ 2+\left(-\frac{7}{4}\right)^2=x^2-\frac{7}{2}x+\left(-\frac{7}{4}\right)^2 $$

$$ 2+\frac{49}{16}=x^2-\frac{7}{2}x+\frac{49}{16} $$

$$ \frac{32}{16}+\frac{49}{16}=x^2-\frac{7}{2}x+\frac{49}{16} $$

$$ \frac{81}{16}=x^2-\frac{7}{2}x+\frac{49}{16} $$

Rewrite the quadratic side as a square by factoring. The form is \(\left(x+\frac{b}{2}\right)^2\).

$$ \frac{81}{16}=\left(x-\frac{7}{4}\right)^2 $$

Square root both sides. Remember the ±.

$$ \pm\sqrt{\frac{81}{16}}=\sqrt{\left(x-\frac{7}{4}\right)^2} $$

$$ \pm\frac{9}{4}=x-\frac{7}{4} $$

Solve for x.

$$ \frac{7}{4}\pm\frac{9}{4}=x $$

x = 4, \(\mathbf{-\frac{1}{2}}\)

Example 4: Solve by Completing the Square

Solve x2 − 4x + 13 = 0 by completing the square.

Solution

Rewrite the quadratic so the x terms are on one side of the = and the constant is on the other.

x2 − 4 = −13

The leading coefficient is already 1 because it is just x2.

Complete the square by adding \(\left(\frac{b}{2}\right)^2\) to both sides.

$$ x^2-4x+\left(-\frac{4}{2}\right)^2=-13+\left(-\frac{4}{2}\right)^2 $$

x2 − 4x + 4 = −13 + 4

x2 − 4x + 4 = −9

Rewrite the quadratic side as a square by factoring. The form is \(\left(x+\frac{b}{2}\right)^2\).

(x − 2)2 = −9

Square root both sides. Remember the ±.

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

x − 2 = ±3i

Solve for x.

x = 2 ± 3i

Choosing a Method to Solve

To decide what method to use to solve quadratic equations, it is usually fastest to try factoring or square roots first.

  1. If x appears only once and it is squared, solve by taking square roots.

  2. If both x2 and x appear,
    1. Try solving by factoring.
    2. If it cannot be factored quickly, solve by completing the square.

Writing Quadratics in Standard Form

For some applications, such as finding maximums or minimums, if is convenient to write a quadratic function in standard (vertex) form. To do that, complete the square.

Write a General Quadratic Function in Standard Form

To write a general quadratic function in standard form (sometimes called vertex form) by completing the square.

  1. Start with general form, y = ax2 + bx + c.
  2. Group the terms with the x.
  3. Factor out any number in front of the x2.

  4. Add \(\left(\frac{b}{2}\right)^2\) to both sides. (Add it inside the group on the right.)
  5. Rewrite as a perfect square.
  6. Solve for y.

Example 5: Write Quadratic in Standard Form

Rewrite y = x2 + 2x + 4 in standard form.

Solution

Start with general form.

y = x2 + 2x + 4

Group the terms with the x.

y = (x2 + 2x) + 4

Factor out any number in front of the x2. There is nothing to factor out.

Add \(\left(\frac{b}{2}\right)^2\) to both sides. (Add it inside the group on the right.)

$$ y+\left(\frac{2}{2}\right)^2=\left(x^2+2x+\left(\frac{2}{2}\right)^2\right)+4 $$

y + 1 = (x2 + 2x + 1) + 4

Rewrite as a perfect square.

y + 1 = (x + 1)2 + 4

Solve for y.

y = (x + 1)2 + 3

Example 6: Write Quadratic in Standard Form

Rewrite y = 2x2 − 12x + 22 in standard form.

Solution

Start with general form.

y = 2x2 − 12x + 22

Group the terms with the x.

y = (2x2 − 12x) + 22

Factor out any number in front of the x2.

y = 2(x2 − 6x) + 22

Add \(\left(\frac{b}{2}\right)^2\) to both sides. (Add it inside the group on the right.) Add \(2\left(\frac{b}{2}\right)^2\) on the left because the one on the right is inside the parentheses which is multiplied by 2.

$$ y+\mathbf{2}\left(\frac{-6}{2}\right)^2=\mathbf{2}\left(x^2-6x+\left(\frac{-6}{2}\right)^2\right)+22 $$

y + 18 = 2(x2 − 6x + 9) + 22

Rewrite as a perfect square.

y + 18 = 2(x − 3)2 + 22

Solve for y.

y = 2(x − 3)2 + 4

Practice Problems

114 #9, 11, 21, 23, 27, 31, 33, 35, 37, 39, 41, 43, 45, 51, 55, Mixed Review = 20

    Mixed Review

  1. (3-03) Solve by graphing \(0=-\frac{1}{2}x^2-x+4\).
  2. (3-03) Solve by square roots (x + 4)2 + 25 = 0.
  3. (3-02) Solve by factoring x2 − 5x + 6 = 0.
  4. (3-01) Evaluate \(\sqrt{-72}\).
  5. (1-02) Solve by substitution \(\left\{ \begin{align} y &= x^2 - 2 \\ y &= 2x - 2 \end{align} \right.\)

Answers

  1. −4, 2
  2. −4 ± 5i
  3. 2, 3
  4. \(6\sqrt{2}i\)
  5. (0, −2), (2, 2)