Algebra 2 by Richard Wright
Are you not my student and
has this helped you?
Objectives:
SDA NAD Content Standards (2018): AII.4.1, AII.4.2, AII.5.1, AII.6.3
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.
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.
If a quadratic is in the form x2 + bx + c, then to make a perfect square
$$ c = \left(\frac{b}{2}\right)^2 $$
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
To solve a quadratic equation 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
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, \(\mathbf{-\frac{1}{2}}\)
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
To decide what method to use to solve quadratic equations, it is usually fastest to try factoring or square roots first.
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.
To write a general quadratic function in standard form (sometimes called vertex form) by completing the square.
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
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
114 #9, 11, 21, 23, 27, 31, 33, 35, 37, 39, 41, 43, 45, 51, 55, Mixed Review = 20
Mixed Review