Lesson Objectives
• Demonstrate an understanding of how to Solve a Quadratic Equation by Completing the Square
• Demonstrate an understanding of how to write a Quadratic Equation in Standard Form: ax2 + bx + c = 0
• Learn how to derive the Quadratic Formula
• Learn about the possible solutions using the discriminant: b2 - 4ac

In our last lesson, we learned how to solve a quadratic equation by Completing the Square. Although this method allows us to solve any quadratic equation, it is a quite tedious procedure. Luckily, we have an easier method that can be used to solve any quadratic equation. When we write our quadratic equation in standard form: $$ax^2 + bx + c=0$$ We can use the Quadratic Formula to find our solution or solutions. We do this by plugging in for a, b, and c in our quadratic formula: $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Where does the quadratic formula come from? Essentially, we start out with our quadratic equation in standard form and we work through the completing the square method: $$ax^2 + bx + c=0$$ Step 1) Subtract c away from each side of the equation. This will place the ax2 and bx terms on one side and the constant on the other: $$ax^2 + bx=-c$$ Step 2) Divide each part by a, the coefficient of x2. This will give us a coefficient of 1 on our squared term: $$\frac{ax^2}{a}+ \frac{b}{a}x=\frac{-c}{a}$$ $$x^2 + \frac{b}{a}x=\frac{-c}{a}$$ Step 3) Complete the square. We will add one-half of the coefficient of the first-degree term squared to both sides of the equation. This will create a perfect square trinomial on the left side, that can be factored into a binomial squared: $$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2=\frac{-c}{a}+ \left(\frac{b}{2a}\right)^2$$ Simplify: $$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}=\frac{-c}{a}+ \frac{b^2}{4a^2}$$ $$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}=\frac{-4ac}{4a^2}+ \frac{b^2}{4a^2}$$ $$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}=\frac{b^2 - 4ac}{4a^2}$$ Step 4) Solve the equation using the square root property. We will begin by factoring the left side into a binomial squared: $$\left(x + \frac{b}{2a}\right)^2=\frac{b^2 - 4ac}{4a^2}$$ $$\sqrt{\left(x + \frac{b}{2a}\right)^2}=\pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$ $$x + \frac{b}{2a}=\pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$ $$x + \frac{b}{2a}=\pm \frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}}$$ $$x + \frac{b}{2a}=\pm \frac{\sqrt{b^2 - 4ac}}{2a}$$ $$x=- \frac{b}{2a}\pm \frac{\sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

### The Discriminant: b2 - 4ac

The discriminant for the quadratic formula is the part under the square root symbol:
b2 - 4ac
The discriminant will tell us if we have two solutions, only one solution, or no real solution.
If b2 - 4ac = 0 » There is one solution. This is because the plus or minus part is no longer relevant. We would have plus or minus 0, which is just zero.
If b2 - 4ac > 0 » There are two solutions.
If b2 - 4ac < 0 » There are no real solutions. This is because we will be taking the square root of a negative number. We will learn how to deal with this scenario using imaginary numbers in our Algebra 2 course. For now, we will write our answer as "no real solution".

### Solving Equations with the Quadratic Formula

• Write the equation in standard form: ax2 + bx + c = 0
• Record the values for a, b, and c
• Plug into the quadratic formula and simplify
Let's look at a few examples.
Example 1: Solve each equation. $$5x^2 + 18x - 4=10x$$ Step 1) Write the equation in standard form. We will subtract 10x away from each side: $$5x^2 + 8x - 4=0$$ Step 2) Record the values for a, b, and c
a = 5, b = 8, c = -4
Step 3) Plug into the quadratic formula and simplify: $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-8 \pm \sqrt{8^2 - 4(5)(-4)}}{2(5)}$$ $$x=\frac{-8 \pm \sqrt{64 - (-80)}}{2(5)}$$ $$x=\frac{-8 \pm \sqrt{144}}{10}$$ $$x=\frac{-8 \pm 12}{10}$$ We will have two solutions for x: $$x=\frac{-8 + 12}{10}=\frac{4}{10}=\frac{2}{5}$$ $$x=\frac{-8 - 12}{10}=\frac{-20}{10}=-2$$ Our solutions: $$x=\frac{2}{5}\hspace{.5em}or \hspace{.5em}x=-2$$ Example 2: Solve each equation. $$4x^2 - 8x - 2=6$$ Step 1) Write the equation in standard form. We will subtract 6 away from each side: $$4x^2 - 8x - 8=0$$ Step 2) Record the values for a, b, and c
a = 4, b = -8, c = -8
Step 3) Plug into the quadratic formula and simplify: $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(-8)}}{2(4)}$$ $$x=\frac{8 \pm \sqrt{64 - (-128)}}{8}$$ $$x=\frac{8 \pm \sqrt{192}}{8}$$ $$x=\frac{8 \pm \sqrt{4 \cdot 4 \cdot 4 \cdot 3}}{8}$$ $$x=\frac{8 \pm 8\sqrt{3}}{8}$$ $$x=\frac{8(1 \pm 1\sqrt{3})}{8}$$ $$\require{cancel}x=\frac{\cancel{8}(1 \pm 1\sqrt{3})}{\cancel{8}}$$ $$\require{cancel}x=1 \pm \sqrt{3}$$ We will have two solutions for x: $$x=1 + \sqrt{3}\hspace{.5em}or \hspace{.5em}1 - \sqrt{3}$$ Example 3: Solve each equation. $$3x^2 + 16=-8x + 9$$ Step 1) Write the equation in standard form. We will subtract 9 away from each side: $$3x^2 + 7=-8x$$ We will add 8x to each side: $$3x^2 + 8x + 7=0$$ Step 2) Record the values for a, b, and c
a = 3, b = 8, c = 7
Step 3) Plug into the quadratic formula and simplify: $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-8 \pm \sqrt{8^2 - 4(3)(7)}}{2(3)}$$ $$x=\frac{-8 \pm \sqrt{64 - 84}}{6}$$ $$x=\frac{-8 \pm \sqrt{-20}}{6}$$ We have the square root of a negative number, so we won't have any real number solutions. Again, when the discriminant is less than zero, we will be taking the square root of a negative number, which will result in an answer of "no real solution".

#### Skills Check:

Example #1

Solve each equation. $$4x^{2}+ 4x=3$$

A
$$x=-14, 8$$
B
$$x=-2 \pm 7$$
C
$$x=-\frac{3}{2}, 11$$
D
$$x=\frac{-5 \pm \sqrt{205}}{10}$$
E
$$x=-\frac{3}{2}, \frac{1}{2}$$

Example #2

Solve each equation. $$9x^{2}- 12x=19$$

A
$$x=\pm 12$$
B
$$x=\pm 6$$
C
$$x=-6 \pm \sqrt{55}$$
D
$$x=\frac{2 \pm \sqrt{23}}{3}$$
E
$$x=-2 \pm \sqrt{5}$$

Example #3

Solve each equation. $$6x^{2}- 21=-11x$$

A
$$x=-7, 9$$
B
$$x=\frac{1}{2}, \frac{4}{3}$$
C
$$\text{No Real Solution}$$
D
$$x=-3, \frac{7}{6}$$
E
$$x=\frac{5 \pm \sqrt{102}}{7}$$         