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 which 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, which 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".