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: ax
^{2}+ bx + c = 0 - Learn how to derive the Quadratic Formula
- Learn about the possible solutions using the discriminant: b
^{2}- 4ac - Learn how to Solve Quadratic Equations using the Quadratic Formula

## How to Solve a Quadratic Equation using the Quadratic Formula

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:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
### The Discriminant: b

The discriminant for the quadratic formula is the part under the square root symbol:

b

The discriminant will tell us if we have two solutions, only one solution, or no real solution.

If b

If b

If b

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".

### Deriving the Quadratic Formula

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 ax^{2}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 x^{2}. 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: b^{2} - 4ac

The discriminant for the quadratic formula is the part under the square root symbol:
b

^{2}- 4acThe discriminant will tell us if we have two solutions, only one solution, or no real solution.

If b

^{2}- 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 b

^{2}- 4ac > 0 » There are two solutions.If b

^{2}- 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: ax
^{2}+ bx + c = 0 - Record the values for a, b, and c
- Plug into the quadratic formula and simplify

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".

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