The Quadratic Formula, Part 1

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One of the great things about quadratic equations is that there's a formula we can use to get the solutions to an equation just by plugging in the quadratic polynomial part's coefficients. Say we have a quadratic equation that looks like:

ax² + bx + c = 0

In a real world problem, a, b and c are all going to be numbers. If we take those numbers and substitute them into this equation, we'll get the solutions that we're looking for:

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

Example 1

Solve the equation x² + 3x - 4 = 0.

First, notice that the equation is already set equal to zero. That's an important first step when using the Quadratic Formula, just like it was when we were solving by factoring.

Now we need to pick out the a, b and c values that we'll substitute into the formula. Remember that the a value is the number in front of the x² term, the b value is the number in front of the x term and the c value is the constant at the end. This means that, for our equation, we have:

a = 1b = 3c = -4

If we substitute those values into the formula, we get:

$$x = \frac {-3 \pm \sqrt {3^2 - 4\cdot 1 \cdot (-4)}} {2 \cdot 1}$$ $$x = \frac {-3 \pm \sqrt {9 + 16}} {2}$$ $$x = \frac {-3 \pm \sqrt {25}} {2}$$ $$x = \frac {-3 \pm 5} {2}$$ $$x = \frac {-3 + 5} {2}, \frac {-3 - 5} {2}$$ $$x = \frac {2} {2}, \frac {-8} {2}$$ $$x = 1, -4$$

Example 3

Solve the equation x² + 10x + 20 = -5.

In this example, we can't just jump right in to the formula because the equation isn't equal to 0. We have to start off by adding 5 to both sides:

x² + 10x + 25 = 0

Now we can pull out the values of a, b and c.

a = 1b = 10c = 25

If we substitute those values into the formula, we get:

$$x = \frac {-10 \pm \sqrt {10^2 - 4\cdot 1 \cdot 25}} {2 \cdot 1}$$ $$x = \frac {-10 \pm \sqrt {100 - 100}} {2}$$ $$x = \frac {-10 \pm \sqrt {0}} {2}$$ $$x = \frac {-10} {2}$$ $$x = -5$$

Example 2

Solve the equation 2x² - 5x - 2 = 0.

First, we need to pick out the a, b and c values that we'll substitute into the formula. Remember that the a value is the number in front of the x² term, the b value is the number in front of the x term and the c value is the constant at the end. This means that, for our equation, we have:

a = 2b = -5c = -2

If we substitute those values into the formula, we get:

$$x = \frac {-(-5) \pm \sqrt {(-5)^2 - 4\cdot 2 \cdot (-2)}} {2 \cdot 2}$$ $$x = \frac {5 \pm \sqrt {25 + 16}} {4}$$ $$x = \frac {5 \pm \sqrt {41}} {4}$$

Since $\sqrt{41}$ can't be reduced or simplified any further. That's our final answer. It's a little messy with both a square root and a fraction left in it but these problems are often going to come out that way.

Example 4

Solve the equation x² + 2x + 5 = 0.

Since the equation is already set equal to zero we can start by picking out the values we'll need for the formula.

a = 1b = 2c = 5

If we substitute those values into the formula, we get:

$$x = \frac {-2 \pm \sqrt {2^2 - 4\cdot 1 \cdot 5}} {2 \cdot 1}$$ $$x = \frac {-2 \pm \sqrt {4 - 20}} {2}$$ $$x = \frac {-2 \pm \sqrt {-16}} {2}$$

Now we've run into a problem. We can't take the square root of a negative number so, at this point, we have to conclude that the equation doesn't have any real number solutions. (It does have complex solutions but we'll save that for the next section.)

Video Lectures

Deriving the Quadratic Formula

The Quadratic Formula is a high point in introductory algebra classes. It gives us a general formula that can be used to solve any quadratic equations. In this lecture, I'm going to derive the equation starting from a general quadratic equation.

Solving Equations with the Quadratic Formula, 1

Now that we have our new Quadratic Formula in hand, we can go through some examples of how it's used and look at some special cases you should be aware of.

Dynamic Practice - Using the Quadratic Formula