Quadratic formula and examples

A quadratic equation is any equation that can be written as \(ax^2+bx+c=0\), for some numbers \(a\), \(b\), and \(c\), where \(a\) is nonzero. The quadratic formula is one method of solving this type of question. Below, we will look at several examples of how to use this formula and also see how to work with it when there are complex solutions.

Table of Contents

  1. Basic examples of applying the formula
  2. When there are complex solutions (involving \(i\))
  3. Summary


advertisement

Examples of applying the quadratic formula

Looking at the formula below, you can see that \(a\), \(b\), and \(c\) are the numbers straight from your equation. Applying this formula is really just about determining the values of \(a\), \(b\), and \(c\) and then simplifying the results.

image showing the quadratic formula and the values used in the formula

But, it is important to note the form of the equation given above. If your equation is not in that form, you will need to take care of that as a first step. Let’s take a look at a couple of examples.

Example

Solve: \(x^2-x=6\)

Solution

Before we do anything else, we need to make sure that all the terms are on one side of the equation. As you can see above, the formula is based on the idea that we have 0 on one side.

Subtract 6 from both sides to get:

\(x^2-x-6=0\)

Now that we have it in this form, we can see that:

\(a=1\) , \(b= -1\) , \(c=-6\)

Why are \(b\) and \(c\) negative? The formula is based off the form \(ax^2+bx+c=0\) where all the numerical values are being added and we can rewrite \(x^2-x-6=0\) as \(x^2 + (-x) + (-6) = 0\). To keep it simple, just remember to carry the sign into the formula.

Once you have the values of \(a\), \(b\), and \(c\), the final step is to substitute them into the formula and simplify.

\(\begin{align}x &= \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)} \\ &=\dfrac{1\pm\sqrt{1+24}}{2} \\ &=\dfrac{1\pm\sqrt{25}}{2}\end{align}\)

At this stage, the plus or minus symbol (\(\pm\)) tells you that there are actually two different solutions:

\(\begin{align} x &= \dfrac{1+\sqrt{25}}{2}\\&=\dfrac{1+5}{2}\\&=\dfrac{6}{2}\\&=3\end{align}\)

and

\(\begin{align} x &= \dfrac{1- \sqrt{25}}{2}\\ &= \dfrac{1-5}{2}\\ &=\dfrac{-4}{2}\\ &=-2\end{align}\)

Therefore the final answer would be:

\(x= \bbox[border: 1px solid black; padding: 2px]{3}\) , \(x= \bbox[border: 1px solid black; padding: 2px]{-2}\)

This particular quadratic equation could have been solved using factoring instead, and so it ended up simplifying really nicely. Often, there will be a bit more work – as you can see in the next example.

Example

Solve: \(2x^2+2x-7=0\)

Solution

This time we already have all the terms on the same side. So, we just need to determine the values of \(a\), \(b\), and \(c\).

\(a=2\), \(b=2\) , \(c=-7\)

Applying the formula:

\(\begin{align}x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-2\pm\sqrt{(2)^2-4(2)(-7)}}{2(2)}\\ &=\dfrac{-2\pm\sqrt{4+56}}{4} \\ &=\dfrac{-2\pm\sqrt{60}}{4}\\ &=\dfrac{-2\pm 2\sqrt{15}}{4}\end{align}\)

Notice that 2 is a FACTOR of both the numerator and denominator, so it can be cancelled.

\(x =\dfrac{-1\pm\sqrt{15}}{2}\)

This answer can not be simplified anymore, though you could approximate the answer with decimals. Therefore the final answer is:

\(x=\bbox[border: 1px solid black; padding: 2px]{\dfrac{-1+\sqrt{15}}{2}}\) ,

\(x=\bbox[border: 1px solid black; padding: 2px]{\dfrac{-1-\sqrt{15}}{2}}\)

Complex Solutions

When using the quadratic formula, it is possible to find complex solutions – that is, solutions that are not real numbers but instead are based on the imaginary unit, \(i\). Recall the following definition:

\(i = \sqrt{-1}\)

If a negative square root comes up in your work, then your equation has complex solutions which can be written in terms of \(i\).

Example

Solve: \(x^2-2x+5=0\)

Solution

Just as in the previous example, we already have all the terms on one side. So, we will just determine the values of \(a\), \(b\), and \(c\) and then apply the formula.

\(a=1\), \(b=-2\) , \(c=5\)

Applying the formula:

\(\begin{align}x &=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2(1)}\\ &=\dfrac{2\pm\sqrt{4-20}}{2} \\ &=\dfrac{2\pm\sqrt{-16}}{2}\end{align}\)

A negative value under the square root means that there are no real solutions to this equation. However, there are complex solutions. Using the definition of \(i\), we can write:

\(\sqrt{-16} = 4i\)

This allows us to keep simplifying.

\(\begin{align} x &=\dfrac{2\pm 4i}{2}\\ &=1 \pm 2i\end{align}\)

The solutions to this quadratic equation are:

\(x= \bbox[border: 1px solid black; padding: 2px]{1+2i}\) , \(x = \bbox[border: 1px solid black; padding: 2px]{1 – 2i}\)

There are three cases with any quadratic equation: one real solution, two real solutions, or no real solutions (complex solutions). Each case tells us not only about the equation, but also about its graph as each of these represents a zero of the polynomial.

advertisement

Summary

Understanding the quadratic formula really comes down to memorization. As long as you can check that your equation is in the right form and remember the formula correctly, the rest is just arithmetic (even if it is a little complicated).

Subscribe to our Newsletter!

We are always posting new free lessons and adding more study guides, calculator guides, and problem packs.

Sign up to get occasional emails (once every couple or three weeks) letting you know what's new!

Copyright 2010- 2017 MathBootCamps | Privacy Policy