I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 2-7 : Quadratic Equations : A Summary

In the previous two sections we’ve talked quite a bit about solving quadratic equations. A logical question to ask at this point is which method should we use to solve a given quadratic equation? Unfortunately, the answer is, it depends.

If your instructor has specified the method to use then that, of course, is the method you should use. However, if your instructor had NOT specified the method to use then we will have to make the decision ourselves. Here is a general set of guidelines that *may* be helpful in determining which method to use.

- Is it clearly a square root property problem? In other words, does the equation consist ONLY of something squared and a constant. If this is true then the square root property is probably the easiest method for use.
- Does it factor? If so, that is probably the way to go. Note that you shouldn’t spend a lot of time trying to determine if the quadratic equation factors. Look at the equation and if you can quickly determine that it factors then go with that. If you can’t quickly determine that it factors then don’t worry about it.
- If you’ve reached this point then you’ve determined that the equation is not in the correct form for the square root property and that it doesn’t factor (or that you can’t quickly see that it factors). So, at this point you’re only real option is the quadratic formula.

Once you’ve solved enough quadratic equations the above set of guidelines will become almost second nature to you and you will find yourself going through them almost without thinking.

Notice as well that nowhere in the set of guidelines was completing the square mentioned. The reason for this is simply that it’s a long method that is prone to mistakes when you get in a hurry. The quadratic formula will also always work and is much shorter of a method to use. In general, you should only use completing the square if your instructor has required you to use it.

As a solving technique completing the square should always be your last choice. This doesn’t mean however that it isn’t an important method. We will see the completing the square process arise in several sections in later chapters. Interestingly enough when we do see this process in later sections we won’t be solving equations! This process is very useful in many situations of which solving is only one.

Before leaving this section we have one more topic to discuss. In the previous couple of sections we saw that solving a quadratic equation in standard form,

\[a{x^2} + bx + c = 0\]we will get one of the following three possible solution sets.

- Two real distinct (
*i.e.*not equal) solutions. - A double root. Recall this arises when we can factor the equation into a perfect square.
- Two complex solutions.

These are the ONLY possibilities for solving quadratic equations in standard form. Note however, that if we start with rational expression in the equation we may get different solution sets because we may need avoid one of the possible solutions so we don’t get division by zero errors.

Now, it turns out that all we need to do is look at the quadratic equation (in standard form of course) to determine which of the three cases that we’ll get. To see how this works let’s start off by recalling the quadratic formula.

\[x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]The quantity \({b^2} - 4ac\) in the quadratic formula is called the **discriminant**. It is the value of the discriminant that will determine which solution set we will get. Let’s go through the cases one at a time.

- Two real distinct solutions. We will get this solution set if \({b^2} - 4ac > 0\). In this case we will be taking the square root of a positive number and so the square root will be a real number. Therefore, the numerator in the quadratic formula will be –\(b\) plus or minus a real number. This means that the numerator will be two different real numbers. Dividing either one by 2\(a\) won’t change the fact that they are real, nor will it change the fact that they are different.
- A double root. We will get this solution set if \({b^2} - 4ac = 0\). Here we will be taking the square root of zero, which is zero. However, this means that the “plus or minus” part of the numerator will be zero and so the numerator in the quadratic formula will be –\(b\). In other words, we will get a single real number out of the quadratic formula, which is what we get when we get a double root.
- Two complex solutions. We will get this solution set if \({b^2} - 4ac < 0\). If the discriminant is negative we will be taking the square root of negative numbers in the quadratic formula which means that we will get complex solutions. Also, we will get two since they have a “plus or minus” in front of the square root.

So, let’s summarize up the results here.

- If \({b^2} - 4ac > 0\) then we will get two real solutions to the quadratic equation.
- If \({b^2} - 4ac = 0\) then we will get a double root to the quadratic equation.
- If \({b^2} - 4ac < 0\) then we will get two complex solutions to the quadratic equation.

- \(13{x^2} + 1 = 5x\)
- \(6{q^2} + 20q = 3\)
- \(49{t^2} + 126t + 81 = 0\)

All we need to do here is make sure the equation is in standard form, determine the value of \(a\), \(b\), and \(c\), then plug them into the discriminant.

a \(13{x^2} + 1 = 5x\) Show Solution

First get the equation in standard form.

\[13{x^2} - 5x + 1 = 0\]We then have,

\[a = 13\hspace{0.25in}\hspace{0.25in}b = - 5\hspace{0.25in}\hspace{0.25in}c = 1\]Plugging into the discriminant gives,

\[{b^2} - 4ac = {\left( { - 5} \right)^2} - 4\left( {13} \right)\left( 1 \right) = - 27\]The discriminant is negative and so we will have two complex solutions. For reference purposes the actual solutions are,

\[x = \frac{{5 \pm 3\sqrt 3 \,i}}{{26}}\]b \(6{q^2} + 20q = 3\) Show Solution

Again, we first need to get the equation in standard form.

\[6{q^2} + 20q - 3 = 0\]This gives,

\[a = 6\hspace{0.25in}\hspace{0.25in}b = 20\hspace{0.25in}\hspace{0.25in}c = - 3\]The discriminant is then,

\[{b^2} - 4ac = {\left( {20} \right)^2} - 4\left( 6 \right)\left( { - 3} \right) = 472\]The discriminant is positive we will get two real distinct solutions. Here they are,

\[x = \frac{{ - 20 \pm \sqrt {472} }}{{12}} = \frac{{ - 10 \pm \sqrt {118} }}{6}\]c \(49{t^2} + 126t + 81 = 0\) Show Solution

This equation is already in standard form so let’s jump straight in.

\[a = 49\hspace{0.25in}\hspace{0.25in}b = 126\hspace{0.25in}\hspace{0.25in}c = 81\]The discriminant is then,

\[{b^2} - 4ac = {\left( {126} \right)^2} - 4\left( {49} \right)\left( {81} \right) = 0\]In this case we’ll get a double root since the discriminant is zero. Here it is,

\[x = - \frac{9}{7}\]