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.
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.
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.
So, let’s summarize up the results here.
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.
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,
\[x = \frac{{5 \pm 3\sqrt 3 ,円i}}{{26}}\]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,
\[x = \frac{{ - 20 \pm \sqrt {472} }}{{12}} = \frac{{ - 10 \pm \sqrt {118} }}{6}\]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,
\[x = - \frac{9}{7}\]