# How to Solve Quadratic Equations – Part 2

This post continues from where my last post left off, on how to solve quadratic equations. I explained the general form that a quadratic equation will take, with the key being that there is an x^{2} term present. To solve them without using the quadratic formula, you need to use a bit of factoring methods to come up with the roots. In particular, one common factoring method to use is the grouping method of factoring. Then, once factored, you consider the property that says “two terms multiplied will equal zero only if one or both of those terms is 0.” This may seem like a lot of work, and may sound a bit confusing with all the steps you need to take. But I think with a bit of practice you will come to better appreciate and understand the process you need to follow to arrive at your solution. You will see that you already know the individual steps you need to solve the equation. You just need to become familiar with the order that you use these steps.

Follow along through my example and you will hopefully be able to see what I mean.

Let’s consider the equation x^{2} + 7x + 10 = 0

First, we can identify that there is an x^{2} term (with a non-zero coefficient… 1), so we can say that it is a quadratic equation.

To solve a quadratic equation, we want to determine the roots, or what values make the equation true. To help us to achieve this, we want to rearrange the left side so that it is a product of two terms (or expressions). In this way, we can say that “something times something equals zero”. And since we need one of those “somethings” to be zero if the product is zero, we essentially break this down to “something #1 = 0” and “something #2 = 0”, and by solving these two simpler equations, we will arrive at our roots. So, continuing with our example then, let’s factor it. Review my post on methods of factoring if you need a bit of a refresher!

x2 + 7x + 10 = 0

(x + 2)(x + 5) = 0

This is what we’re looking for: two expressions multiplied together to give zero. Now, we have two equations to work with to find our roots of the quadratic equation. Rewriting, this gives us:

x + 2 = 0 and x + 5 = 0

And quite obviously, these can be solved to show that x = (–2) and (–5). And since we followed that whole process, we can consider these two values to be roots of our original quadratic equation. However, it is VERY IMPORTANT to substitute these values back into the original equation to check! With these values, we can show that:

(–2)2 + 7(–2) + 10 = 0

4 – 14 + 10 = 0…….. this is true. So –2 is for sure one of the roots. I’ll leave –5 for you to verify on your own.

If you find a question and proceed all the way through to find the roots, and you go and plug them back into the original equation, if one of the roots does NOT satisfy the equation, you cannot count it as one of the roots. This sometimes happens when you have an expression in a denominator (eg. (x – 2)), and if you determine through the above steps that your expression gives you a root of 2, by plugging this into your original equation, specifically into the denominator, the denominator will equal 0 and cause the expression to be undefined. Therefore, this root does not satisfy the original equation and you just ignore it.

I hope this has helped to explain the process you need to follow to solve quadratic equations. With practice, they will become second nature. However, despite all of the work required, sometimes it just is not practical or apparent how to factor your quadratic equation. In these cases, you would likely want to rely on the use of the quadratic formula, which I will go over in a future post to explain what it is and how it works. Let me know if this makes sense or if you’d like anything more added.