Solving Trinomial Functions in Quadratic Form
Factoring & Substitution
In the last section we learned how to factor trinomials in quadratic form. If you’re not 100% solid with that topic, please visit that page and practice until you get it! We are building on that knowledge here.
On this page we will learn how to find the real x – intercepts of a trinomial function that is in quadratic form. We will see how to do so in two fashions. First, if the function is factorable, and second, if the function is non-factorable. The factorable versions are pretty straight forward and only apply previous knowledge here in a familiar application.
Find the intercepts for a non-factorable trinomial in quadratic form involves a new and initially tricky, concept and application. The method is called substitution, sometimes u – substitution. It is an application of the transitive property of mathematics.
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In the previous section we learned what a trinomial in quadratic form was, and how to factor it. In this section we are going to see how to solve trinomial equations in quadratic form.
If the trinomial is factorable, then life is easy. Just like you did with a quadratic equation, take each factor and create an equation with the factor equal to zero. Then, solve each equation. Boom, done! Here’s an example.
x4 – x2 – 12 = 0
First, this is in quadratic form because the degree is double of the exponent for the second term, it has three terms, and the last term is a constant. It is factorable because the product of 3 and 4 is 12, and their difference is one.
(x2 – 4)(x2 + 3) = 0
Notice the first term is NOT factored completely. We can factor it further, into linear factors.
(x + 2)(x – 2)(x2 + 3) = 0
Now we can set each fully factor equal to zero and solve each equation.
x + 2 = 0, x – 2 = 0, x2 + 3 = 0
When we solve these factors we find four x – intercepts. Let’s round the square root of 3 to 1.7. This gives us the solutions -2, -1.7, 1.7, and 2.
If you’re solid on solving quadratic equations by factoring, this should be a walk in the park, so far!
There’s always a but, right? But … what if the equation wasn’t factorable, but was a trinomial in quadratic form? Could the quadratic formula be used?
The graph below is of the function f(x) = x4 – 5x2 + 3. It has four real x – intercepts. The graph was created using desmos, and their software will round irrational numbers. It is likely that these decimals we see are approximations.
The question is this: Is there a way we can use the quadratic formula on a trinomial in quadratic form (one that is not quadratic)?
Let’s see how it works.
Our function is f(x) = x4 – 5x2 + 3. The quadratic formula only works for quadratic equations. We are going to do a little bit of mathemagic here. We are going to create a substitution.
Instead of f(x), let’s use f(m). Let’s set the value of m = x2. We started off with x4 – 5x2 + 3. But, now if x2 = m, we can substitute to get m2 – 5m + 3. This is quadratic! We can use the formula to solve this. Miracles never cease!
Once we solve for m, we can substitute those values into the equation m = x2, and solve for x.
Let’s see how it works. Here is the quadratic formula being used for our equation.
Since desmos.com rounded to 3 decimal places, we did, too.
Remember, we set m = x2. We now know what m equals. We need to find the values of x. So, we just plug in our m values into m = x2.
Taking the square root, remembering the positive and negative values for each, we get the following.
x = -2.074, 2.072 and -0.835, 0.835
New Concept and Procedure
The method we are using is called substitution. It’s often called u substitution, but the letter is irrelevant. The logic we are using here is not overly complicated. It’s called the transitive property. In abstract it says that if a = b, and b = c, then we know that a = c. In common ideas, if John has as much money as Sally, and Sally has as much money as Fred, John has as much money as Fred.
Here’s how we’re applying that concept here. We cannot factor the quadratic form trinomial. However, we can set an equal expression that is simpler. We can find the value of the new, simpler expression, and use that to find the more complicated values.
Consider the case of f(x) = 3x6 – 4x3 – 11. At this point we don’t know a formula that can be used to solve a polynomial of degree six. However, because this is quadratic form, we can substitute a value for x3. This works great because x6 = (x3)2. Let’s see a color coded example of this below.
We can solve the equation 3m2 – 4m – 11 = 0 by using the quadratic formula. Once we know the solutions, we can plug those values into the equation x3 = m. Let’s do just that.
Remember, m also equals x3. So, x3 also equals these two numbers. We can set up two equations and now solve for x.
So this polynomial function of degree six has only 2 real roots. In the next section we will learn about the other roots!
