After watching this video lesson, you will be able to solve more complicated quadratic equations by the method of substitution. Learn how substitution makes your problem solving easier.
Super Quadratic Equations
What do you think of when you hear the words ‘quadratic equations’? Do you think of the usual polynomials of degree 2? Equations such as x^2 + 3x + 2 and x^2 – 4x – 12? But, did you know that we can actually have quadratic equations in disguise? Just like we have our super heroes in disguise posing as regular people, so we have quadratic equations turning into super equations.What kind of super equations can we expect to see? We can expect to see equations such as x^4 + 3x^2 + 2 or even (x – 2)^2 – 4(x – 2) – 12.
These are super equations because we see that they are disguised quadratics. The x^4 + 3x^2 + 2 can be rewritten as (x^2)^2 + 3(x^2) + 2. Our quadratic has the x^2 disguise on.
The (x – 2)^2 – 4(x – 2) – 12 has the (x – 2) disguise on. Once we’ve discovered the disguise, we see that they still follow the quadratic form of ax^2 + bx + c.
Making the Substitution
What makes these super equations quadratic equations in disguise is that by making one substitution, we can turn it into a quadratic equation that we can easily solve. Let’s take a look at how we can go ahead and solve the equation x^4 – 3x^2 + 2.
To help us figure out what kind of substitution to make, we can ask ourselves, ‘how can we change the equation so it becomes like the regular quadratic equations that we are so familiar with’? Looking at our equation, we see that if we make the substitution u = x^2, then our equation will turn back to normal. We are substituting with whatever the disguise is. We use another letter for our substitution so that we don’t get confused with the problem’s variable. Let’s see what happens when we make this substitution. Our equation, x^4 – 3x^2 + 2, turns into u^2 – 3u + 2. Hey, that’s a normal quadratic equation! We know how to solve this.
Solving by Substitution
To solve this, we use our skills in solving normal quadratic equations.
For this problem, we can factor to solve. We have (u – 1)(u – 2). To find what our u variables will equal we set each factor equal to 0, and then solve for u. We get u = 1 and u = 2.To finish our problem, we now need to find what our x equals. To do this, we go back to our substitution, u = x^2. We plug this in for the us we found.
We get x^2 = 1 and x^2 = 2. Solving for x, we get x = +/- 1 and x = +/- sqrt(2).In this case, we have a total of four answers.
This is because our super equation has a degree of 4. A good check to make sure we have all the answers that we need is to look at the degree of our super equation, our quadratic equation in disguise. Whatever degree that equation is, is the number of solutions that we will find. For our problem, we end up with x = 1, x = -1, x = sqrt(2), and x = – sqrt(2).
One More Example
Let’s solve (x – 2)^2 – 4(x – 2) – 12.We can make the substitution u = x – 2 since x – 2 is the disguise of this super equation. Our equation then becomes the quadratic u^2 – 4u – 12.
Factoring, we get (u – 6)(u + 2). Solving for u, we get u = 6 and u = -2. Substituting our u = x – 2 back in, we get x – 2 = 6 and x – 2 = -2.
Solving for x now, we get x = 8 and x = 0. And we are done!
Let’s review what we’ve learned:Quadratic equations are polynomials of degree 2. Sometimes, we will see our quadratic equations disguised as super equations such as x^4 + 3x^2 + 2. We know these are disguised quadratics because by making one substitution, we can turn the super equation into a normal quadratic equation that we can easily solve. After we make the substitution, we solve the quadratic equation as we normally do using the skills we have already learned, such as factoring.
Then we replace our substitution to find our final answer in the variable we began with.
After you’ve completed this lesson, you should have the ability to:
- Define quadratic equations
- Explain how to solve super equations/disguised quadratics by substitution