Intro to Quadratic Equations - Online Tutor, Practice Problems & Exam Prep

A quadratic equation, or polynomial of degree 2, is expressed in standard form as a^2x + b^1x + c = 0, where coefficients a, b, and c represent the numerical coefficients of the terms. To solve quadratic equations, one can factor them into binomials, set each factor to zero, and solve for x, often yielding two solutions. Understanding these concepts is crucial for mastering polynomial equations and their applications.

Introduction to Quadratic Equations

Hey, everyone. We've learned to solve a couple of different types of equations, and now we're going to add a new equation to the mix called a quadratic equation. Now, a quadratic equation is going to be a bit more complicated than a linear equation, and it might even seem a bit overwhelming at times. But don't worry, I'm going to walk you through everything you need to know about quadratic equations in the next several videos, starting with what a quadratic equation even is and then going into how to solve them. So let's go ahead and get started.

Now, if you take a linear equation, like 2x − 6 = 0, and you simply add an x2 term, so if I had 3x2 + 2x − 6 = 0, this is now a quadratic equation. A quadratic equation is also called a polynomial of degree 2 because the highest degree or power in my equation is this 2 on my x2 term. Whether you hear it called a quadratic equation or a second-degree polynomial, these two mean the same thing.

You're often going to have to write quadratic equations in standard form. The standard form of a quadratic equation is ax2+ bx +0, where all of my terms are on the same side of the equation and they're all written in descending order of power. I begin with the ax2 term, which has a power of 2, followed by the bx term (imagine an invisible one here, so it has a power of 1), and ending with a constant that has no power. So, I go from 2 to 1 to 0, in descending order of power.

When identifying the coefficients, in 3x2 + 2x − 6, a is the coefficient of the x2 term, so it is 3. b is the coefficient of the x term, which here is positive 2, and c, the constant term, is negative 6. Always pay attention to the signs to correctly identify each coefficient.

Let's look at some other quadratic equations. For our first example, we have 5x2 = x − 3. To put this in standard form, I move all terms to the left side, subtract x, and add 3. This simplifies to 5x2 – x + 3 = 0. This equation is then in standard form and in descending order of power: 2 for x2, 1 for x, and 0 for the constant. The coefficients are a=5, b=−1 (as x implies 1x), and c=3.

In our next example, −2x2 + 53 = 0, the equation is already in standard form, with terms in descending order of power. Here, a=−2, b=0 (since the x term is missing), and c=53. As long as there is an x2 term, it qualifies as a quadratic equation, regardless of whether a, b, or c are fractions or even zero.

That's all for this video, and I'll see you in the next one.

Write the given quadratic equation in standard form. Identify a, b, and c. $-4x^2+x=8$

• A. a = - 4, b = 0, c = - 8
• B. a = - 4, b = 1, c = 8
• C. a = - 4, b = 1, c = - 8
• D. a = 2, b = 1, c = 0

Solving Quadratic Equations by Factoring

Hey everyone. Whenever we solved linear equations, we wanted to find a sum value for x that we could plug back into our original equation to make it true. And we want to do the same thing when solving quadratic equations. We want to find some value for x that makes our equation true. But whenever we solve linear equations, we were able to move some numbers around and then isolate x to get an answer. But what happens if I try to do the same thing with a quadratic equation? Well, looking at this x2-5x+4 over here, I might go ahead and move this 5x to the other side, maybe move my 4 as well leaving me with x2=5x-4. And then you might think, oh, we can just take the square root to get x by itself, which would leave me with x is equal to where do I go from there? What is the square root of 5x-4? So I can't actually solve a quadratic equation the same way I solved a linear equation. And more than that, there are actually often going to be 2 correct values for x when solving a quadratic equation. So how do we find those two values? Well, luckily, this something else that we need to know how to do from the previous chapter, is factoring. So let's go ahead and take a look at how we can factor to solve quadratic equations. Well, we're going to want to factor from standard form and then simply set each factor equal to 0. So if I have a quadratic equation like x2+x-6 is equal to 0 and then I go ahead and factor that into (x+3)(x-2) equals 0, I can simply take each of those individual factors, so x+3 and x-2, set them equal to 0, and then solve for x. And the reason that I can do this is because these factors are multiplied together. So x+3 times x-2 is going to equal 0 and anything times 0 is 0. So if either one of my factors, let's say that x+3 was 0, if I multiply that times anything, that would give me 0. So that might seem a little bit abstract right now. Let's go ahead and take a look at that in action. So in my example here, I want to solve this by factoring and I have x2-9x is equal to negative 20. So our first step here is going to be to write our equation in standard form. So I want to go ahead and get all my terms on the same side in descending order of power. So to do that, I just need to move my negative 20 to the other side, which I can do by adding 20 to both sides. It will cancel on the right side leaving me with x2-9x+20 is equal to 0 and we're done with step 1. Now step 2 is going to be to factor completely. Now remember, there are multiple different methods to factor, so let's go ahead and take a look at what method of factoring we should use for this equation. So we first want to look at how many terms this has. And it has this x2, negative 9x, x, and a positive twenty. So that is 3 terms, which tells me that I need to then say, does it fit a factoring formula? And looking quickly at my factoring formulas here, it doesn't seem to fit a factoring formula either. So my answer is no, which means that I then need to use the a c method to factor this quadratic. So let's go ahead and use the a c method here. So first, we want to look at a and c. So let's go ahead and make sure that we know what a, b, and c are here. So in this case, I have this kind of invisible one and that is my a, then b is negative 9 and c is 20. So if I multiply a and c together, a is just 1, so it's really just c. So a c is 20. So I want to find 2 factors that multiply to 20 and then add to b, which in this case is negative 9. So two factors that multiply to 20 add to negative 9. Since that b is negative, I know that both of my factors have to be negative, so let's think of factors of 20 that are negative. So I have negative 1 and negative 20, and then I have negative 2 and negative 10, and then I have negative 4 and negative 5. Now, negative 1 and negative 20 are going to add to negative 21, which is not negative 9. Negative 2 and negative 10 are going to add to negative 12, which is also not negative 9. But then I have negative 4 and negative 5, which do add to negative 9, so it turns out that negative 4 and negative 5 are my factors here. So once I have figured out what my factors are, I can go ahead and write this as (x-4)(x-5) is equal to 0, and step 2 is done. Now for step 3, I want to set my factors equal to 0 and then solve for x. So let's go ahead and look at each individual factor. So I have x-4. I'm setting that equal to 0, and then x-5 equals 0 as well. So let's go ahead and solve here. To solve for x with this x-4, I just need to add 4 to both sides leaving me with x is equal to 4. And then with x-5, I simply add 5 to both sides and I end up with x equals 5. And these are actually my solutions. I have completed step number 3. My solutions are x equals 4 and x equals 5. Now because we want to find 2 values for x that we could plug back into our equation that makes it true, let's go ahead and make sure that that does happen. So let's first try this x equals 4 and go ahead and plug that in. So if I plug it into my original equation, I will get 4 squared minus 9 times 4 is equal to negative 20. Now 4 squared is 16, 9 times 4 is 36. So 16 minus 36 is equal to negative 20. So negative 20 is equal to negative 20. That's definitely a true statement. I know that 4 is a value for x that makes it true. It's definitely one of my answers. Now we could do the same thing for 5 here but I'm going to leave that up to you. So that is all we need to do to solve quadratic equations by factoring. Let's go ahead and get some practice.

Solve the given quadratic equation by factoring. $3x^2+12x=0$

• A. $x=3,x=4$
• B. $x=0,x=-4$
• C. $x=-3,x=-4$
• D. $x=1,x=4$

Solve the given equation by factoring. $2x^2+7x+6=0$

• A. $x=3,x=\frac67$
• B. $x=-2,x=0$
• C. $x=2,x=\frac32$
• D. $x=-2,x=-\frac32$

Here’s what students ask on this topic:

What is a quadratic equation and how is it different from a linear equation?

A quadratic equation is a polynomial equation of degree 2, typically written in the form $a^{2}x+b^{1}x+c=0$, where $a$, $b$, and $c$ are constants. The key difference from a linear equation, which is of the form $\mathrm{mx}+b=0$, is the presence of the $x^2$ term in a quadratic equation. This makes the graph of a quadratic equation a parabola, whereas the graph of a linear equation is a straight line.

How do you write a quadratic equation in standard form?

To write a quadratic equation in standard form, you need to arrange it as $a^{2}x+b^{1}x+c=0$. This means all terms should be on one side of the equation, with the $x^2$ term first, followed by the $x$ term, and then the constant term. For example, if you have $5x^2=x-3$, you would move $x$ and $3$ to the left side to get $5x^2-x+3=0$.

What are the coefficients in a quadratic equation and how do you identify them?

In a quadratic equation of the form $a^{2}x+b^{1}x+c=0$, the coefficients are the numerical values in front of the variables. The coefficient $a$ is the leading coefficient of the $x^2$ term, $b$ is the coefficient of the $x$ term, and $c$ is the constant term. For example, in the equation $3x^2+2x-6=0$, $3$ is $a$, $2$ is $b$, and $-6$ is $c$.

How do you solve a quadratic equation by factoring?

To solve a quadratic equation by factoring, follow these steps: 1) Write the equation in standard form $a^{2}x+b^{1}x+c=0$. 2) Factor the quadratic expression into two binomials. For example, $x^2-9x+20=0$ factors to $x-4$ and $x-5$. 3) Set each factor equal to zero: $x-4=0$ and $x-5=0$. 4) Solve for $x$ to get the solutions: $x=4$ and $x=5$.

Why do quadratic equations often have two solutions?

Quadratic equations often have two solutions because they are polynomials of degree 2, meaning their graphs are parabolas. A parabola can intersect the x-axis at two points, corresponding to the two solutions of the equation. When you factor a quadratic equation, you set each binomial factor to zero, leading to two potential values for $x$. For example, in the equation $x^2-9x+20=0$, factoring gives $x-4$ and $x-5$, leading to solutions $x=4$ and $x=5$.