Back To CourseMath 101: College Algebra
12 chapters | 94 lessons | 11 flashcard sets
As a member, you'll also get unlimited access to over 55,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed.Free 5-day trial
Luke has taught high school algebra and geometry, college calculus, and has a master's degree in education.
You just passed level one of Furious Fowls, the game that asks you to launch birds across the screen to get back at those pesky pigs that stole your eggs. You learned that we could figure out where our bird would hit the ground by solving the given quadratic equation where it equals 0. That required us to factor the equation and then use the zero product property to determine which two values would give us 0.
But now we're on to level two! It's going to get harder, but your goal will still be the same: use the equation given to you on the screen to make sure that your bird will hit its target on the first try. Alright, so here we go; let's go ahead and make a guess at where we'll need to launch our bird this time. I think this looks about right. Now, let's check our equation to see if this will actually work out or not.
The game tells us that our bird will fly in a parabola given by the equation y = -x^2 + 4x + 7. Since we're again shooting for a pig on the ground, we're curious where this equation equals 0, so let's go ahead and substitute that in. Okay, now we need to try to factor in order to find x. Let's divide out the negative, leaving us with this: 0 = -(x^2 - 4x - 7). Now look for a pair of numbers that have a product of -7 and a sum of -4. Hmmm, this isn't looking good. I don't think this trinomial can be factored. So, what now? In level one this worked every time, so we never had to do anything else! Looks like we're going to have to learn some new skills.
Situations just like this, where we're trying to solve a quadratic that is unfactorable, is what the quadratic formula was made for! It's a little long and messy, but if you can remember it, you'll be able to solve any quadratic equation out there!
You'll have to first remember the standard form of a quadratic equation, y = ax^2 + bx + c. This will tell you what your a, b and c values are, but then it simply becomes a matter of putting the numbers into the right spots. Let's see if we can successfully do this to pass the second level of Furious Fowls.
Going back to the original equation and identifying a, b and c should be our first step. A is in front of the x^2, where there's only a negative symbol. That means that a is just -1. B is the number in front of the xs, which makes it 4, and c is the constant on the end, 7.
Now, we can substitute these values into the quadratic formula and evaluate the resulting expression to find our answers! While this is theoretically pretty straightforward, there are quite a few spots that can easily trip you up. Because this is our first time, let's evaluate this one step by step together and go over some of the common mistakes. We're going to have to remember our order of operations and make sure we also keep track of all our negative signs.
For this problem, then, the first thing we'll have to take care of is the exponent we see. The exponent I see is a 2, which means a square, which means multiply that number by itself, so we do 4 * 4 to get 16. Next in the order of operations will come multiplication. Doing the multiplication on the inside of the square root (4 * -1 * 7) gives us -28, then doing the multiplication on the bottom of the fraction (2 * -1) is simply -2. We're now left with x = (-4 +/- sq. root 16 - -28) / -2, and we're going to focus on the inside of the square root in the top of the fraction.
The inside of the square root in the formula is b^2 - 4ac. This part is one of the hardest parts to evaluate and one of the easiest parts to make a mistake on. It also has a special name; it's called the discriminant. It has some important features that we'll talk about in other lessons, but for this lesson, you just need to know to be careful when evaluating it, specifically for the reason I'm about to mention. Notice that we ended up with a -28 after we did the multiplication, but I still have 16 - -28. Minus a negative is plus a positive, so 16 + 28 gives us 44.
Now that we have the square root of 44, we'd like to check to see if we can take the square root of 44 nicely. Is there a number that, multiplied by itself, gives us 44? Well, 7 * 7 is 49, which is pretty close, but not quite, and 6*6 is 36, so again, not quite. So, the square root of 44 is an irrational number; it's going to be a really long decimal. That means that what we have is actually a pretty good answer, but there are two mistakes that students tend to make without realizing that this is an okay answer. One of those mistakes is to divide the -4 by the -2 to get 2 +/- the square root of 44. This is not allowed because the + root 44 kind of gets in the way of you doing that division.
Another mistake is to divide the -4 and the 44 by the -2 to give us (-4/-2) +/- (square root of 44/-2). The problem with this is that dividing inside a square root like that isn't really allowed. That means that what we have here (x = -4 +/ - the square root of 44/-2) is actually the most mathematically correct answer. But since we're estimating, let's change it to a decimal by rounding the answers off. The square root of 44 is in between 6 and 7; it's actually about 6.633. That means we'll find our two answers by doing -4 + 6.633/-2 and also -4 - 6.633/-2. Doing that gives us our two estimated answers as -1.316 or 5.316. Our -1.316 answer is what would happen if the slingshot broke, so we're not too worried about that one. But over at 5.316 it looks like we're pretty close to our target but maybe a little short.
Maybe if we leveled out our shot a little bit our bird might fly a little further. Maybe like this? Okay, apparently now our equation is y = -3x^2 + 15x + 29, but again we want to know when the bird will hit the ground, or what the roots of the equation are. That means we substitute in y = 0 and begin thinking about which way we want to solve. We could try the factoring method here, but those numbers seem pretty messy, and there's also a good chance it won't work anyways, so let's go straight into the quadratic formula.
This means identifying a, b and c from our standard form quadratic. That makes a equal to -3, b equal to 15 and c equal to 29. Now it's just a matter of carefully substituting these values into the quadratic formula and then following the order of operations to come up with our two answers. Substituting the numbers in will look like this: x= -15 +/- (the square root of 15^2 - 4 (-3) (29)) /2 (-3).
One other common mistake I see students make is to mess up the b value in the front left of the fraction. In the formula, it's a -b, so in this case 15 turns into -15. But often times your b value itself will be negative. When you have a -b and you plug it into the formula, - -b will turn it positive again, so you've got to be careful with that b in front.
Anyway, moving on to evaluating, we can start with the exponent on the inside - the b^2 part of the discriminant - and 15^2 gives us 225. Next, we do 4 * -3 * 29 to get -348, and we do 2 * -3 to get -6 on the bottom. Again, changing a minus negative into plus a positive will make our discriminant equal to 573. The square root of 573 is about 23.937. Doing -15 plus this and dividing by -6 will give us -1.489, and doing -15 minus this divided by -6 will give us 6.489.
We can again focus on the positive answer even though the negative one is a valid mathematical solution, and it looks like this one's going to work! I'm feeling pretty confident; let's let our furious fowl fly... and hey-o! We got it! You've now passed level two of Furious Fowls and are pretty good at solving quadratics, but are you ready for level three?
Let's quickly review what we learned from level two. Not every quadratic equation is factorable, so we need another way to solve them. Enter the quadratic formula. Using a, b and c from the standard form quadratic, we substitute these values into the formula and then use the order of operations to end up with our two answers.
Once you finish this lesson you'll be able to:
To unlock this lesson you must be a Study.com Member.
Create your account
Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
To learn more, visit our Earning Credit Page
Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.
Back To CourseMath 101: College Algebra
12 chapters | 94 lessons | 11 flashcard sets