Hey everyone, it's Justin again. In the last video, you learned that solving an equation
involving quadratics is often as simple as moving everything to one side and factoring. This video,
you'll see that this isn't always the case. There are some pretty non-standard ways that
quadratics can be expressed. By the end of this lesson, you will be able to solve
quadratic equations written in non-standard forms by rewriting them in standard form.
First, we'll review standard form and look at what it means to be in non-standard form.
Then, we'll work through two examples from beginning to end, one of which will involve a
shortcut that you can use in some cases. And we'll end with an example for you to try on your own.
In the last video, you learned that this could be solved by factoring the left
side of the equation. You also learned that sometimes equations have polynomials
on both side. In this case, we zero out all the terms on one side, then factor it.
Sometimes, equation doesn't even seem to involve polynomials at all. Cases like these
are what this video is all about. Don't worry, you've already learned everything you need.
For our first example, we're going to use that equation that we just saw.
The key for solving complicated equations like this, simplify.
We know that if we square this linear binomial, we'll get a second-degree polynomial;
and then we can combine these terms with those to make a simplified quadratic expression.
Let's do it. First, we need to square 2x plus 5. Remember the pattern for squaring binomials?
The first term squared, plus 2 times both terms, plus the second term squared.
So that's the first term squared, plus 2 times both terms,
plus the second term squared. When we multiply these together we get 4x squared plus 20x
plus 25. That was just the 2x plus 5 squared part. We also need to remember the rest of it.
We could see a couple pairs of like terms here, 20x plus x is 21x, and 25 minus 20 is 5.
Then we bring down the rest of it, and we're done simplifying.
Now that it's simplified, it will be much easier to solve by factoring.
There's a leading coefficient but no GCF so we have to use the AC method.
4 times 5 makes 20; So, we need a pair of numbers whose product is 20 but whose
sum is 21. Those numbers are 20 and 1. So we split the middle term using those two numbers.
Now we group the terms into pairs and pull out a GCF. First pair have a GCF of 4x, and when we
divide that out it leaves x plus 5. The second group doesn't have a GCF, except the number that
everything is divisible by, 1. Both of these remainders match, we know this will work.
One of the factors is x plus 5 and the other factor is the GCF's we just pulled out, 4x and 1.
Now, we just have to find when each of these factors is equal to 0.
For the first one, subtract 5 and we get x equals -5 as the solution.
Other one, subtract 1 and then divide by 4. So -1/4 is the other solution. We found two solutions
to that weird, complicated, equation we started with; but most of the steps were the same thing
we did in the last video. The only difference here that we had to simplify the equation first.
Sometimes, instead of having to go through all the steps to simplify, there's a little shortcut. See
how both of these pieces match? We can replace them. Something a bit easier to handle. We just
need to remember that n is a replacement, not what was originally in the equation.
Now we've got a function that can be factored using the AC method,
2 times -15 is -30. So we need two numbers with a product of -30, the sum of -7. How about
-10 and 3. Now that we've broken up the middle term, we can group the terms and pull out GCF's.
The first pair has a GCF of 2n with a remainder of n minus 5. And the second pair has a GCF of 3 with
the same remainder. So then our factors are the matching remainders, n minus 5,
and the GCF's 2n plus 3. Now we just need to set each factor equal to 0 and solve. For this one
we just add 5 and get that n equals 5; And for the other, we subtract 3 from both sides, divide by 2,
get -3/2. It seems like we're done except, we weren't trying to solve for n.
The solution to the original equation has to tell us what x is. Luckily, we made a note of what
n means, and is the same as 2x minus 1. So we can re-replace it then do a little more solving.
If n is 5, then we can add 1 to both sides, and divide by 2, and x ends up being 3. The other one,
we'll have to use some fraction operations. If we add 1 to -3/2,
that's the same as adding 2 halves to it, so we get -1/2, then we divide by 2. In other words,
what's half of a half. We get -1/4. So the solutions for x this time are 3 and-1/4.
Remember this was only a shortcut. The process was shorter but also a little more complicated. If you
had trouble following along with the shortcut feel free to just use the simplifying method instead.
Do you think you could solve this one? Pause the video now, see if you can find the solutions.
First, since we don't have any matching pieces, we can't use the shortcut here,
so we have to simplify. After we expanded the square and we distribute, you should end up
with this. If you didn't get this, check your work and see where your mistake was.
Next, we zero out the right side, and combine like terms on the left, to get 9x squared plus
21x plus 10 equals 0. If you got the previous step right but something went wrong here,
check your work. Now we need to factor. Using the AC method, we know we need numbers whose product
is 90 and whose sum is 21; which leads us to split 21 into 6 and 15. Now we can group and factor and
we get GCF's of 3x and 5; both with remainders of 3x plus 2. Which means the factored form
is 3 x plus 5 times 3 x plus 2. We split both of these factors off and solve, and we get solutions
of -5 over 3 and -2 over 3. If at any point you didn't get the same thing I did, make sure you
look through your work to see where your mistake was. Finding and fixing your mistakes helps you
avoid making them in the future, which is super important when a lot of steps are involved.
Solving quadratic functions can take a lot of work
and careful calculations. Luckily, there's a quicker way to find the solutions,
without all these steps. In the next video, you're going to learn about an
incredibly helpful and short shortcut called the quadratic formula. See you then!