معادلات درجه دوم

سرفصل: بخش ریاضی / سرفصل: جبر، معادلات و نابرابری ها / درس 13

معادلات درجه دوم

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Quadratic Equations

Quadratic equations. You may remember quadratics from our discussions of expression and factoring. A quadratic equation is one of the form a squared plus bx plus c equals zero. In other words, a quadratic expression set equal to zero. Most often these equations have two different solutions. So before we even begin to talk about how to solve these, let me say, that the strategy that we learn for linear equations, try to get all the x’s on one side of the equation, all the constants on the other side of the equation.

That is exactly the strategy you wanna follow for linear equations, but following that for quadratics is an unmitigated disaster. So do not follow the same strategy. When we’re dealing with quadratics is a completely different procedure of equation solving that we follow and that’s what we’re gonna talk about here. The vast majority of quadratic equations you will see on the test will have two solutions and the quadratic coefficient, the coefficient of x squared will equal 1 unless a numerical greatest common factor can be factored out from all three terms.

The vast majority of quadratic equations you will see on the test can be solved by the factoring methods we discussed in the factoring lessons. So if you haven’t seen those lessons, it will be very helpful to watch those before watching this video. Because we are gonna employ, all those factoring strategies here. We will factor a quadratic to a product of linear binomials.

And notice, we’ll be doing this so that this product equals 0. So very important to get things equal to 0 first. Then do this factoring. Because once we have a product that equals 0, we can apply this mathematical law, the Zero Product Property. The Zero Product Property says, if A times B equals 0, if a product of two things equals 0.

Then, either one is 0 or the other is 0. And before we go on, let’s think about this statement for a minute. Keep in mind that the OR that appears in that statement is the mathematical OR. Not the or of colloquial language. Well, what’s the distinction I’m drawing here? The mathematical OR is an inclusive or.

That is, it includes the and case. Thus, A equals 0 or B equals 0, includes three cases. It could be that A equals 0 and B doesn’t. It could be that B equals 0 and A doesn’t 0. Or it could be that they both equal 0 at the same time. You see sometimes in colloquial language, people use the word or to mean exclusive or.

They mean you could do this or that and the implication is that you can’t do both. That’s the exclusive or. That is not how the word or is used in mathematics. The word used in mathematics is the inclusive or which includes the and case. And this is important here and it will become important again in the probability lessons, okay.

Now, we’re ready to solve. So this first thing we’re gonna do is factor. And we can do that from our factoring lessons. Now we’re going to apply the zero product property. So if we’re multiplying two things to get 0, it means either 1 equals 0 or the other equals 0.

Now we have too many equations, we solve them in parallel. And we get that x equals negative 12 or x equals 2, and those are the two solutions to that original equation. Here’s another example. Now before we can apply the zero product property, we have to have something that equals zero.

So, our first step is gonna be to get the quadratic equal to zero, and we’re gonna do this by subtracting 11 from both sides. So, we get x squared minus 3x minus 54 equals zero. So, 54 that’s a tricky number, but notice 6 times 9 is 54 so, if we have a positive 6, and a negative 9 those will add to negative 3. Those are the numbers we need.

Then we apply the zero product property. Set each one of those equal to zero and solve. Again notice that I’m being very careful to write the word OR at each step. Again the word OR is not decoration. The word OR is a valid piece of mathematical equipment here. Suppose we have to solve this.

Hm. This looks a little bit trickier. Well, we’ll start by subtracting 13 from both sides. And now notice we have a greatest common factor of 5. So, we can factor that out. In fact, what we could just do is divide both sides by 5. Because, of course, dividing 0 by 5 is just 0.

So we divide by the greatest common factor. Then we get down to something like this, this is something we can factor. Zero product property and solve for the individual roots and those are the answers. Suppose we have this, now we have a quadratic equaling a quadratic, looks very tricky, and of course the test loves to throw stuff like this at you.

But all you have to do is just get everything on one side. So we’re gonna subtract the x squared, subtract the 10x, and subtract the 16. And this will get us down to, x squared minus 6x plus 9 equals 0. And notice that’s a special pattern. That is a square of a difference. So we can write that immediately as x minus 3 times x minus 3, which means that our only possibility for getting 0 is if x minus 3 equals 0, or in other words, if x equals 3.

So this is an example of a quadratic that has one solution instead of two. Now this one hm,. That’s tricky. If we move the 5 to the other side of the equation, we see a problem. There’s no way that we can square something and get a negative number. We can square.

If we square a positive, we get a positive number, we square 0 we get 0, we square a negative we get a positive, but there’s nothing we can square to get a negative. This is impossible. This is a simple example of a quadratic that has no solutions. So, as I said at the beginning, most quadratics have two solutions, but there’s some that have one, and then there’s some that have none.

And this is a simple example of a quadratic with no solution. For the vast majority of quadratics on the test you will follow this procedure. First of all you’ll get everything on one side of the equation set equal to zero. Then you will divide off any greatest common factor. You will factor it into a product of linear binomials.

And then use the Zero Product Property to create two linear equations and solve each one of them separately. So this is the procedure. Here are some practice problems. Pause the video and solve these on your own and then we’ll talk about these. And these are the solutions.

That covers most of what you need to know about quadratics for the test. On very advanced portions of the Quadrat, of the Quantitative section you may need to know and use the Quadratic Formula. We’ll cover that in a lesson in the Powers and Roots module. So don’t worry about that now. We’ll get to that later.

To solve most quadratic equations on the test. First of all, you need to gather all terms on one side, set equal to zero. Then divide by any numerical greatest common factor. Then factor into a product of two linear binomials. And use the Zero product property to separate into two linear equations and solve.

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