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Course: Staging content lifeboat > Unit 17
Lesson 2: Quadratic odds and ends- Solving a quadratic by factoring
- CA Algebra I: Factoring quadratics
- Algebra II: Quadratics and shifts
- Examples: Graphing and interpreting quadratics
- CA Algebra I: Completing the square
- Introduction to the quadratic equation
- Quadratic equation part 2
- Quadratic formula (proof)
- CA Algebra I: Quadratic equation
- CA Algebra I: Quadratic roots
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Solving a quadratic by factoring
factoring quadratics. Created by Sal Khan.
Want to join the conversation?
- Just checking but in the 1st example (0:26) the graph that Sal draws should be to the left of the y axis? Because the x co-ordinates are both negative right (@3:00)?(5 votes)
- that is correct but he was just showing that it should be a parabola. He wasnt showing how it should really look like.(8 votes)
- How does he get the negative 5(3 votes)
- (x-6)(x+5) = 0, which means that each term in () can be equal to 0 and meet the equality. Therefor, x-6 = 0, and x+5 = 0 are both solutions. Solving for each results in x being equal to both 6 and -5 (after you change signs across the equality)(3 votes)
- what does f(x) stand for? is it the same thing as y= x2 + x + z?(3 votes)
- f(x) is a different way to look at an equation. Your equation of y = x^2 + x + z could be written as f(x, z) = x^2 + x + z. What this is telling you is that you have an equation (also known as a function) that takes as input a x and a z. For example, f(2, 4) would be equal to (2)^2 + (2) + (4) in this instance. However, f(x) can refer to any equation that we could want. f(x) may equal x^2 or x^5 + 3^4 - 84^2 - 1.(2 votes)
- How does adding exponents affect the graph? X^2 creates a parabola, right? So what happens when you have, say, x^3 or x^6? Is there a way to accurately predict how the graph will appear with each exponent?(3 votes)
- Ryujin,
If the exponent is odd, the y values will be negative, when x is negative, so the graph will curve down on the left and up on the right.
If the exponent is even, he graph will curve up on both the left and right.
And the larger the even or odd number the steeper the graph gets.
Here is a graph program that you can change the value in the first line and see how the graph changes with different exponents:
https://www.khanacademy.org/cs/yxabxc/5710211763404800
Change the values in the first three lines of the program and you can see equation your create and the affect on the shape of the graph.
I hope that is of help to you.
P.S. You can also change the exponent in the first line of the program to decimals both above and below 1 and even to negative numbers. If you do the graph on the left side disappears because the domain must be x ≥ 0 for negative or fractional exponents, but it is interesting to see what happens to the left side of the graph with negative or non-integer exponents.(3 votes)
- what is f(x)? it is making me extremely confused :((2 votes)
- Function for the value of x. So if you see f(x) = x^2 +4 and then they ask for f(3) - you just plug the 3 into anywhere the x was and solve.(3 votes)
- what if the problem is not factorable such as -x2+-5x+15 can you simplify any other way?(2 votes)
- thanks petrie! this helped for sure- especially since only a few days after your response my alg. 2 teacher taught us the quadratic formula, so i was ahead of the game (:(1 vote)
- So, does f(x) mean where the parabola croses the x axis?(2 votes)
- f(x), said as "f of x" or 'function of x" is used to represent all the stuff to the right of the equals sign. In the video example, Sal wrote f(x) = x² + 6x + 8. In that case f(x) is x² + 6x + 8. And what is x² + 6x + 8? it is an expression that will produce a result, or output, given an input. So if I write f(2), then I am saying that the input is 2, so now where ever there is an x on the right hand side, replace it with the 2, which if we do that, we get: 2² + 6(2) + 8, which equals 4 + 12 + 8 = 24.
The way to think about it is that f(x)= x² + 6x + 8 is a function that produces an output for a given input. If we input 2, 24 is output, if we input 1, then f(1)=1² + 6(1) + 8 = 1 + 6 + 8 = 15, so the output is 15 if 1 is the input. For another example, lets use zero: f(0) = 0² + 6(0) + 8 = 0 + 0 + 8 = 8.
For more information and background, visit: https://www.khanacademy.org/math/algebra/two-var-linear-equations-and-intro-to-functions/functions-and-function-notation/v/what-is-a-function
Where the parabola crosses the x axis is found by factoring the function, and then determining what input x values will produce an output of 0, that is, we are looking for some x value(s) say, b such that f(b) = 0. We can factor x² + 6x + 8 into (x+2)(x+4), which means that:
f(x) = x² + 6x + 8 = (x+2)(x+4). Now it is easier to see that when x=-2 then f(-2) = (-2+2)(-2+4) = (0)(2) = 0 and that f(-4) = (-4 +2)(-4 + 4) = (-2)(0) = 0. So the x values where the parabola crosses the x axis are -2 and -4. Let's check this with the original definition of the function f(x) = x² + 6x + 8.
f(-2) = (-2)² + (6)(-2) + 8 = 4 - 12 + 8 = 0 and f(-4) = (-4)² + (6)(-4) + 8 = 16 - 24 + 8 = 0.
Do you see how factoring the function made finding the zeros of the function (where the parabola crosses the x axis) much easier?(3 votes)
- How to solve x^2 +2√5 x + 3(3 votes)
- What do you do if there isn't a number for position C?
Example: x^2+ 9x =0(1 vote)- If there isn't a "c" value shown, then c is simply equal to 0! If you think about it, if you add zero, it doesn't change anything, so you might as well not put it. Long story short, c is simply equal to 0.
Also, b might sometimes be equal to 0. What if the equation was x^2-9=0? Then b is zero since multiplying by zero gets 0(3 votes)
- 4:42I think he made a mistake here. since the numbers are adding we should divide each by 2. any problems here?(2 votes)
Video transcript
Welcome to solving a
quadratic by factoring. Let's start doing
some problems. So, let's say I had a function
f of x is equal to x squared plus 6x plus 8. Now if I were to graph f of
x, the graph is going to look something like this. I don't know exactly what it's
going to look like, but it's going to be a parabola and it's
going to intersect the x-axis at a couple of points,
here and here. And what we're going to try
to do is determine what those two points are. So first of all, when a
function intersects the x-axis, that means f
of x is equal to zero. Because this is f of x-axis,
similar to the y-axis. So here f of x is 0. So in order to solve this
equation we set f of x to 0, and we get x squared plus
6x plus 8 is equal to 0. Now this might look like you
could solve it pretty easily, but that x squared term messes
things up and you could try it out for yourself. So we're going to
do is factor this. And we're going to say that x
squared plus 6x plus 8, but this can be written as x plus
something times x plus something. It will still equal that,
except that's equal to 0. Now in this presentation, I'm
going to just show you the systematic or you could say the
mechanical way of doing this. I'm going to give you another
presentation on why this works. And you might want to just
multiply out the answers we get in and multiply out
the expressions and see why it works. And the message we're going
to use is, we look at the coefficient on this x term, 6. And we say what two
numbers will add up to 6. And when those same two
numbers are multiplied you're going to get 8. Well let's just think
about the factors of 8. The factors of 8 are
one to 4 and 8. well 1 times 8 is 8, but 1 plus
8 is 9, so that doesn't work. 2 times 4 is 8, and
2 plus 4 is 6. So that works. So we could just say x plus 2
and x plus 4 is equal to 0. Now if two expressions or two
numbers times each other equals 0, that means that one of those
two numbers or both of them must equal 0. So now we can say that x
plus 2 equals 0, and x plus 4 is equal to zero. Well, this is a very
simple equation. We subtract 2 from both sides
and we get x equals negative 2. And here we get x
equals minus 4. And if we substitute
either of these into the original equation, we'll
see that it works. Minus 2-- so let's just try it
with minus 2 and I'll leave minus 4 up to you --so minus 2
squared plus 6 times minus 2 plus 8. Minus 2 squared is 4, minus
12-- 6 times minus 2 --plus 8. And sure enough that equals 0. And if you did the same thing
with negative 4, you'd also see that works. And you might be saying,
wow, this is interesting. This is an equation that
has two solutions. Well, if you think about it, it
makes sense because the graph of f of x is intersecting the
x-axis in two different places. Let's do another problem. Let's say I had f of x
is equal to 2 x squared plus 20x plus 50. So if we want to figure out
where it intersects the x-axis, we just set f of x equal to 0,
and I'll just swap the left and right sides of the equation. And I get 2x squared plus
20x plus 50 equals 0. Now, what's a little different
this time from last time, is here the coefficient on that x
squared is actually a 2 instead of a 1, and I like
it to be a 1. So let's divide the whole
equation, both the left and right sides, by 2. I get x squared plus
10x plus 25 equals 0. So all I did is I multiplied
1/2 times-- this is the same thing as dividing
by 2 --times 1/2. And of course 0 times 1/2 is 0. Now we are ready to do what
we did before, and you might want to pause it
and try it yourself. We're going to say x plus
something times x plus something is equal to 0 and
those two somethings, they should add up to 10, and
when you multiply them, they should be 25. Let's think about
the factors of 25. You have 1, 5, and 25. Well 1 times 25 is 25. 1 plus 25 is 26, not 10. 5 times 5 is 25, and 5 plus 5
is 10, so 5 actually works. It actually turns out that
both of these numbers are 5. So you get x plus 5 equals
0 or x plus 5 equals 0. So you just have to
really write it once. So you get x equals negative 5. So how do you think
about this graphically? I just told you that these
equations can intersect the x-axis in two places, but
this only has one solution. Well, this solution
would look like. If this is x equals negative 5,
we'd have a parabola that just touches right there, and
then comes back up. And instead of intersecting
in two places it only intersects right there
at x equals negative 5. And now as an exercise just to
prove to you that I'm not teaching you incorrectly, let's
multiply x plus 5 times x plus 5 just to show you that it
equals what it should equal. So we just say that this is the
same thing is x times x plus 5 plus 5 times x plus 5. x squared plus 5x
plus 5x plus 25. And that's x squared
plus 10x plus 25. So, it equals what we
said it should equal. And I'm going to once again do
another module where I explain this a little bit more. Let's do one more problem. And this one I am just
going to cut to the chase. Let's just solve x squared
minus x minus 30 is equal to 0. Once again, two numbers when we
add them they equal-- whats the coefficient here,
it's negative 1. So we could say those two
numbers are a plus b equals minus 1 and a times b
will equal minus 30. Well let's just think about
what all the factors are of 30. 1, 2, 3, 5, 6, 10, 15, 30. Well, something interesting is
happening this time though. Since a times b is negative
30, one of these numbers have to be negative. They both can't be negative,
because if they're both negative then this would
be a positive 30. a times b is negative 30. So actually we're going to have
to say, two of these factors, the difference between them
should be negative 1. Well, if we look at all of
these, all these numbers obviously when you pair them
up, they multiply out to 30. But the only ones that have a
difference of 1 is 5 and 6. And since it's a negative 1,
it's going to be-- and I know I'm going very fast with this
and I'll do more example problems --this would be x
minus 6 times x plus 5 is equal to 0. So how did I think about that? Negative 6 times 5
is negative 30. Negative 6 plus 5
is negative 1. So it works out. And the more and more you do
these practices-- I know it seems a little confusing
right now --it'll make a lot more sense. So you get x equals 6 or
x equals negative 5. I think at this point you're
ready to try some solving quadratics by factoring and
I'll do a couple more modules as soon as you get some
more practice problems. Have fun.