Algebra II: Quadratics and Shifts 32-35, Solving quadratics and graph shifts
Algebra II: Quadratics and Shifts
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- We're on problem 32.
- What are the solutions to the equation 1 plus 1 over x
- squared is equal to 3 over x?
- So at first. This looks like a pretty daunting equation.
- You have these x's in the denominator and x squared in
- the denominator.
- But I think we can simplify it if we can just get rid of
- these x squares in the denominator.
- The easiest way to do that is to multiply
- everything by x squared.
- So let's multiply both sides of this equation by x squared.
- Then we'll get x squared times 1 is x squared, x squared
- times 1 over x squared, that's just 1.
- Then x squared times 3 over x, that's 3x squared over x.
- x squared divided by x is just x, so that is equal to 3x.
- We can subtract 3x from both sides and you get x squared
- minus 3x plus 1 is equal to 0.
- This is a simple quadratic.
- It's not obvious that you can factor it.
- In fact, two numbers when you multiply them equal 1, and
- then when you add them equal minus 3.
- I'm guessing it might even be imaginary.
- It will probably not be imaginary, but it's just a
- strange number.
- So let's use the quadratic equation.
- When in doubt, use the quadratic equation.
- So minus B, this is B right?
- B is this 3 right there, negative 3.
- B is negative 3.
- So minus B is going to be plus 3, plus or minus the square
- root of B squared.
- Minus 3 squared is 9, minus 4 times A, which is 1, times C
- which is 1.
- So it's minus 4, all of that over 2A.
- A is 1 so it's just over 2.
- That is equal to 3/2 plus or minus the square
- root of 5 over 2.
- I just separated these out because I'm looking at the
- choices and it seems like they did that.
- So you could've said that's 3/2 plus square root of 5 over
- 2 or 3/2 minus the square root of 5 over 2.
- I just did that because it seems like that's
- how they write it.
- That is choice A.
- Next problem, 33.
- I think this one actually might be good to copy and
- paste the problem.
- Let me see if I can do this.
- OK, there are two numbers with the following properties.
- Let me write down the properties and let me copy and
- paste it for you.
- OK, I've copied it.
- Let me go here.
- Then I've pasted it for you.
- All right.
- So there's two numbers with the following properties.
- The second number is 3 more than the first number.
- So let's say S for second number and F for first number.
- So the second number is 3 more than the first number.
- So the second number is equal to the first number plus 3.
- That's from statement 1.
- The product of the two numbers is 9 more than the sum.
- So the product of the two numbers, that's S times F is 9
- more, 9 plus their sum, plus S plus F.
- So let's see, we have two equations and two unknowns.
- This is nonlinear because I'm
- multiplying these two variables.
- But I think we should be able to solve them
- one way or the other.
- So let's see, we have what S is equal to.
- So let's just substitute that back into this equation.
- So let's say that S is equal to F plus 3.
- So if we substitute for these S's, we get F plus 3 times F
- is equal to 9 plus F plus 3, right, instead of an S, F plus
- 3 and then plus F.
- Let's see if we can simplify this.
- F times F is F squared plus 3F is equal to 9 plus
- 3 is 12 plus 2F.
- Subtract 2F from both sides, you get F squared plus F is
- equal to 12.
- Subtract 12, you get F squared plus F minus 12 is equal to 0.
- This one looks factorable.
- I don't have to take out the quadratic equation.
- Let's see, this is F plus 4 times F minus 3, right?
- Because when you multiply those, you get negative 12.
- When you add those, you get plus 1, so that is equal to 0.
- So in order for this to be true, one or both of these
- have to be equal to 0.
- So if F plus 4 is equal to 0, that means F could be
- equal to minus 4.
- If F minus 3 is equal to 0, then that says
- that F could be 3.
- So F could be minus 4 or 3.
- Now, S is F plus 3, so if we're dealing with the minus 4
- scenario, if F is equal to minus 4, then what is S?
- Then S is going to be minus 4 plus 3, and S is going to be
- equal to minus 1.
- Then if F is equal to 3, than S is equal to 6.
- So let's see if we see either of these combinations.
- Minus 4, minus 1, that's choice B.
- All right, problem 34.
- Let me see, maybe I should copy and paste these word
- problems so we can see how we parse the problems. So I've
- copied it, then I'll go here, and then pasted it.
- Jenny is solving the equation x squared minus 8x equals 9 by
- completing the square.
- What number should be added to both sides of the equation to
- complete the square?
- So x squared minus 8x is equal to 9.
- I wrote it with space for a reason.
- When you're completing the square, you're trying to turn
- the left-hand side of this equation into some type of a
- perfect square.
- So if it's a perfect square, I have two numbers.
- It's the same number that when you add them together, you get
- minus 8, and when you square them, you should get something
- else, right?
- So what's half of minus 8?
- Half of minus 8 is minus 4.
- So minus 4 squared is 16.
- So if I add 16 to both sides, I'm all set.
- Why did that work?
- Well, now it's a perfect square.
- This is now x minus 4 squared is equal to 9 plus 16 is 25.
- They're not even asking us to solve it.
- They just want to know what we had to add to both sides.
- So it's 16, D.
- Remember the whole logic here, and I've done a few videos on
- completing the squares, is what number do I add here to
- make this a perfect square?
- You say, OK, I have a minus 8x, so I take half of this
- number, because the same number added to itself twice
- is going to become minus 8.
- I take half of that number, then I squared it.
- So half of minus 8 is minus 4.
- If you square it, you get the 16.
- So add 16 to both sides, you get this.
- You can actually solve for this.
- x minus 4 is plus or minus 5.
- You keep going.
- That's actually where the quadratic equation comes from.
- Anyway, next problem.
- 16 was choice number D.
- I'm going to copy and paste this entire problem here.
- Let's go here and paste it here.
- OK, which of the following most accurately describes the
- translation of the graph y is equal to x plus 3 squared
- minus 2 to the graph y equals x minus 2 squared plus 2?
- So the y translation tends to be pretty easy to figure out.
- Let me just draw some example graph.
- So if I had the graph x squared, the graph x squared
- looks something like this.
- Let's see if I can draw it.
- The graph x squared looks something like this, right?
- And it intersects.
- When x is equal to 0, we're at our minimum point.
- And any other value increases in both directions.
- The graph of x squared plus 2, you're shifting up.
- This is the graph of x squared plus 2.
- You would shift it up by 2.
- The graph of x squared minus 2, you would shift down by 2.
- This would be x squared plus 2.
- This would be x squared minus 2.
- So the shift in the y direction is very easy to see.
- So if we're going from something minus 2 to plus 2
- we're going to be shifting it up 4, right?
- So that's always the easy one to just
- eyeball and figure out.
- So we're definitely going to be shifting from minus 2 to 2.
- So it's up 4.
- It's either going to be choice A or choice D.
- The left/right shift is often a little bit more hard for
- people to visualize or to at least internalize, but I'll
- give you an attempt.
- Let's just go back to this.
- This is the graph of x squared, this yellow line
- right there.
- That's the graph of x squared.
- Let me ask you a question.
- What is the graph of x minus 3 squared?
- So does this shift it down 3 to the negative direction or 3
- to the positive direction?
- Your intuition might say, oh, I'm subtracting 3.
- When I did minus 2, I shifted down.
- But it's actually the opposite here.
- Because you have to think about for what value of x am I
- going to have a 0 squared here?
- That happens with x is equal to 3.
- So you can think of it this way.
- Now, when we're at this point, when x is equal to 3, it's the
- same thing as this point when we have just x squared.
- When you put 3 in here, this whole expression becomes zero.
- As you get above 3, that's like going above zero.
- As you go below 3, that's like going below zero.
- So this graph will just get shifted to the right by 3.
- That's x minus 3 shifts to the right by 3.
- x plus three would go in the other direction, because when
- x is minus 3, that's when it would equal to zero.
- I haven't written that down.
- So let's think about this.
- We're going from x plus 3, so if this is x squared, x plus 3
- would look something-- let me do it in a different color. x
- plus 3 is actually shifted to the left.
- The way I always think about it, there's two ways
- to think about it.
- The y shift is intuitive and the x shift might not be.
- If you have a plus 3 here, you're actually shifting in
- the downward direction.
- The way to actually think about the intuition is when
- will this whole expression equal 0?
- This whole expression equals 0 when x is equal to minus 3.
- So that's the point at which you're getting 0 squared.
- When I'm drawing these graphs, I'm not
- doing the y shift here.
- So this is going to be shifted to the left 3.
- This is going to be shifted to the right by 2.
- So if this is shifted to the left 3 and this is shifted to
- the right by 2, to go from this to this, you're shifting
- to the right by 5.
- So the actual graph x plus 3 squared minus 2
- is going to be here.
- Then to go here, you have a plus 2, so you're shifting the
- graph up by 4, and then you're going to x minus 2.
- So this graph right here is going to be up here.
- So you're shifting up by 4 and then you're shifting to the
- right by 5.
- Actually, even if you're confused with your shifting
- left or right, you can just say the difference between
- plus 3 and minus 2 is 5, and 5 is only there.
- But you should hopefully understand the problems a
- little bit deeper than that.
- Anyway, I'll see you in the next video.
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