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# Algebra II: Quadratics and shifts

## Video transcript

we're on problem 32 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 Squared's in 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 and then we'll get CX squared times 1 is x squared x squared times 1 over x squared well that's just 1 and then x squared times 3 over X that's 3x squared over X x squared or divided by X is just X so that is equal to 3x we could subtract 3x from both sides and you get x squared minus 3x plus 1 is equal to 0 this is a simple quadratic and it's not obvious that you can factor it in fact yeah two numbers when you multiply them equal 1 and that when you when you add them equal minus 3 I'm guessing it might even be imaginary when I but not be imaginary but is just a strange number so let's use the quadratic equation when in doubt use a quadratic equation so minus B this is B right B is this 3 right there the night negative 3 right 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 wants us minus 4 all of that over 2a a is 1 so it's just over 2 C that is equal to 3 halves plus or minus is that the square root of 5 over 2 I just separated these out because look any of the choices and it seems like they did that so you could say that's three halves plus square root of 5 over 2 or 3 halves minus the square root of 5 over 2 and I just did that because it seems like that's how they write it and that is choice a next problem 33 33 I think this one actually might be good to copy and paste the problem let me see if I can do this okay there are two numbers with the following properties let me write down the properties and let me copy and paste it from for you okay I've copied it let me go here and then I've pasted it for you all right now so the second number so there's two numbers of the file purpose the second number is three more than the first number so let's say s for second number and F for first number so the second number the second number is three more than the first number so a second number is equal to the first number plus three that's from statement one and then the product of the two numbers is nine more than the sum so the product of the two numbers product of the two numbers that's s times F s times F is nine more nine plus their sum plus s plus F plus C we have two equations in two unknowns this is not linear because I'm multiplying these two variables but I think we should be able to 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 see if we can supply this F times F is f squared plus 3 F is equal to 9 plus 3 is 12 plus 2 F plus 2 F let's see subtract two 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 and this one looks factorable I don't have to take out the quadratic equation let's see is f c/f plus four times F minus three right because when you multiply those you get negative twelve and you add those you get plus one so that is equal to zero so in order for this to be true one of one or both of these have to be equal to zero so if F plus four is equal to zero f plus four is equal to zero that means that F is f could be equal to minus four if F minus three is equal to zero then that says that F could be three so through F could be minus four or three now s is f plus three so if we're dealing with the minus four scenario if F is equal to minus four then what is s then s is going to be minus four plus 3 and s is going to be equal to minus one and that if F is equal to 3 then s is equal to 6 so let's see if we see either of these combinations minus 4 minus 1 that's choice B excellent all right problem 34 problem 34 let me see maybe I should copy and paste these these word problems so we can see how we parse the problems so I've copied it now let me go here 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 and 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 right so if it's a perfect square I have two numbers and it's the same number that when you add them together 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 and if I add 6 so minus 4 squared is 16 so if I add 16 to both sides I'm all set and why did that work well now it's a perfect square this is now X minus 4 squared is equal to 9 plus 625 they're not even asking us to solve it they just want we had to add to both sides so it's 16 D and remember the whole logic here and I've done a few videos on completing the square is what number do I add here to make this a perfect square and you say okay I have a minus 8x so I take half of this number right because the same number add it to itself twice is going to become minus 8 I take half of that number and then I square it so half of minus 8 is minus 4 and you square it you get the 16 so I had 16 to both sides you get this and you could actually solve for this you know X minus 4 is plus or minus 5 and you keep going and that's actually where the quadratic equation comes from anyway next problem 16 was choice number D all right so they say actually I'm gonna copy and paste this entire problem here let's go here you paste it here ok which of the following most accurately describes a translation of the graph ok y is equal to X plus 3 squared minus 2 to the graph y equals X minus 2 squared plus 2 so this translations and all that that's often so the y translation it tends to be pretty pretty easy to to figure out because if I have some let me just draw some example graph so if I had the graph x squared the graph x squared looks something like this see if I can draw it the graph x squared looks something like this right and there's sex when x is equal to 0 we're at our minimum point right and any other value increases in these both direction the graph of x squared plus 2 you're shifting up right this is the graph of x squared plus 2 you would shift it up by 2 and the graph of x squared minus 2 you would shift down by 2 right this would be x squared plus 2 and 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 if we're going from something minus 2 to plus 2 so we're going from minus 2 to plus 2 we're going to be shifting it up for 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 so 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 to at least internalize but let's we'll give you an attempt if this is the graph let's just go back to this this is the graph of x squared this yellow line right there right that's the graph of x squared now let me ask you a question what is the graph of x - I don't know X minus 3 squared so does this shift it down 3 2 3 2 the negative direction or 3 to the positive direction your intuition might say oh I'm subtracting 3 when I had when I did that you know 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 right and that happens when 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 right because when you put 3 in here this whole expression becomes 0 and as you get above 3 that's like going above 0 and as you go below 3 that's not going below 0 so this graph will just get shifted to the right by 3 right that's X minus 3 shifts to the right by 3 X plus 3 would go in the other direction because when X is minus 3 that's when it would equal is your ivan run that down so let's think about this we're going from x plus 3 right so if this is x squared x plus 3 is actually it would look something let me do it in a different color X plus 3 is actually shifted to the left and the way I always think about there's two ways to think about it but the wide shift isn't intuitive and the F shift might not be if you have a plus 3 here you're actually shifting in the downward direction and the way to actually think about that 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 and I actually didn't I'm when I'm drawing these graphs I'm not doing the y shift here right so this is going to be shifted to the left 3 this is going to be shifted to the right by 2 so this shifted to the left three and this is shifted to the right by two to go from this to this you're shifting to the right by 5 right so you know X plus 3 squared is here so these grant the actual graph X plus 3 squared minus 2 is your it's going to be here right and then your 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 completely confused whether you're shifting left or right you get to say okay the difference between plus 3 and minus 2 is 5 and fives only there but you should you should hopefully understand the problems a little bit deeper than that anyway I'll see you in the next video