Below are the steps we discussed in class, in case some of you had trouble copying them down:
1. Rearrange the equation in a way that it becomes an argument of finding missing length on a diagram that has a square and a rectangle joined together to form a bigger rectangle whose total area is a given number.
2. After drawing the diagram next to the equation, cut the rectangular portion of the diagram in half and relocate it to the bottom of the square portion.
3. Add the missing piece to the diagram such that a square is completed and add the area of that missing piece to both sides of the algebraic equation.
4. Now you can just factor the left side of your equation (the one with variables in it) and simplify the right side of your equation (the numerical side) and continue to solve for x as you saw in class.
I will post answers to the problems on the weekend so that you can try and check your work before Monday. Please don't take too many short cuts and always draw the diagram or you will needlessly suffer with inaccurate results. You should use the blog to post questions and share your thoughts with each other.
Well sir, in part (c) i'm getting a little confused. It says x square - 9x. What my question is that when we have an addition sign in between we know that x square+ something is the area but here its x square minus something = area. Can we just simply work it out with a negative sign before the 9 times x ?
ReplyDeleteExcellent question Hira! It does seem strange when we have a negative sign there, because we then have to consider making a negative length. So, your confusion is actually right on!
ReplyDeleteNow here is the explanation:
You see, the ancients would never deal with a problem like part (c) and it was in the Islamic period, around the 9th century C.E. that a mathematician started dealing with such cases. Now you don't have to worry about how he did it because it turns out that if you stick to the diagram I showed you and just label one of the lengths as -9 then you can still solve this equation algebraically.
You just have to be okay with labeling that length as -9 and continue precisely as you have learned in class (which I think is what you were thinking of doing).
Hope this helps.
Indeed , that is what confused me but my question is answered now, thank you. :)
ReplyDeletei want to ask two things.
ReplyDeletecan we leave the statement as e.g (something + something)2 = something
and i am having a problem in part 4 with the diagram as i dont know how to divide the rectangle as the diagram is not square + rectangle as 3x2 cant be a square as one side would have to be 3x and the other x. so what do i do
Okay, the answer to Falah's first question is as follows:
ReplyDeleteWhy would you just leave it as (x+something)squared=something? The whole point was to solve for x and so you should go ahead and square root both sides and get the values for x that solve the equation. However, what you don't need to worry about at this stage is getting a decimal approximation by simplifying the answers that you get.
Now my response to your second question:
The trouble you are having with the last question (or part (d)) is very valid! Al-khwarismi had dealt with problems such as these using different geometric models than the one I have taught you. But he had to because there was not a wonderful symbolic algebra system at that time. So, what you might consider doing is somehow getting rid of that coefficient of x squared, i.e. the 3. See if you can factor out just the 3 and then divide both sides by it...
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ReplyDeleteDid anyone get x=(underoot15 -2)for ques 2
ReplyDeleteAnd sir i want to add something in hira's question,ur saying that we should take the side length as -9 but as ur notes say that we have to divide -9 into two equal parts so if we do that we wont get -9 in any way cuz (-9/2*-9/2=18/2=9) so i was wondering can we take one side length as -9/2 and the other as 9/2?
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ReplyDeletenevermind my last comment, hira answered it :)
ReplyDelete