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From users who are members of Transition to university :
Christian LawsonPerfect  said  Ready to use  3 years, 8 months ago 
Hannah Aldous  said  Needs to be tested  3 years, 9 months ago 
Elliott Fletcher  said  Has some problems  3 years, 9 months ago 
Vicky Hall  said  Has some problems  3 years, 9 months ago 
History
Christian LawsonPerfect 3 years, 8 months ago
Saved a checkpoint:
Whoops  the prompt had the constant term as $c \times d$ instead of $b \times d$. Thanks for pointing this out!
Christian LawsonPerfect 3 years, 8 months ago
Gave some feedback: Ready to use
Christian LawsonPerfect 3 years, 8 months ago
Gave some feedback: Has some problems
Katherine Tomlinson 3 years, 8 months ago
Has some problems. I was given the equation 24x^2  37x  8 and it said that factorised this was (3x+5)(8x1).
Christian LawsonPerfect 3 years, 8 months ago
Gave some feedback: Ready to use
Christian LawsonPerfect 3 years, 8 months ago
Saved a checkpoint:
Removed a lot of rubbish which must have come from a previous question, and renamed the variables so they make sense.
It's important that the coefficients of $x$ in the two factors aren't the same  otherwise, the order of the factorisation would be ambiguous.
Elliott Fletcher 3 years, 9 months ago
Published this.Hannah Aldous 3 years, 9 months ago
Gave some feedback: Needs to be tested
Christian LawsonPerfect 3 years, 9 months ago
Gave some feedback: Has some problems
Christian LawsonPerfect 3 years, 9 months ago
Saved a checkpoint:
The statement says you can factorise as $(ax+m)(bx+n)$ but then part a asks you to factorise in the form $a(x+m)(x+n)$. Should this be a separate question, since spotting a common factor of all the coefficients is a fairly simple corollary of factorising a quadratic with leading coefficient 1?
In fact, part b is the same! I was shown $8x^2+240x+1600=0$, which factorises as $8(x+10)(x+20) = 0$. The expected answer had roots $5/4$ and $5/2$, which is equivalent to $8x^2+30x+25=0$. So is it the displayed equation that's wrong?
The marking for part a is wrong  I think the wrong variable is used somewhere, but it's not obvious where.
Part b could begin by asking you to find the factorisation.
In part c, I got $2x^2+13x+20$, so a particularly pedantic student might want to leave the second gapfill empty, since the coefficient of $x$ in the second part is $1$. Can you make sure that the coefficient of $x^2$ in the equation is greater than $2$?
I got $5/3$ as one of my roots in part d, so I've enabled "allow the student to enter a fraction".
Parts c and d could use the same equation  why not? That would make a nice, selfcontained question.
So, in summary:
 Fix the marking in parts a and b.
 If they use the same equation rather than a new one in each part, I think parts (a,b) make a selfcontained question, and (c,d) another.
Hannah Aldous 3 years, 9 months ago
Gave some feedback: Needs to be tested
Elliott Fletcher 3 years, 9 months ago
Gave some feedback: Has some problems
Elliott Fletcher 3 years, 9 months ago
I think the questions here are good and will be reasonably challenging for the students, there are just a few errors that i noticed.
Main Parts
a) the correct numbers that you put into the answer box are marked correct, but the expected answer comes up as somethinwe g else which would be incorrect. For example, i was given the equation 2x^2+12x+10 =0, which factorises to 2(x+5)(x+1) = 0, which is marked correct, however the expected answer comes up as 2(x+4)(x+3) = 0.
b) Here the correct answer is marked incorrect,
i had 8x^2+112x+320=0 which factorises to 8(x+10)(x+4) = 0. Thus x1 = 10 and x2=4. However this is marked wrong and the expected answer is x1=5 and x2= 1, which wouldn't be right.
c) good
d) i would write "in order to calculate the possible values of x that satisfy the equation" instead of "in order to calculate the possible values of x".
Again, the correct answer is marked incorrect. The expected answer shows different values which would be incorrect.
Advice
a)
I don't think you need a comma after "with" on the first line here.
I think there should be a 2 in front of the brackets in the final answer, as you have done in the question itself.
b)
I don't think you need a comma after "with" on the first line here.
I think you need a full stop after the factorised equation on the 7th line.
c)
The equation here shows the wrong coefficient of x in the second factor, for example for
6x^2+5x+1 = 0, you have
"This means our factorised equation must take the form (3x+a)(3x+b) = 0" whereas the equation should be "(3x+a)(2x+b) =0".
On that note i think it should be 3b+2a = 5 instead of 3a+2b=5.
I would put the multiplications in brackets here when you write when you multiply the values of a and b by 3 and 2.
d)
I think you should just write "we need to find two values that add together to make 5 and multiply to make 6" as one line instead of two.
Typo: "mutipy" should be "multiply"
I think this part needs more explanation for how you obtain each of the numbers for the factorised expression, something like you have done in part c)
Although the way you find the final answer does make sense and does work, i don't really see why you need to find the previously mentioned values like 3 and 2 that add and multiply to make a certain number to do this. I think you'd be better answering this question in the same way as you answered part c.
Sorry for all the feedback!
Hannah Aldous 3 years, 9 months ago
Gave some feedback: Needs to be tested
Vicky Hall 3 years, 9 months ago
Gave some feedback: Has some problems
Vicky Hall 3 years, 9 months ago
The statement is exactly the same as the other quadratic equations questions, but this time its not true that $ax^2+bx+c$ factorises to $(x+m)(x+n)$  the $x$s need coefficients.
I think there should be two more parts to this question, a question at the beginning that only wants to students to factorise, and a question at the end that doesn't give them one of the $x$ coefficients.
It would also be very helpful to tell students in the statement that we can sometimes divide through by the $x^2$ coefficient to obtain a simpler equation, but sometimes the coefficent is not a factor of all terms so we can't. (I know you show this in the advice but it would be nice for the student to see this before they try the question as otherwise they will start looking for factors of the existing numbers).
Hannah Aldous 3 years, 9 months ago
Gave some feedback: Needs to be tested
Hannah Aldous 3 years, 10 months ago
Gave some feedback: Has some problems
Hannah Aldous 3 years, 10 months ago
Gave some feedback: Needs to be tested
Hannah Aldous 3 years, 10 months ago
Gave some feedback: Has some problems
Chris Graham 3 years, 10 months ago
In part a), i),ii) should be in italics, and I would start a new line afterwards.
You need some string restrictions in part (a), for example I can enter the expression into the gap as given, and get full marks. See the bottom of the Numbas tutorial.
In (b) and (c) "values of x in the following equation" would be better expressed as "values of x which satisfy the following equation".
In the advice, put a),b)... on a new line, and preferably set the style using format>Formats>Headings>Heading 4
The advice is not easy to scan, and unfortunately students will not take care to read the whole thing. Help them out by placing any important equations on a new line, so that the student can easily see how the solution develops. And as a general rule, use display style for any equations on their own line.
In part (d) of the advice you obtain the possible values of x, however this is not asked in the question. Did you intend to include this? If you do so you will need to think about the precision that you would like, and formatting in the advice (see e.g. dpformat on the jme reference page).
Hannah Aldous 3 years, 10 months ago
Gave some feedback: Needs to be tested
Hannah Aldous 3 years, 10 months ago
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