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Algebra Homework Question
30 Answers
Year 9 son has an algebra test tomorrow and is happy with most apart from questions on magic numbers. I have never come across these?
Question is solve the following equations using magic numbers a+b and a times b
apologies I can't figure out the algebraic x on my keyboard so have substituted k instead
2ksquared + 10k +5
2ksquared + 10k + 10 =0
Many thanks if anyone can shed a light on these for us
Question is solve the following equations using magic numbers a+b and a times b
apologies I can't figure out the algebraic x on my keyboard so have substituted k instead
2ksquared + 10k +5
2ksquared + 10k + 10 =0
Many thanks if anyone can shed a light on these for us
Answers
Now to try and work out how in blazes this works out in terms of "magic numbers". This post is in two parts: part one is my trying to make sure I've understood what's going on, and part two (after the asterisks) is a more coherent summary (
23:34 Mon 19th Feb 2018
I don't think I've ever heard the term "magic number" in this context before. As neither of the two equations you've presented leads to what I'd call an "easy" answer, I'm not sure how helpful any "magic numbers" approach would be.
I did find the following video, and perhaps that would be helpful to explain the "magic numbers" method. But I'd need the notation clarified before I could offer any help myself.
I did find the following video, and perhaps that would be helpful to explain the "magic numbers" method. But I'd need the notation clarified before I could offer any help myself.
The Answerbank font does not lend itself easily to replication of mathematical formulae but this may be of some assistance (or not!).
https:/
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The method he may have used is that covered here in the bit that refers to Babylonian mathematicians
https:/
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Thank you all so far. My (and his elder sisters) response was to use the normal quadratic formula but he has been told on some questions he has to use magic numbers which are the formulas a+b and a times b.
The one from his notes show 2 k squared + 3 k -2 = 0
He has worked this out to be
a times b = -4
a+b = 3
a = -1 and b = 4 with the workings out being 2 ( (k-1) (k + 4) ) = 0
(k-1) ( k+4)=0
Which has been marked correct.
But in the example above on my first question a times b and a + b I can't work out if he does 2 time 10 and 2 add 5 or 2 times 5 and 2 add 10.
I'm not sure I'm making much sense here so if anyone is still following this and can help I will be very grateful
The one from his notes show 2 k squared + 3 k -2 = 0
He has worked this out to be
a times b = -4
a+b = 3
a = -1 and b = 4 with the workings out being 2 ( (k-1) (k + 4) ) = 0
(k-1) ( k+4)=0
Which has been marked correct.
But in the example above on my first question a times b and a + b I can't work out if he does 2 time 10 and 2 add 5 or 2 times 5 and 2 add 10.
I'm not sure I'm making much sense here so if anyone is still following this and can help I will be very grateful
Hope it went okay for him. Did you get chance to check whether these two questions and the example you gave were set out correctly?
I've never heard the term magic numbers before but basically when factorising quadratics in the form x²+ bx+ c = 0 you do find the pair of numbers that multiply together to make c and add together to make b
I've never heard the term magic numbers before but basically when factorising quadratics in the form x²+ bx+ c = 0 you do find the pair of numbers that multiply together to make c and add together to make b
Jim n FF
isnt this question about
sum of roots ( a + B ) is x coefficient
product of roots (ab) is the number ? ( constant)
the only problem is that if you try to solve ab=5 A+B=10 or whatever - you get back to your original equation
year 9 I thought was a nine year old
but he's 14 ( sorry left skool in 1968)
and this relates to the findamental theorem of algebra which I would have thought is OK for someone of that age
the theorem says a quadratic has two roots and a cubic has three....
isnt this question about
sum of roots ( a + B ) is x coefficient
product of roots (ab) is the number ? ( constant)
the only problem is that if you try to solve ab=5 A+B=10 or whatever - you get back to your original equation
year 9 I thought was a nine year old
but he's 14 ( sorry left skool in 1968)
and this relates to the findamental theorem of algebra which I would have thought is OK for someone of that age
the theorem says a quadratic has two roots and a cubic has three....
It seems like the video I posted was on the right lines. I have to say from the outset that I'm seriously unimpressed that it's called "a times b" and "a plus b" when they typically refer to different quantities altogether. The point of notation is to be clear what you mean, not to hide it behind unnamed symbols and undefined quantities.
I'm also slightly perplexed because, as FF notes, it's not clear that the "magic numbers" route will help on either of the problems you gave. The idea is to factorise the equation neatly into whole-number brackets but that doesn't always work and here it's not clear that it will.
Still, thanks for your example. I'll post some more comments in a separate post, I just wanted to rant at the apparently sloppy teaching here...
I'm also slightly perplexed because, as FF notes, it's not clear that the "magic numbers" route will help on either of the problems you gave. The idea is to factorise the equation neatly into whole-number brackets but that doesn't always work and here it's not clear that it will.
Still, thanks for your example. I'll post some more comments in a separate post, I just wanted to rant at the apparently sloppy teaching here...
So:
2 k squared + 3 k -2 = 0
First, again as FF notes, the example is wrong... (sigh, what even is the point of an example if it's done wrongly?)
Secondly, I honestly can't see the point in trying to reproduce an explanation that looks to me to lead to the wrong answer sometimes and doesn't really explain what's going on, so I'm going to provide my own explanation. Below, I use ^2 to mean "squared" and * to mean "times", and I've indicated each multiplication explicitly.
1. Let's suppose that you have already factorised the equation a*x^2+b*x+c. You'll get two brackets, each linear in x, of the form (p*x+q)*(r*x+s). Now what we do is expand the brackets again! We find that:
(p*x+q)*(r*x+s) = (px)*(rx)+(px)*(s)+(q)*(rx)+(q)*(s)
= (p*r)*x^2+(p*s)*x+(q*r)*x+(q*s)
= (p*r)*x^2+(p*s+q*r)*x+(q*s)
2. Now we already know, though, that:
(p*x+q)*(r*x+s)=a*x^2+b*x+c
There's an important theorem that says that two polynomials are the same if they have the same coefficients (which are the numbers a, b and c in a*x^2+b*x+c). So we conclude that:
p*r=a
q*r+p*s=b
q*s=c
3. If we are going to do anything with these equations, then we had best hope that the numbers p, q, r and s are whole numbers. (If not, then we revert to the mundane approach of the quadratic formula.) So now we start trialling whole numbers that divide a and c and compare them with b to see if we can get a match.
4. Now for an example, then!
2*k^2 + 3*k - 2 = 0
5. We see that (in my notation) a=2, b=3, c=-2 (don't forget the minus sign!). Therefore, in terms of the "solution" (p*k+q)*(r*k+s):
p*r=2
q*r+p*s=3
q*s=-2
6. Let's write out all the possible whole number factors of 2, which are (1)*(2), (-1)*(-2) only, and all the possible factors of -2, which are (-1)*(2) and (1)*(-2) only in this case.
7. Let's match these up to values p, q r and s and see if the combination p*s + q*r equals 3:
Guess: p=1, r=2, q=1,s=-2
p*s + q*r = (1)*(-2)+(1)*(2) = -2 + 2 = 0 FALSE
Guess: p=2, r=1, q=1,s=-2
p*s + q*r = (2)*(-2)+(1)*(1) = -4 + 1 = -3 FALSE
Guess: p=2, r=1, q=-1,s=2
p*s + q*r = (2)*(2)+(-1)*(1) = 4 - 1 = 3 TRUE
Yes, it works!
8. So we put these values of p, q r and s into the form (p*k+q)*(r*k+s) to give:
2*k^2 + 3*k - 2 = (2*k -1)*(1*k+2) = (2k-1)(k+2)
9. Finally, we are solving 2*k^2 + 3*k - 2 = 0, which, as we have shown, is the same thing as solving (2k-1)(k+2) = 0. In words this is the same as saying, basically, "something times something else is zero". But this is ONLY true if one of the two things you are multiplying together is itself zero!. Hence:
(2k-1)(k+2) = 0 is the same as saying 2k-1 = 0 OR k+2 = 0.
10. Rearrange, and you get k = 1/2 OR k = -2, as your two solutions. And we're done!
2 k squared + 3 k -2 = 0
First, again as FF notes, the example is wrong... (sigh, what even is the point of an example if it's done wrongly?)
Secondly, I honestly can't see the point in trying to reproduce an explanation that looks to me to lead to the wrong answer sometimes and doesn't really explain what's going on, so I'm going to provide my own explanation. Below, I use ^2 to mean "squared" and * to mean "times", and I've indicated each multiplication explicitly.
1. Let's suppose that you have already factorised the equation a*x^2+b*x+c. You'll get two brackets, each linear in x, of the form (p*x+q)*(r*x+s). Now what we do is expand the brackets again! We find that:
(p*x+q)*(r*x+s) = (px)*(rx)+(px)*(s)+(q)*(rx)+(q)*(s)
= (p*r)*x^2+(p*s)*x+(q*r)*x+(q*s)
= (p*r)*x^2+(p*s+q*r)*x+(q*s)
2. Now we already know, though, that:
(p*x+q)*(r*x+s)=a*x^2+b*x+c
There's an important theorem that says that two polynomials are the same if they have the same coefficients (which are the numbers a, b and c in a*x^2+b*x+c). So we conclude that:
p*r=a
q*r+p*s=b
q*s=c
3. If we are going to do anything with these equations, then we had best hope that the numbers p, q, r and s are whole numbers. (If not, then we revert to the mundane approach of the quadratic formula.) So now we start trialling whole numbers that divide a and c and compare them with b to see if we can get a match.
4. Now for an example, then!
2*k^2 + 3*k - 2 = 0
5. We see that (in my notation) a=2, b=3, c=-2 (don't forget the minus sign!). Therefore, in terms of the "solution" (p*k+q)*(r*k+s):
p*r=2
q*r+p*s=3
q*s=-2
6. Let's write out all the possible whole number factors of 2, which are (1)*(2), (-1)*(-2) only, and all the possible factors of -2, which are (-1)*(2) and (1)*(-2) only in this case.
7. Let's match these up to values p, q r and s and see if the combination p*s + q*r equals 3:
Guess: p=1, r=2, q=1,s=-2
p*s + q*r = (1)*(-2)+(1)*(2) = -2 + 2 = 0 FALSE
Guess: p=2, r=1, q=1,s=-2
p*s + q*r = (2)*(-2)+(1)*(1) = -4 + 1 = -3 FALSE
Guess: p=2, r=1, q=-1,s=2
p*s + q*r = (2)*(2)+(-1)*(1) = 4 - 1 = 3 TRUE
Yes, it works!
8. So we put these values of p, q r and s into the form (p*k+q)*(r*k+s) to give:
2*k^2 + 3*k - 2 = (2*k -1)*(1*k+2) = (2k-1)(k+2)
9. Finally, we are solving 2*k^2 + 3*k - 2 = 0, which, as we have shown, is the same thing as solving (2k-1)(k+2) = 0. In words this is the same as saying, basically, "something times something else is zero". But this is ONLY true if one of the two things you are multiplying together is itself zero!. Hence:
(2k-1)(k+2) = 0 is the same as saying 2k-1 = 0 OR k+2 = 0.
10. Rearrange, and you get k = 1/2 OR k = -2, as your two solutions. And we're done!
I'm sure that's right jim- it doesn't seem to fit with the solution that was marked correct though in the post at 18:52 yesterday.
I think that example is one that can be done by inspection- write it as x²+ 1.5x -1 =0
The two values with a product of -1 and a sum of -1.5 are 2 and -0.5
So (x+2)(x-0.5) = 0
Solutions are, as you say, x= -2 and 0.5. That seems ifferent to the answer marked a s correct in his son's book.
Anyway, unfortunately that simple method doesn't work for 2k² + 10k + 10 =0 (or the simpler version k²+5k+5=0) because you get surds in the solution which are difficult to work out in your head
I think that example is one that can be done by inspection- write it as x²+ 1.5x -1 =0
The two values with a product of -1 and a sum of -1.5 are 2 and -0.5
So (x+2)(x-0.5) = 0
Solutions are, as you say, x= -2 and 0.5. That seems ifferent to the answer marked a s correct in his son's book.
Anyway, unfortunately that simple method doesn't work for 2k² + 10k + 10 =0 (or the simpler version k²+5k+5=0) because you get surds in the solution which are difficult to work out in your head
Now to try and work out how in blazes this works out in terms of "magic numbers". This post is in two parts: part one is my trying to make sure I've understood what's going on, and part two (after the asterisks) is a more coherent summary (hopefully!).
1. As far as I can see the point is that the "magic numbers" in my language would be the two numbers A = q*r and B = p*s, so that, in terms of the "standard" equation I used:
(p*x+q)*(r*x+s)=a*x^2+b*x+c
A*B = q*r*p*s = (p*r)*(q*s) = a*c
and
A + B = q*r + p*s = b as before.
(I call them A and B not a and b because that way you don't get a stupid clash with the standard notation ax^2 + bx + c = 0 that's almost universal for quadratic equations).
2. Then, you would get the following, from your example 2 k^2 + 3k -2 = 0 :
A + B = 3 (middle coefficient)
A*B = (2)*(-2) = -4 (product of first and last coefficients).
3. Now guess solutions to these equations, by finding the possible factors of -4 and checking if they sum to give 3:
-4 = (4)*(-1): sum = 4 -1 = 3. TRUE.
That didn't take long, so the MAGIC NUMBERS are 4 and -1.
4. Now write "(x + A)(x+B)", which here would be:
(x + 4) (x - 1)
But do not set this equal to anything: as in, literally write it on the page.
5. Now, I think what you have to do is basically a little bit of fudging to make sure this is actually the right answer. Take the coefficient of x^2 and write in three separate places: in front of the two brackets, and under both A and B (as written on the page). Here, then, you'd write:
2k^2 + 3k -2 = 0
a=2, A = 4, B = -1:
2k^2 + 3k -2 = (a)(k + A/a)(k + B/a)
= (2)(k + (4)/(2))(k + (-1)/2 )
= 2 (k + 2)(k - (1/2) )
Then, as before, this equals zero when either k+2=0 or k-1/2 = 0, which is the same as saying k = -2 or k = 1/2.
Phew!
* * * * * * * * * * * *
Summary process for the "magic number" technique, as far as I can see, is this:
1. Take your quadratic equation and compare it with the form ax^2 + bx + c = 0.
2. Identify the values of a, b and c.
3. write A + B = b (for your value of b), and A*B = a*c (for your values of a and c).
4. Find all pairs of factors of your answer to a*c. These are "possible magic number pairs "A,B".
5. Check all possible sums A+B and see if they match B. If no, repeat until you find a match.
6. If you cannot find a match A+B= b for whole numbers then use the quadratic formula.
7. If you CAN find a match then write (x+A)(x+B), taking care to remember to include any minus signs attached to A and/or B.
8. finally, write a(x+A/a)(x+B/a) and this is your "true" factorisation.
9. The solutions are x = -A/a and x = -B/a.
* * * * * * * * * * * * *
1. As far as I can see the point is that the "magic numbers" in my language would be the two numbers A = q*r and B = p*s, so that, in terms of the "standard" equation I used:
(p*x+q)*(r*x+s)=a*x^2+b*x+c
A*B = q*r*p*s = (p*r)*(q*s) = a*c
and
A + B = q*r + p*s = b as before.
(I call them A and B not a and b because that way you don't get a stupid clash with the standard notation ax^2 + bx + c = 0 that's almost universal for quadratic equations).
2. Then, you would get the following, from your example 2 k^2 + 3k -2 = 0 :
A + B = 3 (middle coefficient)
A*B = (2)*(-2) = -4 (product of first and last coefficients).
3. Now guess solutions to these equations, by finding the possible factors of -4 and checking if they sum to give 3:
-4 = (4)*(-1): sum = 4 -1 = 3. TRUE.
That didn't take long, so the MAGIC NUMBERS are 4 and -1.
4. Now write "(x + A)(x+B)", which here would be:
(x + 4) (x - 1)
But do not set this equal to anything: as in, literally write it on the page.
5. Now, I think what you have to do is basically a little bit of fudging to make sure this is actually the right answer. Take the coefficient of x^2 and write in three separate places: in front of the two brackets, and under both A and B (as written on the page). Here, then, you'd write:
2k^2 + 3k -2 = 0
a=2, A = 4, B = -1:
2k^2 + 3k -2 = (a)(k + A/a)(k + B/a)
= (2)(k + (4)/(2))(k + (-1)/2 )
= 2 (k + 2)(k - (1/2) )
Then, as before, this equals zero when either k+2=0 or k-1/2 = 0, which is the same as saying k = -2 or k = 1/2.
Phew!
* * * * * * * * * * * *
Summary process for the "magic number" technique, as far as I can see, is this:
1. Take your quadratic equation and compare it with the form ax^2 + bx + c = 0.
2. Identify the values of a, b and c.
3. write A + B = b (for your value of b), and A*B = a*c (for your values of a and c).
4. Find all pairs of factors of your answer to a*c. These are "possible magic number pairs "A,B".
5. Check all possible sums A+B and see if they match B. If no, repeat until you find a match.
6. If you cannot find a match A+B= b for whole numbers then use the quadratic formula.
7. If you CAN find a match then write (x+A)(x+B), taking care to remember to include any minus signs attached to A and/or B.
8. finally, write a(x+A/a)(x+B/a) and this is your "true" factorisation.
9. The solutions are x = -A/a and x = -B/a.
* * * * * * * * * * * * *
In answer to jourdain: I think it should be doable for an average year-nine student, assuming the teacher explained things clearly enough. I am not convinced that the teacher in this case has made things clear, but then again maybe I didn't either. I find it easier to teach, though, when you can talk to the student in question and pause/clarify steps as necessary, rather than (as here) throw a whole lot of text at them and cross your fingers that it all makes sense without further explanation.
Incidentally, applying the "magic numbers" approach to both 2k^2 + 10k +5=0 and 2k^2 + 10k +10=0 (ie, the original problems) end up failing at the fifth step. There are no whole numbers A and B such that A*B = (2)*(5) = 10 and A+B = 10, and none such that A*B = (2)*(10) = 20 and A+B = 10 either (try it and see!). This makes me wonder what the point of the two examples is, since they don't illustrate the "magic number" method at all.
Incidentally, applying the "magic numbers" approach to both 2k^2 + 10k +5=0 and 2k^2 + 10k +10=0 (ie, the original problems) end up failing at the fifth step. There are no whole numbers A and B such that A*B = (2)*(5) = 10 and A+B = 10, and none such that A*B = (2)*(10) = 20 and A+B = 10 either (try it and see!). This makes me wonder what the point of the two examples is, since they don't illustrate the "magic number" method at all.
