- #1
- 17,876
- 1,660
- Homework Statement
- I don't know why people show worked examples like this:
https://www.youtube.com/watch?v=oNFbxHQpDxA
[i]I am posting this more for discussion for homework helpers, and hoping it will help people struggling to understand the maths. I'd also be interested in students at this level comparing the approach below with the one in the video to see what they think.[/i]
The problem being demonstrated is
"simplify ##\sqrt{53 - 10\sqrt{6}}## -- no calculator"
The video demonstration simply states a step-by-step with no reasoning.
The result is that anyone unfamiliar with the type of puzzle, ie. the sort of person who may be interested in the video, will remain clueless as to how to solve other forms of this puzzle.
It's not even the most elegant way to demonstrate the solution, which is the usual failing in school maths potted-proofs.
I suspect the demonstrator does not know the reasoning behind what is being shown, but has simply recognized the general problem and memorized an algorithm. A common issue with the way maths has been taught at this level for generations.
Here's my approach ...
- Relevant Equations
- $$(a+b)^2 = a^2+b^2+2ab$$ + prime numbers.
I'd start with a short discussion of what it means to "simplify" in this context... student needs a little background here, but not a lot. ie. we want no nesting square-roots if we can help it.
In this working I am illustrating how I'd approach explaining how to approach these sorts of problems to a student. ie. making a better video.
Starting with the end in mind.
The upshot is that the expression can be simplified if: ##53−10\sqrt{6}=(a+b)^2##, then that is a complete square, and the simplified version is just ##a+b##.
So the problem becomes "find ##a## and ##b##."
Expand the brackets so we get something that kinda looks like the expression we want:
$$a^2+b^2+2ab=53−10\sqrt{6}$$
Now we start guessing!
Guessing is a valid approach to solving maths problems quickly, and this should be normalized for students. They can explore!
We could try guessing ##2ab=−10\sqrt{6}##
Why not the other way?
We-ell, if ##a^2+b^2=−10\sqrt{6}## we need to turn ##5\sqrt{6}## into a sum of two terms ... that could happen like this: ##a^2=b^2=−5\sqrt{6}## so ##a=b=\sqrt{−5\sqrt{6}}## which is not simple! Students are encouraged to try for other combinations.
Quickly we find that this approach, at best, looks like it will be hard ... why do harder work than we have to? Start on the easier looking one, vis. $$2ab=−106⟹ab=−5\sqrt{6}$$
That means that ##a^2+b^2=53## ... so we have simultaneous equations.
Could brute-force it that way, but this is represented as a problem that can be worked by hand in an interview setup, so it should be simple and we want to show off a bit.
After all, the interviewer is not giving you the problem because they want to know the answer, they know the answer: they are doing this to see how you reason.
So guess ... there are not many ways to guess.
The easiest way to split ##-5\sqrt{6}## into two factors would be ##a=−5## and ##b=6##, so that's the choice for 1st guess.
Doing that means ##a^2+b^2=25+6=31\neq 53##: so this guess was incorrect.
No worries, there are always more guesses.
Aside: notice that I could have also had ##a = 5## and ##b = -\sqrt{6}##?
The position of the minus sign is arbitrary. That means there will be two correct answers. (Student can check ... could there be four correct answers?)
To keep going, we need to find another way to separate ##5\sqrt{6}## into factors.
If the student knows about prime factorization, then they can do that here ... if not, then just note that 5 is prime, so it's only factors are 5 and 1,so we are not going to be separating the 5. We could try non-integer factors of 5... but that sounds hard, so leave that for if all else fails and we brute-force the simultaneous equations. However: 6 can be factored into 2 and 3 (hurray) ... so ##−5\sqrt{6}=−5\sqrt{2}\sqrt{3}##. This makes for an easy next guess -- just following our noses:
$$a=−5\sqrt{2},\;b=3$$
(The next guess would be ##a=-5\sqrt{3}## etc. Does it matter which square-root we put the 5 on? Ex. for students.)
This choice for a and b means: ##a^2+b^2 = 25\times 2 + 3 = 53##... which is true!
We also get the same result for ##a=5\sqrt{2}## and ##b = -\sqrt{3}## , the choice of where to put the minus sign is arbitrary!
Thus
$$\sqrt{53 - 10\sqrt{6}} = \sqrt{3} - 5\sqrt{2}$$
... is sufficient for the interview (you only need one).
In fact, it won't simplify further, which we can see from the very small numbers with no further sub-factors (ie. all numbers are single-digit prime-numerals).
<fanfare>
But wait, there's more we can do here...
What happens if the problem cannot be simplified?
That would be a reasonable student question, so let's illustrate it: and trick question do show up in interviews. After all, the question statement did not say it could be simplified any further.
Lets say there was a 6 instead of a 10:
Then ##ab = 3\sqrt{6}##, so the guesses run:
##a = 3,\; b= \sqrt{6} \implies a^2+b^2 = 9 + 6 = 15 \to## false
##a = 3\sqrt{2},\; b= \sqrt{3} \implies a^2+b^2 = 18 + 3 = 21 \to## false
##a = 3\sqrt{3},\; b= \sqrt{2} \implies a^2+b^2 = 27 + 2 = 29 \to## false
... at this point, stop and look at the interviewer and say, "I suspect that this is already in it's simplest form, I'll have to brute force it through simultaneous equations to prove it though." At which point, the interviewers says, "You're hired."
The question says not to use a calculator. What happens if we do?
Just plugging the numbers in gives 8.8031186194 ... which is not the simplification needed: it's an approximation. We know there are irrational numbers involved... so we could waste a lot of time dividing out square-roots of 2 and 3 and 5 (likely candidates) to see if we get something recognizable as a rational... but you'd never hit on the actual answer by trial an error (not during the interview).
An interviewer could just allow calculators, but require exactness. When the interviewee whips out their calculator they get marks off, but there may be merit in seeing how long it takes them to realize this approach is useless and try something else.
I mean, someone may have a programmable calculator that can do the problem... in which case, the interviewer just says, "Thank you, don't call us: we'll call you." Maybe they'll be the best candidate just on preparedness alone?
But wait, there's more!The vid could have continued with examining the problem type itself:
In the general approach you recognize the problem is of form:
$$x+y\sqrt{z} = (a+b)^2$$
... and we want to find a,b in terms of x,y,z.
$$a^2+b^2 = x\quad ab = \frac{1}{2}y\sqrt{z}$$
... right away all the hard stuff is in the second one, which is why the demo focuses on that part and plays around with it.
... solve the general case using simultaneous equations:
$$b = \pm y\sqrt{z/(4x-1)}\quad
a = \pm y^2z /2\sqrt{(4x-1)}$$
... and the student can check my working to see if I have included some deliberate "mistakes" to trip them up.
Conclusion
The worked example in the link above goes straight to a prime factorization at the start, but only of one term, then groups the factors, with no explanation of why they would want to do that...
They leave the reveal that they are forming the complete square to the end... as some sort of dramatic reveal maybe? When they do, it is not clear how they knew they'd need to do that at the beginning.
This makes it look like the demonstrator is magic.
It also gives struggling students the impression that maths is just too hard for them: they are not smart enough.
I think the approach above would make a better video: more instructive, and more engaging for students ... if I had the setup to do it I would make it.
If someone wants to try that, please put a link in the comments and link from the video to here. Thanks.
In this working I am illustrating how I'd approach explaining how to approach these sorts of problems to a student. ie. making a better video.
Starting with the end in mind.
The upshot is that the expression can be simplified if: ##53−10\sqrt{6}=(a+b)^2##, then that is a complete square, and the simplified version is just ##a+b##.
So the problem becomes "find ##a## and ##b##."
Expand the brackets so we get something that kinda looks like the expression we want:
$$a^2+b^2+2ab=53−10\sqrt{6}$$
Now we start guessing!
Guessing is a valid approach to solving maths problems quickly, and this should be normalized for students. They can explore!
We could try guessing ##2ab=−10\sqrt{6}##
Why not the other way?
We-ell, if ##a^2+b^2=−10\sqrt{6}## we need to turn ##5\sqrt{6}## into a sum of two terms ... that could happen like this: ##a^2=b^2=−5\sqrt{6}## so ##a=b=\sqrt{−5\sqrt{6}}## which is not simple! Students are encouraged to try for other combinations.
Quickly we find that this approach, at best, looks like it will be hard ... why do harder work than we have to? Start on the easier looking one, vis. $$2ab=−106⟹ab=−5\sqrt{6}$$
That means that ##a^2+b^2=53## ... so we have simultaneous equations.
Could brute-force it that way, but this is represented as a problem that can be worked by hand in an interview setup, so it should be simple and we want to show off a bit.
After all, the interviewer is not giving you the problem because they want to know the answer, they know the answer: they are doing this to see how you reason.
So guess ... there are not many ways to guess.
The easiest way to split ##-5\sqrt{6}## into two factors would be ##a=−5## and ##b=6##, so that's the choice for 1st guess.
Doing that means ##a^2+b^2=25+6=31\neq 53##: so this guess was incorrect.
No worries, there are always more guesses.
Aside: notice that I could have also had ##a = 5## and ##b = -\sqrt{6}##?
The position of the minus sign is arbitrary. That means there will be two correct answers. (Student can check ... could there be four correct answers?)
To keep going, we need to find another way to separate ##5\sqrt{6}## into factors.
If the student knows about prime factorization, then they can do that here ... if not, then just note that 5 is prime, so it's only factors are 5 and 1,so we are not going to be separating the 5. We could try non-integer factors of 5... but that sounds hard, so leave that for if all else fails and we brute-force the simultaneous equations. However: 6 can be factored into 2 and 3 (hurray) ... so ##−5\sqrt{6}=−5\sqrt{2}\sqrt{3}##. This makes for an easy next guess -- just following our noses:
$$a=−5\sqrt{2},\;b=3$$
(The next guess would be ##a=-5\sqrt{3}## etc. Does it matter which square-root we put the 5 on? Ex. for students.)
This choice for a and b means: ##a^2+b^2 = 25\times 2 + 3 = 53##... which is true!
We also get the same result for ##a=5\sqrt{2}## and ##b = -\sqrt{3}## , the choice of where to put the minus sign is arbitrary!
Thus
$$\sqrt{53 - 10\sqrt{6}} = \sqrt{3} - 5\sqrt{2}$$
... is sufficient for the interview (you only need one).
In fact, it won't simplify further, which we can see from the very small numbers with no further sub-factors (ie. all numbers are single-digit prime-numerals).
<fanfare>
But wait, there's more we can do here...
What happens if the problem cannot be simplified?
That would be a reasonable student question, so let's illustrate it: and trick question do show up in interviews. After all, the question statement did not say it could be simplified any further.
Lets say there was a 6 instead of a 10:
Then ##ab = 3\sqrt{6}##, so the guesses run:
##a = 3,\; b= \sqrt{6} \implies a^2+b^2 = 9 + 6 = 15 \to## false
##a = 3\sqrt{2},\; b= \sqrt{3} \implies a^2+b^2 = 18 + 3 = 21 \to## false
##a = 3\sqrt{3},\; b= \sqrt{2} \implies a^2+b^2 = 27 + 2 = 29 \to## false
... at this point, stop and look at the interviewer and say, "I suspect that this is already in it's simplest form, I'll have to brute force it through simultaneous equations to prove it though." At which point, the interviewers says, "You're hired."
The question says not to use a calculator. What happens if we do?
Just plugging the numbers in gives 8.8031186194 ... which is not the simplification needed: it's an approximation. We know there are irrational numbers involved... so we could waste a lot of time dividing out square-roots of 2 and 3 and 5 (likely candidates) to see if we get something recognizable as a rational... but you'd never hit on the actual answer by trial an error (not during the interview).
An interviewer could just allow calculators, but require exactness. When the interviewee whips out their calculator they get marks off, but there may be merit in seeing how long it takes them to realize this approach is useless and try something else.
I mean, someone may have a programmable calculator that can do the problem... in which case, the interviewer just says, "Thank you, don't call us: we'll call you." Maybe they'll be the best candidate just on preparedness alone?
But wait, there's more!The vid could have continued with examining the problem type itself:
In the general approach you recognize the problem is of form:
$$x+y\sqrt{z} = (a+b)^2$$
... and we want to find a,b in terms of x,y,z.
$$a^2+b^2 = x\quad ab = \frac{1}{2}y\sqrt{z}$$
... right away all the hard stuff is in the second one, which is why the demo focuses on that part and plays around with it.
... solve the general case using simultaneous equations:
$$b = \pm y\sqrt{z/(4x-1)}\quad
a = \pm y^2z /2\sqrt{(4x-1)}$$
... and the student can check my working to see if I have included some deliberate "mistakes" to trip them up.
Conclusion
The worked example in the link above goes straight to a prime factorization at the start, but only of one term, then groups the factors, with no explanation of why they would want to do that...
They leave the reveal that they are forming the complete square to the end... as some sort of dramatic reveal maybe? When they do, it is not clear how they knew they'd need to do that at the beginning.
This makes it look like the demonstrator is magic.
It also gives struggling students the impression that maths is just too hard for them: they are not smart enough.
I think the approach above would make a better video: more instructive, and more engaging for students ... if I had the setup to do it I would make it.
If someone wants to try that, please put a link in the comments and link from the video to here. Thanks.