Can you solve this Cambridge Interview Question? Simplify the Radical | No Calculators Allowed!

[RexxSchneider]

The method of solution for these sort of problems is always going to be by making the expression under the radical sign equal to a perfect square of the form (a+b)^2 (or (a-b)^2 if a minus sign is present). The problem is always in decomposing the mixed term containing a radical in such a way that it becomes the 2ab term in a^2 + b^2 + 2ab (or -2ab, as in this case).

In this video, we have 2ab = 10.sqrt(6) and a^2 + b^2 = 53. The method presented in the video is to decompose the 10.sqrt(6) into 2 . (5.sqrt(2)) . (sqrt(3)) = 2ab and then felicitously finding that the resulting a^2 + b^2 happens to come to 53. Given that 5.sqrt(6) can be separated into two factors as (1) and (5.sqrt(6)) or (5) and (sqrt(6)) or (5.sqrt(2)) and (sqrt(3)) or (5.sqrt(3)) and (sqrt(2)), etc. we're not being shown how to pick the correct decomposition that satisfies the other part.

The general method to do that is to solve the symmetric simultaneous equations 2ab = 10.sqrt(6) and a^2 + b^2 = 53 and that can be done by noting that b = 10.sqrt(6)/2a = 5.sqrt(6)/a and then substituting for b in the quadratic, giving:
a^2 + (5.sqrt(6)/a)^2 = 53
a^2 + 25.6/a^2 - 53 = 0
a^4 -53a^2 + 150 = 0
Using the quadratic formula to solve for a^2:
a^2 = (53 +/- sqrt(53^2 - 600) ) / 2 = (53 +/- sqrt(2209) ) / 2 = (53 +/- 47)/2 = 50 or 3
So a = sqrt(50) or sqrt(3). As you can check by substituting into b = 5.sqrt(6)/a, the symmetry of the system of equations means that if a = sqrt(50) then b = sqrt(3), and vice-versa.

So we have 53 - 10.sqrt(6) = (sqrt(50) - sqrt(3))^2 Since the result is conventionally taken to be positive, that means that sqrt( 53 - 10.sqrt(6) ) = sqrt(50) - sqrt(3), which of course is the same as 5.sqrt(2) - sqrt(3).

[davidboeger]

I plugged it into a calculator, and that seemed to really simplify the whole thing.

[onehourmusicbc]

Just learnt that few months back and got it straight in my mind. There is a really simple way for these kinds of problems

[Shikuesi]

It would be good to have a precise objective definition of *simplification* in this context, to know what we're supposed to be working towards.

[TheLastWalenta]

I feel like the answer is just as complicated as the question.

[mysockC]

Nicely explained. Still seems somewhat useless outside of an exam as it’s still in a form that has radicals. That said if you knew the first few square roots the simplified form is a bit easier to get a direct estimate without a calculator.

[victorburnett]

I can't imagine what meaningful skill or quality of the interviewee this question tests for. You might as well solve chess puzzles for the interview.

[HowardARoark]

Square up x = (a * SQRT(2) + b * SQRT(3)) and choose a, b so that this equals 53 - 10 * SQRT(6). 4 possible values of (a, b) arise, giving two possible x's : ie. plus or minus 5 * SQRT(2) - SQRT(3). It's kind of like working with the golden ratio, or with subfields of R involving rationals combined with surds.

[shawnbrown]

As an engineering student, we like to estimate.
So here it goes
Sqrt(6)~=2.5 I guessed, 2^2 was too small and 3^2 was too large.
53-(10*2.5)=28
Sqrt(28)~=5.25 again I guessed, 5^2 was close and 5.5^2 would likely be too much.
Estimation: 5.25
Calculator: ~5.33
Percent error: ~1.5%
Not bad

[aneekbhattacharyac]

Fantastically solved.Greetings from India

[chrissmith]

I always understood applied maths in school, but never really understood pure maths or why it was important. Thanks for this and it would help me and perhaps others why unless it is just important for its own sake?

[l.w.paradis]

Wow, all the terms are small primes. Didn't see that coming. 😊

[fedr]

wow, at an interview in Cambridge, they solve typical school exercises. amazing!

[MichaelRothwell]

In the video it was stated incorrectly that √(x²) = x.
Whilst it is true that (√x)² = x, the "cancellation" doesn't work in the other direction and in fact √(x²) = |x|. To finish the solution correctly, we need to check that 5√2-√3>0, which is of course true.

[faisalwho]

When I saw this, my thought process was
2*2 is 4, 3*3 is 9, so split the difference and I get square root of 6 as roughly 2.5. Multiply that by 10 I get 25. 53 is roughly 50, subtract that with 25 you get 25. 5 * 5 is 25, so the answer is around 5

[ivornworrell]

Outstanding! Please remember though, professor, that in English we do not say "revert back to...", we say "revert to...", because the verb revert means "to go back", so "revert back to..." is a redundant statement.Salaam

[robertsullivan]

I stuck with this for 4:00 minutes then stopped pretending I had any idea of what he was doing. This left me with the question what job would require me to have to solve this whatever it is😊

[davidmata]

Since 49 and 4 are perfect squares, I would have imagined that rewriting 53 as 49+4 would've helped simplify it. But that doesn't seem to be the case.

[XBGamerX]

if you have addition inside a square root take this as a positive fact rather than giving up because there might be factorization that may be hiding and it might usually involve squares so you can exploit this fact and later remove the square root and take the absolute value if necessary

[markphc]

Well, that worked out conveniently! What are the odds?