I got nothing | Mathematics | Page 1

trail oar Aug 4, 2024, 8:45 PM

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I dunno

quartz novaBOT Aug 4, 2024, 8:45 PM

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trail oar I dunno

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muted tapir Aug 4, 2024, 8:46 PM

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Have you tried doing anything?

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@trail oar

trail oar Aug 4, 2024, 8:47 PM

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igot ab=a+b

muted tapir Aug 4, 2024, 8:48 PM

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That might be useful, what about just substituting what you have?

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With a³+b³

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Opening and simplifying

trail oar Aug 4, 2024, 8:49 PM

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still need to find ab

muted tapir Aug 4, 2024, 8:49 PM

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Oh hmm

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Do you have anything written out or do you want me to just do the work for you

trail oar Aug 4, 2024, 8:51 PM

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i don't have my phone near me atm so cant take photo

muted tapir Aug 4, 2024, 8:51 PM

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Ah fair

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Lets just denote $u=\frac{x}{y}$ for ease

normal summitBOT Aug 4, 2024, 8:51 PM

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yoavmal

trail oar Aug 4, 2024, 8:52 PM

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(ab)(15-ab)=a^3+b^3

muted tapir Aug 4, 2024, 8:52 PM

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Then we get $b=1+\frac 1u$

normal summitBOT Aug 4, 2024, 8:52 PM

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yoavmal

muted tapir Aug 4, 2024, 8:52 PM

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And $a=1+u$

normal summitBOT Aug 4, 2024, 8:52 PM

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yoavmal

muted tapir Aug 4, 2024, 8:53 PM

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trail oar (ab)(15-ab)=a^3+b^3

So working with that we can try hmm

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Oh idea

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$(a+b)²=a^2+2ab+b^2$

normal summitBOT Aug 4, 2024, 8:54 PM

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yoavmal

muted tapir Aug 4, 2024, 8:55 PM

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No, not very useful on its own

trail oar Aug 4, 2024, 8:55 PM

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ya

muted tapir Aug 4, 2024, 8:57 PM

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$(1+u)^2+\qty(1+\frac 1u)^2=1+2u+u^2+1+\frac 2u+\frac{1}{u^2}=15$

normal summitBOT Aug 4, 2024, 8:57 PM

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yoavmal

muted tapir Aug 4, 2024, 8:58 PM

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A graphic approach might be useful

trail oar Aug 4, 2024, 8:59 PM

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tbh im not that good with graphs

muted tapir Aug 4, 2024, 8:59 PM

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What intersections of 1+u and 1+1/u are at radius 15?

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As in

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(1+u,1+1/u)

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That's the graph of 1/x, shifted up and right

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By 1 unit on each axis

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Hmm

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Can't find anything particularly insightful

trail oar Aug 4, 2024, 9:01 PM

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could it be done in a non-graphival eay?

muted tapir Aug 4, 2024, 9:02 PM

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It probably can solve it but I don't see a way without actually plotting it

muted tapir Aug 4, 2024, 9:02 PM

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trail oar could it be done in a non-graphival eay?

Probably, yes

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How did you get to (ab)(15-ab)=a³+b³?

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It doesn't work out for me

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It should be noted that due to the symmetry of the question, if u solves it, 1/u solves it

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So there's probably like 4 solutions

trail oar Aug 4, 2024, 9:12 PM

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a^3+b^3= (a+b)(a^2+b^2-ab)

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= (ab)(15-ab)

muted tapir Aug 4, 2024, 9:12 PM

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trail oar a^3+b^3= (a+b)(a^2+b^2-ab)

Ah that makes sense

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So, we can probably solve for u

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ab(15-ab)

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Maybe a sum is easier

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Oh yeah from earlier

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(a+b)²=a²+2ab+b²

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a+b=ab so

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(ab)²=a²+2ab+b²

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ab(ab-2)=15

trail oar Aug 4, 2024, 9:17 PM

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ya

muted tapir Aug 4, 2024, 9:17 PM

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That is something

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And we need ab(15-ab)

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That's nearly there

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How about we continue playing with it

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Oh wait

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There's only two solutions for the product ab

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If we say q=ab

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q²-2q-15=0

trail oar Aug 4, 2024, 9:19 PM

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q=3,-5

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-5 isnt possible as a,b are positive

muted tapir Aug 4, 2024, 9:20 PM

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Aight

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Then we're done

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Lets verify this

trail oar Aug 4, 2024, 9:20 PM

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but ans is given as 50

muted tapir Aug 4, 2024, 9:22 PM

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3(15+3)=54

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q²=15+2q

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,w solve q^2-2q-15

normal summitBOT Aug 4, 2024, 9:22 PM

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muted tapir Aug 4, 2024, 9:23 PM

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Aha

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Well there we go

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5(15-5)

trail oar Aug 4, 2024, 9:24 PM

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thanks @muted tapir

sharp folioBOT Aug 4, 2024, 9:24 PM

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@trail oar has given 1 rep to @muted tapir

muted tapir Aug 4, 2024, 9:24 PM

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Np, it's kind of nice

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The question that is

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Sometimes questions like this are just annoying

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That one isn't so

trail oar Aug 4, 2024, 9:25 PM

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its fun solving it

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+close

#I got nothing