Doubt on a math problem

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AVY2024

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AVY2024

#1 h Apr 8, 2025, 11:09 AM

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Solve for x and y given that xy=923, x+y=84

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fruitmonster97

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fruitmonster97

#2 h Apr 8, 2025, 11:39 AM

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We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

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sp0rtman00000

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sp0rtman00000

#3 h Apr 11, 2025, 11:04 AM

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Yous must solve the following second degree equation: t^2-84t+932=0

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CatsRule222

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CatsRule222

#4 h Apr 11, 2025, 2:36 PM

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I agree with sp0rtman00000;
It will be easier to change it to a second degree polynomial and then use the quadratic formula, which is: (-b+-Sqrt(b^2-4ac))/2a

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fruitmonster97

2492 posts

fruitmonster97

#5 h Apr 11, 2025, 2:56 PM • 1 Y

Y by DDCN_2011

CatsRule222 wrote:

I agree with sp0rtman00000;
It will be easier to change it to a second degree polynomial and then use the quadratic formula, which is: (-b+-Sqrt(b^2-4ac))/2a

That's what I did, except instead of using the quadratic formula or factoring I used Po-Shen Loh's recently(ish) found method which combines vieta's and some common sense.

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jasperE3

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jasperE3

#6 h Apr 11, 2025, 5:47 PM • 1 Y

Y by Solocraftsolo

fruitmonster97 wrote:

We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

$(x,y)=(71,13)$

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Soupboy0

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Soupboy0

#7 h Apr 11, 2025, 5:59 PM

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$x = 1434$ by 1434 number theory lemma. Now, define a sigma number $\mathcal{J}$ from the set $\mathcal{S}$ and such that such that any number can now be expressed as ultra-complex, e.g. $a+bi+c\mathcal{J}$ . Now, using the chicken jockey steve theorem, we find there are $2$ distinct solutions. These solutions actually alter the given value of $x$ . The $2$ solutions are $4+37\sqrt{3}i+27\sqrt[3]{3} \mathcal{J}$ and $26.53+34\sqrt{2} i +14 \mathcal{J}.$ These $2$ solutions alter the quantum space limit, and change the values of $x$ to be $13$ and $71$ . Now, using the newly derived quintic formula yields $y=13, 71$ . Therefore, the $2$ solutions are $(13, 71)$ and $(71, 13)$

This post has been edited 1 time. Last edited by Soupboy0, Apr 11, 2025, 5:59 PM

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sadas123

1306 posts

sadas123

#8 h Apr 13, 2025, 1:15 AM

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Soupboy0 wrote:

$x = 1434$ by 1434 number theory lemma. Now, define a sigma number $\mathcal{J}$ from the set $\mathcal{S}$ and such that such that any number can now be expressed as ultra-complex, e.g. $a+bi+c\mathcal{J}$ . Now, using the chicken jockey steve theorem, we find there are $2$ distinct solutions. These solutions actually alter the given value of $x$ . The $2$ solutions are $4+37\sqrt{3}i+27\sqrt[3]{3} \mathcal{J}$ and $26.53+34\sqrt{2} i +14 \mathcal{J}.$ These $2$ solutions alter the quantum space limit, and change the values of $x$ to be $13$ and $71$ . Now, using the newly derived quintic formula yields $y=13, 71$ . Therefore, the $2$ solutions are $(13, 71)$ and $(71, 13)$

I agree with you

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maxamc

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maxamc

#9 h Apr 13, 2025, 3:39 AM

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Since $3|7$ , $13|71$ (add $10$ to the left hand side and $1$ to the other, difference is $3 \cdot 3=9$ , then Euclidian Algorithm), and $71|13$ (commutativity of the divides relation). Thus our answers are $(13,71)$ and all other permutations.

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evt917

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evt917

#10 h Apr 13, 2025, 4:08 AM

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AVY2024 wrote:

Solve for x and y given that xy=923, x+y=84

me: bash

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Yiyj1

1266 posts

Yiyj1

#11 h Apr 13, 2025, 6:50 AM

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evt917 wrote:

AVY2024 wrote:

Solve for x and y given that xy=923, x+y=84

me: bash

me: runs a program that tests every single pair (x,y)

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giratina3

546 posts

giratina3

#12 h Apr 16, 2025, 3:04 AM

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Yiyj1 wrote:

evt917 wrote:

AVY2024 wrote:

Solve for x and y given that xy=923, x+y=84

me: bash

me: runs a program that tests every single pair (x,y)

me: crashes out

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derekwang2048

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derekwang2048

#13 h Apr 16, 2025, 3:19 AM

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wait is this not trivial by quadratic formula
if you're fine with squaring 84 and 4*923

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giratina3

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giratina3

#14 h May 5, 2025, 11:47 PM

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derekwang2048 wrote:

wait is this not trivial by quadratic formula
if you're fine with squaring 84 and 4*923

This is why we don't use the quadratic formula whenever we see massive numbers.

You're incredibly prone to making an arithmetic mistake.

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Soupboy0

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Soupboy0

#15 h May 6, 2025, 2:35 AM

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do you know what else is massive

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giratina3

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giratina3

#17 h May 6, 2025, 11:46 AM

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Soupboy0 wrote:

do you know what else is massive

I don’t know. For some reason, I thought of hairstyles when I heard “massive”.

This post has been edited 1 time. Last edited by giratina3, May 6, 2025, 11:47 AM

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maxamc

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maxamc

#18 h May 6, 2025, 3:19 PM

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giratina3 wrote:

derekwang2048 wrote:

wait is this not trivial by quadratic formula
if you're fine with squaring 84 and 4*923

This is why we don't use the quadratic formula whenever we see massive numbers.

You're incredibly prone to making an arithmetic mistake.

I always use the quadratic formula on huge numbers.

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LXC007

41 posts

LXC007

#19 h May 7, 2025, 5:21 PM

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giratina3 wrote:

Soupboy0 wrote:

do you know what else is massive

I don’t know. For some reason, I thought of hairstyles when I heard “massive”.

I can hear the “Low taper fade” right now

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Capybara7017

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Capybara7017

#20 h May 7, 2025, 9:07 PM

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jasperE3 wrote:

fruitmonster97 wrote:

We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

$(x,y)=(71,13)$

they aren't wrong

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sadas123

1306 posts

sadas123

#21 h May 7, 2025, 10:06 PM

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Capybara7017 wrote:

jasperE3 wrote:

fruitmonster97 wrote:

We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

$(x,y)=(71,13)$

they aren't wrong

It doesn't matter which order you put in but just use subsitiution then difference of squares and then just use algebra and solve for the final answer.

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