It's Saturday again, and thus time to reveal the answer. Congrats to @chess1048576 who got (part of) the answer!
-------------------------------------
Using the fact that xy, xz and yz describe a cuboid with the room diagonal 7, we can form the equation
((xy)^2) + ((yz)^2) + ((xz)^2)) = 49
using Pythagoras. So we get a system of equations:
x + y + z = -4
(x^2) + (y^2) + (z^2) = 14
((xy)^2) + ((yz)^2) + ((xz)^2)) = 49
Squaring the first equation gives us
(x^2) + (y^2) + (z^2) + 2(xy + xz + yz) = 16
As we know, (x^2) + (y^2) + (z^2) = 14, which we can use to reduce the equation to
14 + 2(xy + xz + yz) = 16
From this, we can see:
14 + 2(xy + xz + yz) = 16
2(xy + xz + yz) = 2
xy + xz + yz = 1
If we square this equation, we get
((xy)^2) + ((yz)^2) + ((xz)^2)) + 2((x^2)yz + x(y^2)z + xy(z^2)) = 1
((xy)^2) + ((yz)^2) + ((xz)^2)) + 2xyz(x + y + z) = 1
Using the equations ((xy)^2) + ((yz)^2) + ((xz)^2)) = 49 and x + y + z = -4, we can reduce this to
49 + 2xyz * (-4) = 1
Furthermore, we can see:
49 + 2xyz * (-4) = 1
-8xyz = -48
xyz = 6
Now we know the values for x + y + z, xy + xz + yz and xyz. According to Vieta, if x, y and z are solutions to a cubic equation, then the cubic equation that x, y and z solve is
(t^3) – (x + y + z)(t ^2) + (xy + xz + yz)t – xyz = 0
where t is the variable that we are looking for when solving this equation.
Using this knowledge, we can form the cubic equation
(t^3) + 4(t^2) + t – 6 = 0
where x, y and z are solutions to this equation. In other words, the solution to our original problem is the solution to this cubic equation. Solving this cubic equation can, for example, be done with the cubic formula. However, I doubt that anyone would read this, and it would be very difficult to write it clearly here as the symbols I can use are limited to simple ones. So, I will leave it to the reader to solve this equation and give you the answer here. Note that we need all solutions, which means there are more solutions than just one, and all of the following satisfy all the conditions mentioned in the question:
x = 1, y = -2, z = -3
x = 1, y = -3, z = -2
x =-2, y = 1, z = -3
x = -2, y = -3, z = 1
x = -3, y = 1, z = -2
x = -3, y = -2, z = 1
This is the complete answer. You might have noticed that with these solutions, this would mean that the cuboid would have negative side lengths, and you are correct. Keep in mind, however, that this is only theoretical and was only meant to help you get a third equation. I didn’t realize that I was going to make a broken cuboid when creating this question.
-----------------------------------------------
This week, let’s get a bit more complex:
Find all possible solutions to the equation
sin(x) + cos(x) = e
------------------------------------------------
Pro Question: For anyone who thinks he’s too smart for the above question... well I don’t have anything really harder, but something more general:
Given the equation
sin(x) + cos(x) = a + bi
where a and b are real numbers and i represents the imaginary unit. Find all possible solutions for x and write the solution in the form
x = u + vi
where u and v are real numbers expressed through a and b.
Good luck everyone!
@ReChesster said in #31:
> It's Saturday again, and thus time to reveal the answer. Congrats to @chess1048576 who got (part of) the answer!
> -------------------------------------
> Using the fact that xy, xz and yz describe a cuboid with the room diagonal 7, we can form the equation
>
> ((xy)^2) + ((yz)^2) + ((xz)^2)) = 49
>
> using Pythagoras. So we get a system of equations:
>
> x + y + z = -4
>
> (x^2) + (y^2) + (z^2) = 14
>
> ((xy)^2) + ((yz)^2) + ((xz)^2)) = 49
>
> Squaring the first equation gives us
>
> (x^2) + (y^2) + (z^2) + 2(xy + xz + yz) = 16
>
> As we know, (x^2) + (y^2) + (z^2) = 14, which we can use to reduce the equation to
>
> 14 + 2(xy + xz + yz) = 16
>
> From this, we can see:
>
> 14 + 2(xy + xz + yz) = 16
>
> 2(xy + xz + yz) = 2
>
> xy + xz + yz = 1
>
> If we square this equation, we get
>
> ((xy)^2) + ((yz)^2) + ((xz)^2)) + 2((x^2)yz + x(y^2)z + xy(z^2)) = 1
>
> ((xy)^2) + ((yz)^2) + ((xz)^2)) + 2xyz(x + y + z) = 1
>
> Using the equations ((xy)^2) + ((yz)^2) + ((xz)^2)) = 49 and x + y + z = -4, we can reduce this to
>
> 49 + 2xyz * (-4) = 1
>
> Furthermore, we can see:
>
> 49 + 2xyz * (-4) = 1
>
> -8xyz = -48
>
> xyz = 6
>
> Now we know the values for x + y + z, xy + xz + yz and xyz. According to Vieta, if x, y and z are solutions to a cubic equation, then the cubic equation that x, y and z solve is
>
> (t^3) – (x + y + z)(t ^2) + (xy + xz + yz)t – xyz = 0
>
> where t is the variable that we are looking for when solving this equation.
> Using this knowledge, we can form the cubic equation
>
> (t^3) + 4(t^2) + t – 6 = 0
>
> where x, y and z are solutions to this equation. In other words, the solution to our original problem is the solution to this cubic equation. Solving this cubic equation can, for example, be done with the cubic formula. However, I doubt that anyone would read this, and it would be very difficult to write it clearly here as the symbols I can use are limited to simple ones. So, I will leave it to the reader to solve this equation and give you the answer here. Note that we need all solutions, which means there are more solutions than just one, and all of the following satisfy all the conditions mentioned in the question:
>
> x = 1, y = -2, z = -3
> x = 1, y = -3, z = -2
> x =-2, y = 1, z = -3
> x = -2, y = -3, z = 1
> x = -3, y = 1, z = -2
> x = -3, y = -2, z = 1
>
> This is the complete answer. You might have noticed that with these solutions, this would mean that the cuboid would have negative side lengths, and you are correct. Keep in mind, however, that this is only theoretical and was only meant to help you get a third equation. I didn’t realize that I was going to make a broken cuboid when creating this question.
> -----------------------------------------------
> This week, let’s get a bit more complex:
> Find all possible solutions to the equation
> sin(x) + cos(x) = e
> ------------------------------------------------
> Pro Question: For anyone who thinks he’s too smart for the above question... well I don’t have anything really harder, but something more general:
>
> Given the equation
>
> sin(x) + cos(x) = a + bi
>
> where a and b are real numbers and i represents the imaginary unit. Find all possible solutions for x and write the solution in the form
>
> x = u + vi
>
> where u and v are real numbers expressed through a and b.
>
> Good luck everyone!
Oh my god my brain hurts every saturday im outta here
@ReChesster the pro answer
We can express sin(x) and cos(x) in terms of exponential functions using Euler’s formula:
eix=cos(x)+i∗sin(x)
So, we can rewrite the equation as:
eix+e−ix=a+bi
This is a quadratic equation in terms of e^{ix}. We can solve this quadratic equation to find the solutions for e^{ix}.
Once we have the solutions for e^{ix}, we can find x by taking the natural logarithm.
Am i right
@RishikaPlaysChess said in #33:
> @ReChesster the pro answer
>
> We can express sin(x) and cos(x) in terms of exponential functions using Euler’s formula:
> eix=cos(x)+i∗sin(x)
> So, we can rewrite the equation as:
> eix+e−ix=a+bi
> This is a quadratic equation in terms of e^{ix}. We can solve this quadratic equation to find the solutions for e^{ix}.
> Once we have the solutions for e^{ix}, we can find x by taking the natural logarithm.
>
> Am i right
Not quite, but you have the right idea.
It is indeed intended that you solve a quadratic equation in terms of e^(ix). I am a bit confused about how you wrote your terms, but I think it's supposed to mean e^(ix) + e^(-ix) = a +bi, right? This isn't correct, as e^(ix) = cos(x) + i*sin(x), but we need cos(x) + sin(x) (without the i). However, you can actually use e^(ix) + e^(-ix) to come up with a formula for sin(x) and cos(x) that you can use to create a quadratic equation.
Bro which grade r u?
From my view u will be like (Q find x Answer- 100)
And also i kinda tried im just 6th grade (i kinda banged my keyboard for some formulas)
@RishikaPlaysChess said in #35:
> Bro which grade r u?
> From my view u will be like (Q find x Answer- 100)
> And also i kinda tried im just 6th grade (i kinda banged my keyboard for some formulas)
Yeah, I thought about answering like this ;)
I'm just seventh grade xD
I learnt all that stuff when I was in second and third grade (when I was 12 to 13 years old) because I loved Math so much, and I still love it ;)
@ReChesster said in #36:
> Yeah, I thought about answering like this ;)
> I'm just seventh grade xD
> I learnt all that stuff when I was in second and third grade (when I was 12 to 13 years old) because I loved Math so much, and I still love it ;)
bruhhh btw in our country second grade is at age 7
@RishikaPlaysChess said in #37:
> bruhhh btw in our country second grade is at age 7
Oh, I didn't include primary school. In that case, let's say it was my sixth and seventh year I had been going to school and now it's my eleventh year :) (I'm 16 now) Also, it's quite impressive you already know about complex numbers at 11
@ReChesster said in #38:
> Oh, I didn't include primary school. In that case, let's say it was my sixth and seventh year I had been going to school and now it's my eleventh year :) (I'm 16 now) Also, it's quite impressive you already know about complex numbers at 11
Yeah instead of 6th grade portion they teach us 11th grade things (we r currently on exponents) so makes sense
@RishikaPlaysChess said in #39:
> Yeah instead of 6th grade portion they teach us 11th grade things (we r currently on exponents) so makes sense
Bruh, where do you live?
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