Find q so that x^4-40x^2+q=0 has 4 solutions forming an arithmetic progression. German Math Olympiad

Here's a quartic problem on the German math olympiad. We want to find q so that the equation x^4-40x^2+q=0 has 4 solutions forming an arithmetic progression. Mathematical Olympiad.

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#blackpenredpen #math #calculus #apcalculus

[emiltonklinga]

My solution was: "They were probably nice giving whole number roots, what two perfect squares add up to 40?"

[pushkaryennawar]

Another way of solving the question is as follows:
You can assume the roots of the original equation to be:
(a-3d), (a-d), (a+d), (a+3d) where 'a' is the first term of A.P. and '2d' is the common difference.
Further, you can solve for 'd' by 'Vieta's relations'.
BTW, LOTS OF LOVE FROM INDIA!

[leftclickben]

Me: he's never going to finish it all on that little board...
blackpenredpen: *starts writing out even more intermediate steps*

[ich_mag_zuege]

This is the way I solved it:
At first I also assumed that u is equal to x^2, which gives you the following equation.
u^2-40u+q=0
If you then solve for u, you get this equation.
u=20+/-sqrt(400-q)
Now I used the fact, that x is the square root of u, which means, that the solutions for x aren't just a, b, c and d but -b, -a, a and b. Because the solutions have to form an A. P. b-a has to be equal to a-(-a):
b-a=a-(-a)
b-a=2a
b=3a
Now we know, that the four solutions for x are -3a, -a, a and 3a. The solutions for u have to be the squares of these four numbers, because I assumed, that x^2 is equal to u. If you square the four numbers you get a^2 and 9a^2. If we now come back to the formula for u, we know that 20+/-sqrt(400-q) is either a^2 or 9a^2 depending on whether you add or subtract sqrt(400-q). Because a^2 is positive you get 9a^2 if you add and a^2 if you subtract. Now you have two equations:
20+sqrt(400-q)=9a^2; 20-sqrt(400-q)=a^2
Now you can multiply the right equation by 9 to get 180-9sqrt(400-q)=9a^2. Then you can combine this equation with the left equation to get the following.

Now you can finally solve for q:
sqrt(400-q)=160-9sqrt(400-q)
10sqrt(400-q)=160
sqrt(400-q)=16
400-q=256
144-q=0
144=q

[OriginalSuschi]

blackpenredpen in quarantine day 2000: Speaking to a Pikachu

[FX]

what i did before watching:
let the roots be a, a+b, a+2b, a+3b
then * can be written as (x-a)(x-a-b)(x-a-2b)(x-a-3b)
by comparing coefficient of x^3, we get (-4a-6b)=0, b=-2a/3
so * can be re-written as (x-a)(x-a/3)(x+a/3)(x+a)
now compare the coefficient of x^2, we get a^2(-10/9) = -40, a=6
therefore, * is (x-6)(x-2)(x+2)(x+6)
q = (-6)(-2)(2)(6) = 144

[MrDerpinati]

when you realise the pikachu is a microphone

*_biggest maths anime betrayal in history_*

[MrKryos]

Beautiful problem! Thanks for all you do, I am looking forward to watching new videos :D

[heliocentric]

I solved it as follows:
The function is even so if r is the smallest positive root then -r is as well. Also 3r and -3r are roots (in order to form AP). It follows that:
Eq1: 81r^4 -360r^2 +q=0
Eq2: r^4 -40r^2 +q=0
(Eq1)-(Eq2) implies 80r^2 (r-2)(r+2)=0, Thus r=2 (positive)
So our original equation of the question becomes (x+6)(x+2)(x-2)(x-6)=0
Now it's clear that the constant term is 144 (which is q)

[GourangaPL]

ok so let's try to solve it before i watch:
t = x^2
t^2 - 40t + q = 0
for 2 different solutions
(-40)^2 - 4*1*q > 0
1600 - 4q > 0
q < 400

for both to be positive from Viete's formulas
-(-40) / 1 > 0 (their sum)
and
q / 1 > 0 (their product)


So my guess is 0 < q < 400

[easyelectronics]

Your understanding of mathematics is really brilliant ❤️i would like to make my brain think like you💕

[samanosvasilias]

I don't even use this kind of math in my classes anymore, I just watch the videos can you make them fun and I also love math, but mainly because of you.

[hellothere]

Love your videos! Your a great inspiration!

[anasghazialgifari]

More math olympiad video like this please!

[jofx]

Don't solve x, solve q
*Proceed with solving d first*

[rohitbandi]

I just made q=mn, and then made the equation into (x^2-m)(x^2-n)=0, then solved that to get +-sqrtm and +-sqrtn. Then I just did sqrtm-sqrtn = sqrtn-(-sqrtn). This gave m = 9n, plug that into m+n=40, get n=4, and m=36, so q=144. This solution took about 2 minutes to solve. Also, whether n > m or m > n doesn't affect the solution, as either way one of m or n will equal 9n or 9m.

[indianshooter]

love u bro I try learning a lot from u. Keep it up and once again love u ❤️

[sea]

If you start with a polynomial (x-a)(x-b)(x-c)(x-d) you have a+b+c+d=0
since it is an arithmetic progression, you can then prove that the roots have the form a, a/3, -a/3 and -a

so the polynomial becomes (x2-a2)(x2-a2/9)

and you get a2 = 36
and a4/9 = 144 = q

[aydencutmore]

You could also use the sum of the roots of a quartic equation 1, 2 and 4 at a time to figure it out, which I found much easier

[mohammadelsayed]

Man this is so much beautiful 🥺 thank you ❤️