European Exam Preparation - Can you solve this equation?

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[@user-nz6tm4bh6y]

There is an infinite number of solutions if it is not specified that a and bare integer

[@quantumbuddha777]

He is definitely treating this problem as if a and b are integers even though it is never actually specified. Otherwise there are an infinite number of solutions and a+b can range
from -inf to -1-sqrt(17) and
from -1+sqrt(17) to inf
These values are local max and min respectively. It becomes a fun calculus problem when looked at in this way.

[@ronbannon]

There is no min/max. Howver, a can not equal -1/2, and b can not equal -1/2.

[@quantumbuddha777]

​@@ronbannonthere are local max/min, just no global max/min. Solve for a in terms of b and graph a+b. You'll find the graph is split with a vertical asymptote at -1/2, as you assert. The left hand curve has a local max and the right hand curve has a local min.

[@mikekomaransky8334]

In the video, the numeral "1" is written four distinctly different ways in the span of 11 seconds, starting at 4:36

[@shinjonmal8936]

Can you atleast understand? The number 1 can be written hundreds of ways in different curly or straight fonts

[@johnlv12]

I don't understand the question. Do a and b have to be integers?

[@deekay2091]

I think because 17 is prime, they must be. Two fractions cannot be multiplied to a prime can they?

[@twwc960]

@@deekay2091 Of course they can. (85/3)*(3/5)=17. There are an infinite number of solutions to this problem. For example, a=1, b=7/3 satisfies the first equation and in that case a+b=10/3. He missed an infinite number of solutions, in fact, by assuming a and b are integers.

[@shinjonmal8936]

This is just 2 solutions plotting a max and min point for forming a line in a graph. Once you draw the line, you can get as many solutions as you please

[@gabriellahavas8229]

@@deekay2091
There is infinit way to multiply fractions, and get a prime as a result.
(34/3)*(3/2)=17
(170/999) *(999/10)
The solution in the video only applies, if we specify from the beginning, that we are lokking for integer solutions. But they forgot to mention that at the beginning.

[@user-tf6cz4pn7b]

I watched the video twice, trying to find the place where the condition says that the result must be an integer

[@TheEulerID]

Me too.

[@pantelostzimas7490]

You could also ask what is a^2+b^2 to make it harder

[@patriciaceli1536]

Nice!

[@arekkrolak6320]

Despite 17 being prime it can be a product of infinitely many multiplications (like 2b + 1 = 4 and 2a + 1 = 17/4)

[@DebdasBandyopadhyay-yq5jg]

Ok, understood the mathematics!

[@sie_khoentjoeng4886]

We can use a spesific number for a variable, and then another variable is depend on this variable...
Let a=1, then
a+2ab+b=8
1+2b+b=8
3b=8-1=7 and b=7/3
Here a+b= 1+7/3=10/3
For a=2, 5b=6 or b=6/5
.. and so on..

[@TheEulerID]

Did I miss the part where the problem was only meant to produce integer solutions for a+b? If you are going to set problems, then it ought to include all the conditions. As stated, there is an infinity of solutions.

[@tombufford136]

At a quick glance. a and b are clearly 'interchangeable' with one solution a=a , b= b and a second solution a=b, b = a. if a = b then 2a^2+ 2a - 8 = 0. Using the quadratic formula this can be solved. (b +/- sqrt( b^2-4ac))/ 2a where a =1, b =1 and c= -4. Then (b +/- sqrt(17))/2 then 2.562 and -1.562. Checking this -1.562 does give a solution then a = -1.562 and b = - 1.562

[@twwc960]

Yes, in fact there are an infinite number of solutions that he missed. His mistake was to assume a and b are integers.

[@streptococcuspneumoniae-ix1ve]

b = 0 then 3a = 8 a = 8/3
then a+b = 8/3

[@berniegreen5401]

Neat

[@hayallerimverenklerim9733]

Fun fact: The answer is actually 8-2ab

[@mauriziograndi1750]

Why so long to say so little. Einstein was much smarter with his equation.

[@user-pi1ez7ju5x]

wrong