European Exam Preparation - Can you solve this equation?
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- Опубліковано 15 жов 2024
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There is an infinite number of solutions if it is not specified that a and bare integer
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
Yes, in fact there are an infinite number of solutions that he missed. His mistake was to assume a and b are integers.
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..
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.
There is no min/max. Howver, a can not equal -1/2, and b can not equal -1/2.
@@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.
I watched the video twice, trying to find the place where the condition says that the result must be an integer
Me too.
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.
I don't understand the question. Do a and b have to be integers?
I think because 17 is prime, they must be. Two fractions cannot be multiplied to a prime can they?
@@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.
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
@@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.
In the video, the numeral "1" is written four distinctly different ways in the span of 11 seconds, starting at 4:36
Can you atleast understand? The number 1 can be written hundreds of ways in different curly or straight fonts
Despite 17 being prime it can be a product of infinitely many multiplications (like 2b + 1 = 4 and 2a + 1 = 17/4)
You could also ask what is a^2+b^2 to make it harder
Fun fact: The answer is actually 8-2ab
Nice!
Ok, understood the mathematics!
Neat
Why so long to say so little. Einstein was much smarter with his equation.
wrong