Tricky question, I mean, it"s not hard to follow your explanation but not in a million years would I possibly come up by myself with this approach to solve the problem, so well done and I miss your videos, hope you come back soon with new videos... Cheers and hope you're doing well
Since it’s between two perfect squares, can I not get this solution by finding the difference between 4y^4+…+1 and the other expression in the inequality and setting that value equal to 0 so then the expressions become equal. This yields the same solutions
Setting the expression 4y^4+4y^3+4y^2+4y+1 to either (2y+y)^2 or (2y+y+1)^2 does get the solutions, but the expression is not necessarily bounded by those squares. It could potentially equal some other square number. Hence the case y=1; the expression is 17, not between 3^2 and 4^2, but also not a square.
My idea was to add 1 to both sides and express both sides as n^k-1 = (n-1)(n^(k-1)+...+n+1) and multiplying both sides by the denominator. It automatically rules out x=1 and y=1, but it entirely misses the y=2 case, bcs it's unnoticable in the product
pls solve challenges and thrills of pre-college mathematics quadratic equation question number 28....the question If a,b are positive integers and 1/x+ 1(x-a)+ 1/(x-b) =0 has real roots then prove that one of the roots lie between a/3 and 2/a3 and the other one between -b/3 and -2b/3. I will be very thankful if you do before sept 3 as i have IOQM exam....i googled this up but i was not satisfied. Thank you
Tricky question, I mean, it"s not hard to follow your explanation but not in a million years would I possibly come up by myself with this approach to solve the problem, so well done and I miss your videos, hope you come back soon with new videos... Cheers and hope you're doing well
Wow! Good question, good solution!
very nice. I love the logic how you solve whole number solution questions.
Since it’s between two perfect squares, can I not get this solution by finding the difference between 4y^4+…+1 and the other expression in the inequality and setting that value equal to 0 so then the expressions become equal. This yields the same solutions
Setting the expression 4y^4+4y^3+4y^2+4y+1 to either (2y+y)^2 or (2y+y+1)^2 does get the solutions, but the expression is not necessarily bounded by those squares. It could potentially equal some other square number. Hence the case y=1; the expression is 17, not between 3^2 and 4^2, but also not a square.
Great sir, love from India
My idea was to add 1 to both sides and express both sides as n^k-1 = (n-1)(n^(k-1)+...+n+1) and multiplying both sides by the denominator. It automatically rules out x=1 and y=1, but it entirely misses the y=2 case, bcs it's unnoticable in the product
Very good! but the case y=1 had already been ruled out by the second difference.
No way!
pls solve challenges and thrills of pre-college mathematics quadratic equation question number 28....the question
If a,b are positive integers and 1/x+ 1(x-a)+ 1/(x-b) =0 has real roots then prove that one of the roots lie between a/3 and 2/a3 and the other one between -b/3 and -2b/3.
I will be very thankful if you do before sept 3 as i have IOQM exam....i googled this up but i was not satisfied. Thank you
But when (x, y)= (0,0)...(0= integer), the equation still hold true ??
Yes, that is when y = -1, look at 7:00.