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Albert1
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$a,b$ are all integers
(1) $if : \,\, 5a^2+5ab+5b^2=7a+14b$
please find all solutions of $(a,b)$
(1) $if : \,\, 5a^2+5ab+5b^2=7a+14b$
please find all solutions of $(a,b)$
thanks ,your answer is quite right(Yes)MarkFL said:My solution:
Applying the quadratic formula, we find:
\(\displaystyle b=\frac{(14-5a)\pm\sqrt{14^2-3(5a)^2}}{10}\)
Now we require:
\(\displaystyle 14^2-3(5a)^2\ge0\)
which implies:
\(\displaystyle a\in\{-1,0,1\}\)
Case 1: \(\displaystyle a=-1\)
\(\displaystyle b=\frac{19\pm11}{10}\)
The integral value is \(\displaystyle b=3\)
Hence, $(a,b)=(-1,3)$ is a solution.
Case 2: \(\displaystyle a=0\)
\(\displaystyle b=\frac{14\pm14}{10}\)
The integral value is \(\displaystyle b=0\)
Hence, $(a,b)=(0,0)$ is a solution.
Case 3: \(\displaystyle a=1\)
\(\displaystyle b=\frac{9\pm11}{10}\)
The integral value is \(\displaystyle b=2\)
Hence, $(a,b)=(1,2)$ is a solution.
not correct as belowAlbert said:my solution:
let x=2a+b,and y=a+2b
$ \therefore a=\dfrac {2x-y}{3},\,\,y=\dfrac {2y-x}{3}$
$5(a^2+ab+b^2)=7a+14b,\,\,becomes:$
$5(x^2-xy+y^2)=21y$
and we get :$y=5---(1),\,\, x^2-xy+y^2=21---(2)$
from (1)(2) (x,y)=(1,5)and (4,5)
and the corresponding (a,b)=(-1,3) and (1,2)
of course the third solution of (a,b)=(0,0)
$5(x^2−xy+y^2)=21y----(@)$kaliprasad said:not correct as below
$5(x^2−xy+y^2)=21y$
and we get :y=5−−−(1),$x^2−xy+y^2=21$−−−(2)
is not strictly correctit could be
y=5k−−−(1),$x^2−xy+y^2=21k$−−−(2)
Albert said:$5(x^2−xy+y^2)=21y----(@)$
we get :$21y\geq 5xy$ $(AP\geq GP)$
$\therefore x\leq 4$ (x,y being integers)
if x=4 ,y=5k
from (@) :$16-20k+25k^2=21k$
$16+25k^2=41k$
k must =1
$21y=5(x^2-xy+y^2)\geq 5(2xy-xy)=5xy$kaliprasad said:could you explain me how
we get :21y≥5xy (AM≥GM) AM and GM of what ?
I an sorry that I could not understand
"Find all solutions" refers to finding all possible combinations or values of a and b that satisfy a given equation or problem. It involves determining the range of values for a and b that would make the equation or problem true.
To find all solutions of an equation, you can use algebraic manipulation, substitution, or graphical methods. You can also use trial and error by plugging in different values for a and b until you find the correct solution.
Yes, there can be limitations to finding all solutions of a problem. Some equations may have an infinite number of solutions, making it impossible to list all of them. Additionally, some solutions may be complex numbers, which may not be relevant to the given problem.
Yes, there can be multiple sets of solutions for a given problem. This means that there can be more than one combination of values for a and b that satisfy the equation or problem. It is important to carefully consider all possible solutions before concluding that a problem has only one set of solutions.
If you are confident that you have considered all possible combinations of values for a and b and have found a solution for each one, then you can be sure that you have found all solutions of the problem. However, it is always a good idea to double check your work and make sure that you have not missed any potential solutions.