I substituted x = sqrt(a) and y = sqrt(b), then divided by y^2 and effectively solved for sqrt(a/b) = x/y. Then just squared the result. Worked fine as well.
This may be a hard problem for an introductory round to the math olympiad. As far as math olympiad questions go, this is very easy. Like 5th grade versus honors calculus easy.
Thank you for your feedback! In my opinion, it might be better to rewrite root(725) as 5*root(29) since this form breaks the number into its prime factors.
Hello! Sure, if we divide throughout by a^2, we end up with: 1-27b/a+(b/a)^2 = 0. Then assuming x= a/b, we would still end up with the same quadratic: x^2 -27x +1 = 0
Or, just divide by b in the original equation and make the same substitution, saving a few steps. Solving for x, and recognize that x^2 is just 5x + 1, gives our answer.
I substituted x = sqrt(a) and y = sqrt(b), then divided by y^2 and effectively solved for sqrt(a/b) = x/y. Then just squared the result. Worked fine as well.
Very good solution!
This may be a hard problem for an introductory round to the math olympiad. As far as math olympiad questions go, this is very easy. Like 5th grade versus honors calculus easy.
Great video! Just curious, is it better to rewrite root(725) as 5*root(29) or not?
Thank you for your feedback!
In my opinion, it might be better to rewrite root(725) as 5*root(29) since this form breaks the number into its prime factors.
Divide Step 2 by b^2 instead
Indeed!
wouldnt is be easier to devide by a^2 isntead of ab?
Hello!
Sure, if we divide throughout by a^2, we end up with: 1-27b/a+(b/a)^2 = 0. Then assuming x= a/b, we would still end up with the same quadratic: x^2 -27x +1 = 0
Or, just divide by b in the original equation and make the same substitution, saving a few steps. Solving for x, and recognize that x^2 is just 5x + 1, gives our answer.