This is really an interesting video showing how simple tricks could be used to solve Olympiad problems on permutation and combination. Please subscribe to my channel for more interesting videos. @begechorimaths2975
Пікірлер: 4
@sarantis40kalaitzis48Ай бұрын
Squring Both Sides. We get a^2+b^2+2ab=36ab so a^2+b^2=34ab. We DIVIDE Both Sides with b^2>0, hence (a^2/b^2)+(b^2/b^2)=34ab/b^2 so (a/b)^2+1=34*(a/b). Setting a/b=x we get x^2+1=34x so x^2-34x+1=0 so x=(34+ -sqrt(34^2-4))/2. Using trick of Difference of squares 34^2-2^2=(34+2)*(34-2)=36*32 so x= (34+ -6*sqrt32)/2= 17+ - 3sqrt32= 17+ - 3*sqrt16*sqrt2=17+ -12*sqrt2.
@pmac_Ай бұрын
Easier to divide original equation by b giving: x² + 1 = 6 sqrt(x). Square both sides gives: x² - 34x +1 = 0 Simple.
@begechorimaths2975Ай бұрын
@@pmac_ Thank you!
@juanalfaro7522Ай бұрын
Let a/b = z. Then z+1 = 6*sqrt (z) --> (z+1) ^2 = 6^2 * z -> z^2 + 2z + 1 = 36z -> z^2 - 34z + 1 = 0 -> z^2 - 34z + 289 = 288 -> (z-17) ^2 = 2*14^2 -> z=a/b = 17 +/- 12*sqrt (2) [I.e., there are 2 solutions]