Menu Close

Question-49279

Tinku Tara 0 Comments
Pin




Question Number 49279 by Tawa1 last updated on 05/Dec/18
Commented by MJS last updated on 05/Dec/18
x−4=u 2x−5=v u^(2017) +u^(2015) +u=−(v^(2017) +v^(2015) +v) ⇒ v=−u ⇔ u+v=0 3x−9=0 x=3 we can put u^n +u^(n−2) +u+v^n +v^(n−2) +v=0 and test for n=2k+1; k∈N^★ if u^(2k+1) +u^(2k−1) +u+v^(2k+1) +v^(2k−1) +v=0 has got real solutions I tested for k={1, 2, 3, 4, 5} and found no real solutions except v=−u but I can′t prove it
x−4=u2x−5=vu2017+u2015+u=−(v2017+v2015+v)⇒v=−u⇔u+v=03x−9=0x=3wecanputun+un−2+u+vn+vn−2+v=0andtestforn=2k+1;k∈N★ifu2k+1+u2k−1+u+v2k+1+v2k−1+v=0hasgotrealsolutionsItestedfork={1,2,3,4,5}andfoundnorealsolutionsexceptv=−ubutIcan′tproveit
Answered by afachri last updated on 05/Dec/18
i think the first solution is : (2x − 5) + (x−4) + (2x − 5) + (x − 4) + 3x − 9 = 0 9x − 27 = 0 ⇒ x = 3 check if x = 3 : (2x − 5)^(2017) + (x − 4)^(2017) + (2x − 5)^(2015) + (x − 4)^(2015) + (3x − 9) = 0 1^(2017) + (−1)^(2017) + 1^(2015) + (−1)^(2015) + 0 = 0 1 − 1 + 1 − 1 + 0 = 0 i have no ideas about any other solution. i am just a rookie one. please correct me gently if i′m wrong and teach me please.
ithinkthefirstsolutionis:(2x−5)+(x−4)+(2x−5)+(x−4)+3x−9=09x−27=0⇒x=3checkifx=3:(2x−5)2017+(x−4)2017+(2x−5)2015+(x−4)2015+(3x−9)=012017+(−1)2017+12015+(−1)2015+0=01−1+1−1+0=0ihavenoideasaboutanyothersolution.iamjustarookieone.pleasecorrectmegentlyifi′mwrongandteachmeplease.
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
f(x)=(2x−5)^(2017) +(2x−5)^(2015) +2x−5+(x−4)^(2017) +(x−4)^(2015) +(x−4) critical value (5/2) and 4 f(x)<0 at x=2.5 f(x)>0 at x=4 so from x=2.5 to x=4 sign of f(x) changes so one root of x must lie between 2.5 to 4 now put x=3 f(x)=0 so x=3
f(x)=(2x−5)2017+(2x−5)2015+2x−5+(x−4)2017+(x−4)2015+(x−4)criticalvalue52and4f(x)<0atx=2.5f(x)>0atx=4sofromx=2.5tox=4signoff(x)changessoonerootofxmustliebetween2.5to4nowputx=3f(x)=0sox=3

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *