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Given-f-x-2-x-2f-x-2-3x-2-9x-2-15x-find-the-value-of-f-2017-

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Question Number 137365 by liberty last updated on 02/Apr/21
Given f(x^2 +x)+2f(x^2 −3x+2)= 9x^2 −15x find the value of f(2017).
Givenf(x2+x)+2f(x2−3x+2)=9x2−15xfindthevalueoff(2017).
Answered by MJS_new last updated on 02/Apr/21
f(t)=at+b f(x^2 +x)=ax^2 +ax+b 2f(x^2 −3x+2)=2ax^2 −6ax+4a+2b 3ax^2 −5ax+4a+3b=9x^2 −15x 3a=9 −5a=−15 4a+3b=0 ⇒ a=3∧b=−4 f(x)=3x−4 f(2017)=6047
f(t)=at+bf(x2+x)=ax2+ax+b2f(x2−3x+2)=2ax2−6ax+4a+2b3ax2−5ax+4a+3b=9x2−15x3a=9−5a=−154a+3b=0⇒a=3∧b=−4f(x)=3x−4f(2017)=6047
Commented by liberty last updated on 02/Apr/21
thank you
thankyou
Answered by EDWIN88 last updated on 02/Apr/21
let { ((x^2 +x = a)),((x^2 −3x+2 = b)) :} ⇒ { ((x^2 +x = a)),((2x^2 −6x+4 = 2b)) :} adding two equation gives 3x^2 −5x+4 = a+2b ⇒ 3x^2 −5x = a+2b−4 & 9x^2 −15x = 3a+6b−12 we have f(a)+2f(b) = 3a+6b−12 put a=2017⇒ f(2017)+2f(b)=6b+6039 put b=2017⇒f(a)+2f(2017)=3a+12090 summing the two equation gives 3f(2017)+f(a)+2f(b)=3a+6b+18129 3f(2017)+f(a)+2f(b)_(3a+6b−12) = 3a+6b+18129 ⇔ 3f(2017)= 18129+12=18141 ⇔ f(2017) = ((18141)/3) = 6047
let{x2+x=ax2−3x+2=b⇒{x2+x=a2x2−6x+4=2baddingtwoequationgives3x2−5x+4=a+2b⇒3x2−5x=a+2b−4&9x2−15x=3a+6b−12wehavef(a)+2f(b)=3a+6b−12puta=2017⇒f(2017)+2f(b)=6b+6039putb=2017⇒f(a)+2f(2017)=3a+12090summingthetwoequationgives3f(2017)+f(a)+2f(b)=3a+6b+181293f(2017)+f(a)+2f(b)⏟3a+6b−12=3a+6b+18129⇔3f(2017)=18129+12=18141⇔f(2017)=181413=6047
Commented by liberty last updated on 02/Apr/21
nice
nice

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