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

$G i v e n f (x^{2} + x) + 2 f (x^{2} - 3 x + 2) = 9 x^{2} - 15 x$ $f i n d t h e v a l u e o f f (2017) .$
	Answered by MJS_new last updated on 02/Apr/21

	$f (t) = a t + b$ $f (x^{2} + x) = {a x}^{2} + a x + b$ $2 f (x^{2} - 3 x + 2) = 2 {a x}^{2} - 6 a x + 4 a + 2 b$ $3 {a x}^{2} - 5 a x + 4 a + 3 b = 9 x^{2} - 15 x$ $3 a = 9$ $- 5 a = - 15$ $4 a + 3 b = 0$ $\Rightarrow a = 3 \land b = - 4$ $f (x) = 3 x - 4$ $f (2017) = 6047$
	Commented by liberty last updated on 02/Apr/21

	$t h a n k y o u$
	Answered by EDWIN88 last updated on 02/Apr/21
	$let { ((x^2 +x = a)),((x^2 −3x+2 = b)) :} ⇒ { ((x^2 +x = a)),((2x^2 −6x+4 = 2b)) :} adding two equation gives 3x^2 −5x+4 = a+2b ⇒ 3x^2 −5x = a+2b−4 & 9x^2 −15x = 3a+6b−12 we have f(a)+2f(b) = 3a+6b−12 put a=2017⇒ f(2017)+2f(b)=6b+6039 put b=2017⇒f(a)+2f(2017)=3a+12090 summing the two equation gives 3f(2017)+f(a)+2f(b)=3a+6b+18129 3f(2017)+f(a)+2f(b)_(3a+6b−12) = 3a+6b+18129 ⇔ 3f(2017)= 18129+12=18141 ⇔ f(2017) = ((18141)/3) = 6047$
	$let {\begin{cases} x^{2} + x = a \\ x^{2} - 3 x + 2 = b \end{cases} \Rightarrow {\begin{cases} x^{2} + x = a \\ 2 x^{2} - 6 x + 4 = 2 b \end{cases}$ $a d d i n g t w o e q u a t i o n g i v e s 3 x^{2} - 5 x + 4 = a + 2 b$ $\Rightarrow 3 x^{2} - 5 x = a + 2 b - 4 & 9 x^{2} - 15 x = 3 a + 6 b - 12$ $w e h a v e f (a) + 2 f (b) = 3 a + 6 b - 12$ $p u t a = 2017 \Rightarrow f (2017) + 2 f (b) = 6 b + 6039$ $p u t b = 2017 \Rightarrow f (a) + 2 f (2017) = 3 a + 12090$ $s u m m i n g t h e t w o e q u a t i o n g i v e s$ $3 f (2017) + f (a) + 2 f (b) = 3 a + 6 b + 18129$ $3 f (2017) + \underset{3 a + 6 b - 12}{\underset{⏟}{f (a) + 2 f (b)}} = 3 a + 6 b + 18129$ $\Leftrightarrow 3 f (2017) = 18129 + 12 = 18141$ $\Leftrightarrow f (2017) = \frac{18141}{3} = 6047$
	Commented by liberty last updated on 02/Apr/21

	$n i c e$
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