Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 129090 by Adel last updated on 12/Jan/21

p(x)+p(x+2)=2X^2 +2X+4        p(x)=?

$$\mathrm{p}\left(\mathrm{x}\right)+\mathrm{p}\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{2X}^{\mathrm{2}} +\mathrm{2X}+\mathrm{4}\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{p}}\left(\boldsymbol{\mathrm{x}}\right)=? \\ $$

Answered by MJS_new last updated on 12/Jan/21

p(x)=ax^2 +bx+c  p(x+2)=ax^2 +(4a+b)x+(4a+2b+c)  ⇒  2ax^2 +2(2a+b)x+2(2a+b+c)=2x^2 +2x+4  ⇒ a=1∧b=−3∧c=−1  ⇒  p(x)=x^2 −3x−1

$${p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${p}\left({x}+\mathrm{2}\right)={ax}^{\mathrm{2}} +\left(\mathrm{4}{a}+{b}\right){x}+\left(\mathrm{4}{a}+\mathrm{2}{b}+{c}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{ax}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}{a}+{b}\right){x}+\mathrm{2}\left(\mathrm{2}{a}+{b}+{c}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4} \\ $$$$\Rightarrow\:{a}=\mathrm{1}\wedge{b}=−\mathrm{3}\wedge{c}=−\mathrm{1} \\ $$$$\Rightarrow \\ $$$${p}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1} \\ $$

Commented by Adel last updated on 12/Jan/21

tanks

$$\mathrm{tanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com