Menu Close

If-the-roots-of-9x-2-2x-7-0-are-2-more-than-the-roots-of-ax-2-bx-c-0-then-4a-2b-c-can-be-




Question Number 35534 by jonesme last updated on 20/May/18
If the roots of 9x^2 −2x+7=0 are 2 more  than the roots of ax^2 +bx+c=0, then  4a−2b+c   can be
$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{9}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}=\mathrm{0}\:\mathrm{are}\:\mathrm{2}\:\mathrm{more} \\ $$$$\mathrm{than}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{4}{a}−\mathrm{2}{b}+{c}\:\:\:\mathrm{can}\:\mathrm{be} \\ $$
Answered by ajfour last updated on 20/May/18
⇒ a(x+2)^2 +b(x+2)+c =         9x^2 −2x+7  ⇒  a=9 ,  4a+b=−2 , 4a+2b+c=7  ⇒   b=−38   ,  c=47  ⇒ 4a−2b+c =36+76+47 =159 .
$$\Rightarrow\:{a}\left({x}+\mathrm{2}\right)^{\mathrm{2}} +{b}\left({x}+\mathrm{2}\right)+{c}\:= \\ $$$$\:\:\:\:\:\:\:\mathrm{9}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7} \\ $$$$\Rightarrow\:\:{a}=\mathrm{9}\:,\:\:\mathrm{4}{a}+{b}=−\mathrm{2}\:,\:\mathrm{4}{a}+\mathrm{2}{b}+{c}=\mathrm{7} \\ $$$$\Rightarrow\:\:\:{b}=−\mathrm{38}\:\:\:,\:\:{c}=\mathrm{47} \\ $$$$\Rightarrow\:\mathrm{4}{a}−\mathrm{2}{b}+{c}\:=\mathrm{36}+\mathrm{76}+\mathrm{47}\:=\mathrm{159}\:. \\ $$

Leave a Reply

Your email address will not be published.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