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Question Number 136667 by mathlove last updated on 24/Mar/21
if      f(x)=ax^2 +bx+c      and   f(5)=−3f(2)  an intersiction  point  (−4,0)  an the  X-axis     faind another point   the X-axis
$${if}\:\:\:\:\:\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\:\:\:\:\:{and}\:\:\:{f}\left(\mathrm{5}\right)=−\mathrm{3}{f}\left(\mathrm{2}\right) \\ $$$${an}\:{intersiction}\:\:{point}\:\:\left(−\mathrm{4},\mathrm{0}\right) \\ $$$${an}\:{the}\:\:{X}-{axis}\:\:\:\:\:{faind}\:{another}\:{point}\: \\ $$$${the}\:{X}-{axis} \\ $$$$ \\ $$
Answered by MJS_new last updated on 25/Mar/21
I should have finished it  (1)  25a+5b+c=−3(4a+2b+c) ⇔ 37a+11b+4c=0  (2)  16a−4b+c=0  (1)  ⇒ c=−((37a+11b)/4)  ⇒ (2)  ((27)/4)a−((27)/4)b=0 ⇒ b=a ⇒ c=−12a  ⇒ f(x)=a(x^2 +x−12)=a(x−3)(x+4)  ⇒ zeros are −4 and 3
$$\mathrm{I}\:\mathrm{should}\:\mathrm{have}\:\mathrm{finished}\:\mathrm{it} \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{25}{a}+\mathrm{5}{b}+{c}=−\mathrm{3}\left(\mathrm{4}{a}+\mathrm{2}{b}+{c}\right)\:\Leftrightarrow\:\mathrm{37}{a}+\mathrm{11}{b}+\mathrm{4}{c}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{16}{a}−\mathrm{4}{b}+{c}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\:\Rightarrow\:{c}=−\frac{\mathrm{37}{a}+\mathrm{11}{b}}{\mathrm{4}} \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)\:\:\frac{\mathrm{27}}{\mathrm{4}}{a}−\frac{\mathrm{27}}{\mathrm{4}}{b}=\mathrm{0}\:\Rightarrow\:{b}={a}\:\Rightarrow\:{c}=−\mathrm{12}{a} \\ $$$$\Rightarrow\:{f}\left({x}\right)={a}\left({x}^{\mathrm{2}} +{x}−\mathrm{12}\right)={a}\left({x}−\mathrm{3}\right)\left({x}+\mathrm{4}\right) \\ $$$$\Rightarrow\:\mathrm{zeros}\:\mathrm{are}\:−\mathrm{4}\:\mathrm{and}\:\mathrm{3} \\ $$
Answered by ajfour last updated on 25/Mar/21
25a+5b+c=−3(4a+2b+c)  ⇒  37a+11b+4c=0  (c/a)=−((11)/4)((b/a))−((37)/4)  x=−(b/(2a))±((√(b^2 −4ac))/(2a))  x=−(b/(2a))±(√((1/4)((b/a))^2 −(c/a)))  x=(b/a){−(1/2)±(√((1/4)+((11)/(4(b/a)))+((37)/(4(b/a)^2 )))) }  say  (b/a)=p  (((−4)/p)+(1/2))^2 =(1/4)+((11)/(4p))+((37)/(4p^2 ))  ((16)/p^2 )−(4/p)=((11)/(4p))+((37)/(4p^2 ))  ((27)/(4p^2 ))−((27)/(4p))=0  ⇒  p=1  x={−(1/2)±(√((1/4)+((11)/4)+((37)/4))) }  x=−(1/2)±(7/2)   x=−4 or 3  other point on x-axis is  (3,0)
$$\mathrm{25}{a}+\mathrm{5}{b}+{c}=−\mathrm{3}\left(\mathrm{4}{a}+\mathrm{2}{b}+{c}\right) \\ $$$$\Rightarrow\:\:\mathrm{37}{a}+\mathrm{11}{b}+\mathrm{4}{c}=\mathrm{0} \\ $$$$\frac{{c}}{{a}}=−\frac{\mathrm{11}}{\mathrm{4}}\left(\frac{{b}}{{a}}\right)−\frac{\mathrm{37}}{\mathrm{4}} \\ $$$${x}=−\frac{{b}}{\mathrm{2}{a}}\pm\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${x}=−\frac{{b}}{\mathrm{2}{a}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} −\frac{{c}}{{a}}} \\ $$$${x}=\frac{{b}}{{a}}\left\{−\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{4}\left({b}/{a}\right)}+\frac{\mathrm{37}}{\mathrm{4}\left({b}/{a}\right)^{\mathrm{2}} }}\:\right\} \\ $$$${say}\:\:\frac{{b}}{{a}}={p} \\ $$$$\left(\frac{−\mathrm{4}}{{p}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{4}{p}}+\frac{\mathrm{37}}{\mathrm{4}{p}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{16}}{{p}^{\mathrm{2}} }−\frac{\mathrm{4}}{{p}}=\frac{\mathrm{11}}{\mathrm{4}{p}}+\frac{\mathrm{37}}{\mathrm{4}{p}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{27}}{\mathrm{4}{p}^{\mathrm{2}} }−\frac{\mathrm{27}}{\mathrm{4}{p}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}=\mathrm{1} \\ $$$${x}=\left\{−\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{4}}+\frac{\mathrm{37}}{\mathrm{4}}}\:\right\} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{7}}{\mathrm{2}}\: \\ $$$${x}=−\mathrm{4}\:{or}\:\mathrm{3} \\ $$$${other}\:{point}\:{on}\:{x}-{axis}\:{is} \\ $$$$\left(\mathrm{3},\mathrm{0}\right) \\ $$

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