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Question-141669




Question Number 141669 by bramlexs22 last updated on 22/May/21
Answered by iloveisrael last updated on 22/May/21
(i) f(a)=g(a) ,a>0  ⇒ a+∣2a∣ = −a^2 −(2/3)a+((14)/3)  ⇒3a =−a^2 −(2/3)a+((14)/3)  ⇒3a^2 +9a+2a−14=0  ⇒3a^2 +11a−14=0  (ii) f(c)=g(c); c<0  ⇒−c=−c^2 −(2/3)c+((14)/3)  ⇒3c^2 −3c+2c−14=0  ⇒3c^2 −c−14=0
$$\left({i}\right)\:{f}\left({a}\right)={g}\left({a}\right)\:,{a}>\mathrm{0} \\ $$$$\Rightarrow\:{a}+\mid\mathrm{2}{a}\mid\:=\:−{a}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{a}+\frac{\mathrm{14}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{a}\:=−{a}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{a}+\frac{\mathrm{14}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} +\mathrm{9}{a}+\mathrm{2}{a}−\mathrm{14}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} +\mathrm{11}{a}−\mathrm{14}=\mathrm{0} \\ $$$$\left({ii}\right)\:{f}\left({c}\right)={g}\left({c}\right);\:{c}<\mathrm{0} \\ $$$$\Rightarrow−{c}=−{c}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{c}+\frac{\mathrm{14}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{c}^{\mathrm{2}} −\mathrm{3}{c}+\mathrm{2}{c}−\mathrm{14}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{c}^{\mathrm{2}} −{c}−\mathrm{14}=\mathrm{0} \\ $$$$ \\ $$

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