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Question-157853




Question Number 157853 by emanuelMcCarthy last updated on 28/Oct/21
Commented by cortano last updated on 29/Oct/21
(21)f(2)=−20            16+4p+16=−20            p+8=−5⇒p=−13  (22) f(−a)=−56         2a^3 −3a^2 −13ab+20=−56        2a^3 −3a^2 −13ab+76=0...(1)   f(a)=0    −2a^3 −3a^2 +13ab+20=0...(2)  (1)+(2)⇒−6a^2 =−96  ⇒a=± 4  (1)−(2)⇒4a^3 −26ab+56=0  ⇒b=((2a^3 +28)/(13a))    { ((a=4⇒b=((2.4^3 +28)/(13.4))=((32+7)/(13))=3)),((a=−4⇒b=((2(−4)^3 +28)/(13(−4)))=((32−7)/(13))=((25)/(13)))) :}
$$\left(\mathrm{21}\right){f}\left(\mathrm{2}\right)=−\mathrm{20}\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{16}+\mathrm{4}{p}+\mathrm{16}=−\mathrm{20} \\ $$$$\:\:\:\:\:\:\:\:\:\:{p}+\mathrm{8}=−\mathrm{5}\Rightarrow{p}=−\mathrm{13} \\ $$$$\left(\mathrm{22}\right)\:{f}\left(−{a}\right)=−\mathrm{56} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{13}{ab}+\mathrm{20}=−\mathrm{56} \\ $$$$\:\:\:\:\:\:\mathrm{2}{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{13}{ab}+\mathrm{76}=\mathrm{0}…\left(\mathrm{1}\right) \\ $$$$\:{f}\left({a}\right)=\mathrm{0} \\ $$$$\:\:−\mathrm{2}{a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} +\mathrm{13}{ab}+\mathrm{20}=\mathrm{0}…\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow−\mathrm{6}{a}^{\mathrm{2}} =−\mathrm{96} \\ $$$$\Rightarrow{a}=\pm\:\mathrm{4} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\Rightarrow\mathrm{4}{a}^{\mathrm{3}} −\mathrm{26}{ab}+\mathrm{56}=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{2}{a}^{\mathrm{3}} +\mathrm{28}}{\mathrm{13}{a}} \\ $$$$\:\begin{cases}{{a}=\mathrm{4}\Rightarrow{b}=\frac{\mathrm{2}.\mathrm{4}^{\mathrm{3}} +\mathrm{28}}{\mathrm{13}.\mathrm{4}}=\frac{\mathrm{32}+\mathrm{7}}{\mathrm{13}}=\mathrm{3}}\\{{a}=−\mathrm{4}\Rightarrow{b}=\frac{\mathrm{2}\left(−\mathrm{4}\right)^{\mathrm{3}} +\mathrm{28}}{\mathrm{13}\left(−\mathrm{4}\right)}=\frac{\mathrm{32}−\mathrm{7}}{\mathrm{13}}=\frac{\mathrm{25}}{\mathrm{13}}}\end{cases} \\ $$

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