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Question-195900




Question Number 195900 by Calculusboy last updated on 12/Aug/23
Answered by cortano12 last updated on 13/Aug/23
   [ (1/3)x^3 +(1/2)x^2  ]_(−2) ^b =((16)/3)     [(1/2)x^2 ((2/3)x+1) ]_(−2) ^b = ((16)/3)     (1/2)b^2 ((2/3)b+1)−2(−(1/3))= ((16)/3)      (1/3)b^3 +(1/2)b^2 −((14)/3)=0     2b^3 +3b^2 −28 = 0     (b−2)(2b^2 +7b+14)=0      b = 2      2b^2 +7b+28=0      b=((−7+3i(√7))/4) ; b=((−7−3i(√7))/4)
$$\:\:\:\left[\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:\right]_{−\mathrm{2}} ^{\mathrm{b}} =\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\:\:\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}+\mathrm{1}\right)\:\right]_{−\mathrm{2}} ^{\mathrm{b}} =\:\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{b}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\mathrm{b}+\mathrm{1}\right)−\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)=\:\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{b}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{b}^{\mathrm{2}} −\frac{\mathrm{14}}{\mathrm{3}}=\mathrm{0} \\ $$$$\:\:\:\mathrm{2b}^{\mathrm{3}} +\mathrm{3b}^{\mathrm{2}} −\mathrm{28}\:=\:\mathrm{0} \\ $$$$\:\:\:\left(\mathrm{b}−\mathrm{2}\right)\left(\mathrm{2b}^{\mathrm{2}} +\mathrm{7b}+\mathrm{14}\right)=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{b}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\mathrm{2b}^{\mathrm{2}} +\mathrm{7b}+\mathrm{28}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{b}=\frac{−\mathrm{7}+\mathrm{3i}\sqrt{\mathrm{7}}}{\mathrm{4}}\:;\:\mathrm{b}=\frac{−\mathrm{7}−\mathrm{3i}\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$
Commented by Calculusboy last updated on 13/Aug/23
thanks
$${thanks} \\ $$

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