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Question Number 33899 by mondodotto@gmail.com last updated on 27/Apr/18
express ((4t^2 −28)/(t^4 +t^2 −6)) as a partial fraction.
$$\boldsymbol{\mathrm{express}}\:\frac{\mathrm{4}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{28}}{\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{6}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{partial}}\:\boldsymbol{\mathrm{fraction}}. \\ $$
Answered by MJS last updated on 27/Apr/18
t^4 +t^2 −6=0 t^2 =−(1/2)±(√((1/4)+6))=−(1/2)±(5/2)= −3 ∨ 2 t^4 +t^2 −6=(t^2 +3)(t^2 −2) ((4t^2 −28)/(t^4 +t^2 −6))=(A/(t^2 +3))+(B/(t^2 −2))=((A(t^2 −2)+B(t^2 +3))/(t^4 +t^2 −6)) (A+B)t^2 +(−2A+3B)=4t^2 −28 A+B=4 ∧ −2A+3B=−28 A=4−B −2(4−B)+3B=−28 B=−4 A=8 ((4t^2 −28)/(t^4 +t^2 −6))=(8/(t^2 +3))−(4/(t^2 −2))
$${t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{6}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{6}}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{5}}{\mathrm{2}}=\:−\mathrm{3}\:\vee\:\mathrm{2} \\ $$$${t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{6}=\left({t}^{\mathrm{2}} +\mathrm{3}\right)\left({t}^{\mathrm{2}} −\mathrm{2}\right) \\ $$$$\frac{\mathrm{4}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{28}}{\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{6}}=\frac{{A}}{{t}^{\mathrm{2}} +\mathrm{3}}+\frac{{B}}{{t}^{\mathrm{2}} −\mathrm{2}}=\frac{{A}\left({t}^{\mathrm{2}} −\mathrm{2}\right)+{B}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{6}} \\ $$$$\left({A}+{B}\right){t}^{\mathrm{2}} +\left(−\mathrm{2}{A}+\mathrm{3}{B}\right)=\mathrm{4}{t}^{\mathrm{2}} −\mathrm{28} \\ $$$${A}+{B}=\mathrm{4}\:\wedge\:−\mathrm{2}{A}+\mathrm{3}{B}=−\mathrm{28} \\ $$$${A}=\mathrm{4}−{B} \\ $$$$−\mathrm{2}\left(\mathrm{4}−{B}\right)+\mathrm{3}{B}=−\mathrm{28} \\ $$$${B}=−\mathrm{4} \\ $$$${A}=\mathrm{8} \\ $$$$\frac{\mathrm{4}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{28}}{\boldsymbol{\mathrm{t}}^{\mathrm{4}} +\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{6}}=\frac{\mathrm{8}}{{t}^{\mathrm{2}} +\mathrm{3}}−\frac{\mathrm{4}}{{t}^{\mathrm{2}} −\mathrm{2}} \\ $$
Commented by math1967 last updated on 27/Apr/18
I can also write A(t^2 −2)+B(t^2 +3)=4t^2 −28 putting t^2 =2 A×0+B×5=4×2−28 ∴B=−4 similarly we get A=8
$${I}\:{can}\:{also}\:{write} \\ $$$${A}\left({t}^{\mathrm{2}} −\mathrm{2}\right)+{B}\left({t}^{\mathrm{2}} +\mathrm{3}\right)=\mathrm{4}{t}^{\mathrm{2}} −\mathrm{28} \\ $$$${putting}\:{t}^{\mathrm{2}} =\mathrm{2} \\ $$$${A}×\mathrm{0}+{B}×\mathrm{5}=\mathrm{4}×\mathrm{2}−\mathrm{28} \\ $$$$\therefore{B}=−\mathrm{4}\:{similarly}\:{we}\:{get}\:{A}=\mathrm{8} \\ $$
Answered by math1967 last updated on 27/Apr/18
((8(t^2 −2)−4(t^2 +3))/((t^2 −2)(t^2 +3))) (8/(t^2 +3)) −(4/(t^2 −2))
$$\frac{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}{\left({t}^{\mathrm{2}} −\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{3}\right)} \\ $$$$\frac{\mathrm{8}}{{t}^{\mathrm{2}} +\mathrm{3}}\:−\frac{\mathrm{4}}{{t}^{\mathrm{2}} −\mathrm{2}} \\ $$

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