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Question Number 180069 by Acem last updated on 06/Nov/22
a^2 −b^2 = 4 , ab= 2 , a+b= ?
$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\:\mathrm{4}\:,\:{ab}=\:\mathrm{2}\:\:\:,\:{a}+{b}=\:? \\ $$
Commented by Acem last updated on 07/Nov/22
Have a good day my brothers mr. W and mr. Rasheed
$${Have}\:{a}\:{good}\:{day}\:{my}\:{brothers} \\ $$$$\:{mr}.\:{W}\:{and}\:{mr}.\:{Rasheed} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Nov/22
a^2 −b^2 =4, ab= 2 , a+b= ? a^2 −b^2 =4 (a−b)(a+b)=4 (a−b)^2 (a+b)^2 =4^2 (a^2 +b^2 −2ab)(a^2 +b^2 −2ab)=16 (a^2 +b^2 )^2 −( 2(2) )^2 =16 (a^2 +b^2 )^2 =32 a^2 +b^2 =±4(√2) (a+b)^2 =a^2 +b^2 +2ab=±4(√2) +2(2) =4±4(√2) a+b=2(√(1±(√2) ))
$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{4},\:{ab}=\:\mathrm{2}\:\:\:,\:{a}+{b}=\:? \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{4} \\ $$$$\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{4} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right)=\mathrm{16} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\:\mathrm{2}\left(\mathrm{2}\right)\:\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{32} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\pm\mathrm{4}\sqrt{\mathrm{2}}\: \\ $$$$ \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}=\pm\mathrm{4}\sqrt{\mathrm{2}}\:+\mathrm{2}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\pm\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${a}+{b}=\mathrm{2}\sqrt{\mathrm{1}\pm\sqrt{\mathrm{2}}\:} \\ $$
Commented by Acem last updated on 07/Nov/22
Am sorry again, thank for you
$${Am}\:{sorry}\:{again},\:{thank}\:{for}\:{you} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Nov/22
No matter, it′s a misprint only (not a mistake); I′ve edited my answers.
$${No}\:{matter},\:{it}'{s}\:{a}\:{misprint}\:{only}\:\left({not}\:{a}\right. \\ $$$$\left.{mistake}\right);\:{I}'{ve}\:{edited}\:{my}\:{answers}. \\ $$
Commented by Acem last updated on 06/Nov/22
I am very sorry , a^2 −b^2 = 4 , i don′t know why the =4 didn′t typed though i did... sorry again! Apologies for the inconvenience
$${I}\:{am}\:{very}\:{sorry}\:,\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\:\mathrm{4}\:,\:{i}\:{don}'{t}\:{know}\:{why} \\ $$$$\:{the}\:=\mathrm{4}\:{didn}'{t}\:{typed}\:{though}\:{i}\:{did}…\:{sorry}\:{again}! \\ $$$${Apologies}\:{for}\:{the}\:{inconvenience} \\ $$
Commented by CElcedricjunior last updated on 07/Nov/22
Attention!!! pourquoi mettre 1∓(√2) dans la racine
$${Attention}!!! \\ $$$$\boldsymbol{{pourquoi}}\:\boldsymbol{{mettre}}\:\mathrm{1}\mp\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{{dans}}\:\boldsymbol{{la}}\:\boldsymbol{{racine}} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Nov/22
Can you please translate in English sir!
$${Can}\:{you}\:{please}\:{translate}\:{in}\:\mathcal{E}{nglish}\:{sir}! \\ $$
Commented by Acem last updated on 08/Nov/22
Parce que si a,b∈ Z ⇒ a+b= ∓ 2 (√(1∓(√2))) Mais si a,b∈ R ⇒ a+b= ∓ 2 (√(1+(√2)))
$${Parce}\:{que}\:{si}\:{a},{b}\in\:\mathbb{Z}\:\Rightarrow\:{a}+{b}=\:\mp\:\mathrm{2}\:\sqrt{\mathrm{1}\mp\sqrt{\mathrm{2}}}\: \\ $$$$\:{Mais}\:{si}\:{a},{b}\in\:\mathbb{R}\:\Rightarrow\:{a}+{b}=\:\mp\:\mathrm{2}\:\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\: \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 07/Nov/22
a^2 −b^2 =4 , ab= 2 , a+b= ? (a−b)(a+b)=4 Let a+b=t....(i) a−b=(4/t)......(ii) (i)+(ii): 2a=t+(4/t)......(iii) (i)−(ii): 2b=t−(4/t)......(iv) (iii)×(iv): 4ab=t^2 −((16)/t^2 )=4(2)=8 t^4 −8t^2 −16=0 t^2 =((8±(√(64+64)))/2)=((8±8(√2))/2)=4±4(√5) t=(√(4±4(√2))) a+b=2(√(1±(√2)))
$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{4}\:,\:{ab}=\:\mathrm{2}\:\:\:,\:{a}+{b}=\:? \\ $$$$\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{4} \\ $$$${Let}\:{a}+{b}={t}….\left({i}\right) \\ $$$${a}−{b}=\frac{\mathrm{4}}{{t}}……\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}{a}={t}+\frac{\mathrm{4}}{{t}}……\left({iii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{2}{b}={t}−\frac{\mathrm{4}}{{t}}……\left({iv}\right) \\ $$$$\left({iii}\right)×\left({iv}\right): \\ $$$$\mathrm{4}{ab}={t}^{\mathrm{2}} −\frac{\mathrm{16}}{{t}^{\mathrm{2}} }=\mathrm{4}\left(\mathrm{2}\right)=\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:{t}^{\mathrm{4}} −\mathrm{8}{t}^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$$\:\:\:\:{t}^{\mathrm{2}} =\frac{\mathrm{8}\pm\sqrt{\mathrm{64}+\mathrm{64}}}{\mathrm{2}}=\frac{\mathrm{8}\pm\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{4}\pm\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\:\:\:{t}=\sqrt{\mathrm{4}\pm\mathrm{4}\sqrt{\mathrm{2}}}\: \\ $$$${a}+{b}=\mathrm{2}\sqrt{\mathrm{1}\pm\sqrt{\mathrm{2}}} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Nov/22
a^2 −b^2 =4 , ab= 2 , a+b= ? (a^2 −b^2 )^2 =(4)^2 a^4 +b^4 −2(ab)^2 =16 a^4 +b^4 −2(2)^2 =16 a^4 +b^4 =24 (a^2 +b^2 )^2 −2(ab)^2 =24 (a^2 +b^2 )^2 −2(2)^2 =24 (a^2 +b^2 )^2 =32 a^2 +b^2 =±4(√2) (a+b)^2 −2ab=±4(√5) (a+b)^2 −2(2)=±4(√2) (a+b)^2 =4±4(√2) a+b=2(√(1±(√2) ))
$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{4}\:,\:{ab}=\:\mathrm{2}\:\:\:,\:{a}+{b}=\:? \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} −\mathrm{2}\left({ab}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\:\:\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} −\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\:\:\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\mathrm{24} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({ab}\right)^{\mathrm{2}} =\mathrm{24} \\ $$$$\:\:\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{2}} =\mathrm{24} \\ $$$$\:\:\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{32} \\ $$$$\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\pm\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}=\pm\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}\right)=\pm\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{4}\pm\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${a}+{b}=\mathrm{2}\sqrt{\mathrm{1}\pm\sqrt{\mathrm{2}}\:}\: \\ $$
Commented by Rasheed.Sindhi last updated on 07/Nov/22
Yes sir and if a+b∈R,then a+b=±2(√(1+(√2) )) ⊤hanks sir!
$${Yes}\:\boldsymbol{{sir}}\:{and}\:{if}\:{a}+{b}\in\mathbb{R},{then} \\ $$$${a}+{b}=\pm\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}\:}\: \\ $$$$\top\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{sir}! \\ $$
Commented by mr W last updated on 07/Nov/22
why not a+b=±2(√(1±(√2) )) ?
$${why}\:{not} \\ $$$${a}+{b}=\pm\mathrm{2}\sqrt{\mathrm{1}\pm\sqrt{\mathrm{2}}\:}\:? \\ $$
Commented by mr W last updated on 07/Nov/22
agree!
$${agree}! \\ $$
Commented by Acem last updated on 08/Nov/22
Thanks for the many solutions and for your efforts! By the way i think that there′s ∓ before the last number, a+b= ± 2(√(1±(√2))) if c=a+bi ∈ Z ′See Mr. Frix′s note′
$${Thanks}\:{for}\:{the}\:{many}\:{solutions}\:{and}\:{for}\:{your} \\ $$$$\:{efforts}! \\ $$$$ \\ $$$${By}\:{the}\:{way}\:{i}\:{think}\:{that}\:{there}'{s}\:\mp\:{before}\:{the}\:{last} \\ $$$$\:{number},\:{a}+{b}=\:\pm\:\mathrm{2}\sqrt{\mathrm{1}\pm\sqrt{\mathrm{2}}}\:\:\:\:\:{if}\:{c}={a}+{bi}\:\in\:\mathbb{Z} \\ $$$$'{See}\:{Mr}.\:{Frix}'{s}\:{note}' \\ $$$$\: \\ $$
Commented by Frix last updated on 08/Nov/22
typo I guess... Z={... , −3, −2, −1, 0, 1, 2, 3, ...} you mean C={a+bi∣a, b ∈R∧i=(√(−1))}
$$\mathrm{typo}\:\mathrm{I}\:\mathrm{guess}… \\ $$$$\mathbb{Z}=\left\{…\:,\:−\mathrm{3},\:−\mathrm{2},\:−\mathrm{1},\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…\right\} \\ $$$$\mathrm{you}\:\mathrm{mean}\:\mathbb{C}=\left\{{a}+{b}\mathrm{i}\mid{a},\:{b}\:\in\mathbb{R}\wedge\mathrm{i}=\sqrt{−\mathrm{1}}\right\} \\ $$
Commented by Acem last updated on 08/Nov/22
Exactly Mr. Frix, you are right
$${Exactly}\:{Mr}.\:{Frix},\:{you}\:{are}\:{right} \\ $$
Answered by Acem last updated on 08/Nov/22
a^2 −b^2 = 4 a^4 +b^4 = 16+8∣_(+2a^2 b^2 ) (a^2 +b^2 )^2 = 24+8∣_(+2a^2 b^2 ) a^2 +b^2 = ∓4(√2) (a+b)^2 = ∓4(√2) +4∣_(+2ab) a+b= ∓ 2 (√(1+(√2))) ; a,b∈ R a+b= ∓ 2 (√(1∓(√2))) ; c∈ Z: c= a+ bi; a,b∈ R
$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\:\mathrm{4} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\:\mathrm{16}+\mathrm{8}\mid_{+\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} } \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} =\:\mathrm{24}+\mathrm{8}\mid_{+\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\:\mp\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} =\:\mp\mathrm{4}\sqrt{\mathrm{2}}\:+\mathrm{4}\mid_{+\mathrm{2}{ab}} \\ $$$$\boldsymbol{{a}}+\boldsymbol{{b}}=\:\mp\:\mathrm{2}\:\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\:\:\:;\:{a},{b}\in\:\mathbb{R} \\ $$$${a}+{b}=\:\mp\:\mathrm{2}\:\sqrt{\mathrm{1}\mp\sqrt{\mathrm{2}}}\:\:\:\:;\:{c}\in\:\mathbb{Z}:\:{c}=\:{a}+\:{bi};\:{a},{b}\in\:\mathbb{R} \\ $$

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