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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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