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Question Number 43902 by ajfour last updated on 17/Sep/18
(√(a−b)) + (√(a+b)) = c (√(a−c)) + (√(a+c)) = b Solve for real b, and c ; in terms of real a.
$$\sqrt{{a}−{b}}\:+\:\sqrt{{a}+{b}}\:=\:{c} \\ $$$$\sqrt{{a}−{c}}\:+\:\sqrt{{a}+{c}}\:=\:{b} \\ $$$${Solve}\:{for}\:{real}\:{b},\:{and}\:{c}\:;\:{in}\:{terms}\: \\ $$$${of}\:{real}\:{a}. \\ $$
Commented by MrW3 last updated on 17/Sep/18
a≥2 b=c=2(√(a−1))
$${a}\geqslant\mathrm{2} \\ $$$${b}={c}=\mathrm{2}\sqrt{{a}−\mathrm{1}} \\ $$
Commented by ajfour last updated on 17/Sep/18
This is certainly right sir; no other real answers even, Sir ?
$${This}\:{is}\:{certainly}\:{right}\:{sir};\:\:{no} \\ $$$${other}\:{real}\:{answers}\:{even},\:{Sir}\:? \\ $$
Commented by LYCON TRIX last updated on 17/Sep/18
I′m impressed to see this question I′ll make this question featured in NS7UC
$$\mathrm{I}'\mathrm{m}\:\mathrm{impressed}\:\mathrm{to}\:\mathrm{see}\:\mathrm{this}\:\mathrm{question} \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{make}\:\mathrm{this}\:\mathrm{question}\:\mathrm{featured}\:\mathrm{in}\:\mathrm{NS7UC} \\ $$
Commented by LYCON TRIX last updated on 17/Sep/18
what if I said that Solve for a in terms of b and c (a , b , c ) ∈ R
$$\mathrm{what}\:\mathrm{if}\:\mathrm{I}\:\mathrm{said}\:\mathrm{that}\: \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{a}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}\: \\ $$$$\left(\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}\:\right)\:\in\:\mathrm{R} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Sep/18
b=acos2α c=acos2β (√(2a)) sinα+(√(2a)) cosα=acos2β ((sinα+cosα)/(cos2β)).=(((√a) )/( (√2) )) ((sinβ+cosβ)/(cos2α))=((√a)/( (√2))) ((sinα+cosα)/(cos2β))=((sinβ+cosβ)/(cos2α)) (√(1+sin2α)) cos2α=(√(1+sin2β)) cos2β (√(1+((2t_1 )/(1+t_1 ^2 )))) ×((1−t_1 ^2 )/(1+t_1 ^2 ))=(√(1+((2t_2 )/(1+t_2 ^2 )))) ×((1−t_2 ^2 )/(1+t_2 ^2 )) ((1+t_1 ×1−t_1 ^2 )/((1+t_1 ^2 )^(3/2) ))=((1+t_2 ^2 ×1−t_2 ^2 )/((1+t_2 )^(3/2) )) (((1+t_1 )^2 ×(1−t_1 ))/((1+t_1 ^2 )^(3/2) ))=(((1+t_2 ^ )^2 ×(1−t_2 ))/((1+t_2 )^(3/2) )) (((1+t_1 )/(1+t_2 )))^2 ×((1−t_1 )/(1−t_2 ))=(((1+t_1 ^2 )/(1+t_2 ^2 )))^(3/2) t_1 =tanα t_2 =tanβ contd....
$${b}={acos}\mathrm{2}\alpha\:\:\:{c}={acos}\mathrm{2}\beta \\ $$$$\sqrt{\mathrm{2}{a}}\:{sin}\alpha+\sqrt{\mathrm{2}{a}}\:\:{cos}\alpha={acos}\mathrm{2}\beta \\ $$$$\frac{{sin}\alpha+{cos}\alpha}{{cos}\mathrm{2}\beta}.=\frac{\sqrt{{a}}\:}{\:\sqrt{\mathrm{2}}\:} \\ $$$$\frac{{sin}\beta+{cos}\beta}{{cos}\mathrm{2}\alpha}=\frac{\sqrt{{a}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{{sin}\alpha+{cos}\alpha}{{cos}\mathrm{2}\beta}=\frac{{sin}\beta+{cos}\beta}{{cos}\mathrm{2}\alpha} \\ $$$$\sqrt{\mathrm{1}+{sin}\mathrm{2}\alpha}\:{cos}\mathrm{2}\alpha=\sqrt{\mathrm{1}+{sin}\mathrm{2}\beta}\:\:{cos}\mathrm{2}\beta \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{2}{t}_{\mathrm{1}} }{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}\:\:×\frac{\mathrm{1}−{t}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }=\sqrt{\mathrm{1}+\frac{\mathrm{2}{t}_{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}\:×\frac{\mathrm{1}−{t}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}+{t}_{\mathrm{1}} ×\mathrm{1}−{t}_{\mathrm{1}} ^{\mathrm{2}} }{\left(\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} ×\mathrm{1}−{t}_{\mathrm{2}} ^{\mathrm{2}} }{\left(\mathrm{1}+{t}_{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{\left(\mathrm{1}+{t}_{\mathrm{1}} \right)^{\mathrm{2}} ×\left(\mathrm{1}−{t}_{\mathrm{1}} \right)}{\left(\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{\left(\mathrm{1}+{t}_{\mathrm{2}} ^{} \right)^{\mathrm{2}} ×\left(\mathrm{1}−{t}_{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}_{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\left(\frac{\mathrm{1}+{t}_{\mathrm{1}} }{\mathrm{1}+{t}_{\mathrm{2}} }\right)^{\mathrm{2}} ×\frac{\mathrm{1}−{t}_{\mathrm{1}} }{\mathrm{1}−{t}_{\mathrm{2}} }=\left(\frac{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${t}_{\mathrm{1}} ={tan}\alpha\:\:\:{t}_{\mathrm{2}} ={tan}\beta \\ $$$${contd}…. \\ $$$$ \\ $$
Answered by ajfour last updated on 18/Sep/18
If a is to be found in terms of b and c. squaring first eq. 2a+2(√(a^2 −b^2 )) = c^2 squaring again 4a^2 −4b^2 = c^4 +4a^2 −4ac^2 ⇒ a = (c^2 /4)+(b^2 /c^2 ) similarly from second eq. a = (b^2 /4)+(c^2 /b^2 ) ⇒ a = (c^2 /4)+(b^2 /c^2 ) = (b^2 /4)+(c^2 /b^2 ) ...(i) ⇒ ((c^2 −b^2 )/4) = (((c^2 −b^2 )(c^2 +b^2 ))/(b^2 c^2 )) Two cases arise: If b^2 = c^2 , then a = 1+(b^2 /4) = 1+(c^2 /4) If b^2 ≠ c^2 ⇒ b^2 +c^2 = ((b^2 c^2 )/4) , then using (i) 2a = ((b^2 +c^2 )/4) + ((b^4 +c^4 )/(b^2 c^2 )) ⇒ 2a = ((b^2 +c^2 )/4) + (((b^2 +c^2 )^2 −2b^2 c^2 )/(b^2 c^2 )) 2a = ((b^2 c^2 )/(16))+((b^4 c^4 )/(16b^2 c^2 ))−2 ⇒ a = ((b^2 c^2 )/(16))−1 = ((b^2 +c^2 )/4)−1 .
$$\boldsymbol{{If}}\:\boldsymbol{{a}}\:\boldsymbol{{is}}\:\boldsymbol{{to}}\:\boldsymbol{{be}}\:\boldsymbol{{found}}\:\boldsymbol{{in}}\:\boldsymbol{{terms}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{b}}\:\boldsymbol{{and}}\:\boldsymbol{{c}}. \\ $$$${squaring}\:{first}\:{eq}.\: \\ $$$$\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:=\:{c}^{\mathrm{2}} \\ $$$${squaring}\:{again} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} \:=\:{c}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ac}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{a}\:=\:\frac{{c}^{\mathrm{2}} }{\mathrm{4}}+\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\: \\ $$$${similarly}\:{from}\:{second}\:{eq}. \\ $$$$\:\:\:\:\:\:\:\:\:{a}\:=\:\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{a}\:=\:\frac{{c}^{\mathrm{2}} }{\mathrm{4}}+\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\:=\:\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\:\:…\left(\boldsymbol{{i}}\right) \\ $$$$\Rightarrow\:\:\:\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{4}}\:=\:\frac{\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$${Two}\:{cases}\:{arise}: \\ $$$${If}\:{b}^{\mathrm{2}} =\:{c}^{\mathrm{2}} \:,\:{then} \\ $$$$\:\:\:\:\:\boldsymbol{{a}}\:=\:\mathrm{1}+\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{1}+\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${If}\:{b}^{\mathrm{2}} \neq\:{c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\frac{{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{4}}\:\:,\:{then}\:{using}\:\left({i}\right) \\ $$$$\:\:\:\mathrm{2}\boldsymbol{{a}}\:=\:\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}}\:+\:\frac{{b}^{\mathrm{4}} +{c}^{\mathrm{4}} }{{b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{2}{a}\:=\:\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}}\:+\:\frac{\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\mathrm{2}{a}\:=\:\frac{{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{16}}+\frac{{b}^{\mathrm{4}} {c}^{\mathrm{4}} }{\mathrm{16}{b}^{\mathrm{2}} {c}^{\mathrm{2}} }−\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:{a}\:=\:\frac{{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{16}}−\mathrm{1}\:=\:\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\:. \\ $$

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