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AlgebraQuestion and Answers: Page 302

Question Number 66619 Answers: 1 Comments: 0

solve for x,y∈R ((√(1+x^2 ))/(ln (x+(√(1+x^2 )))))=((√(1+y^2 ))/(ln (y+(√(1+y^2 )))))

$${solve}\:{for}\:{x},{y}\in{R} \\ $$$$\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}=\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\mathrm{ln}\:\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$

Question Number 66546 Answers: 0 Comments: 6

Question Number 66508 Answers: 0 Comments: 3

Question Number 66382 Answers: 0 Comments: 14

Give me any Quintic, i shall solve it. For sure! At^5 +Bt^4 +Ct^3 +Dt^2 +Et+F=0 wont even assume A=1, or B=0. but if A+C+E=B+D+F then my formula dont work but then obviously t=−1 is a root!

$${Give}\:{me}\:{any}\:{Quintic},\:{i}\:{shall}\:{solve} \\ $$$${it}.\:{For}\:{sure}! \\ $$$${At}^{\mathrm{5}} +{Bt}^{\mathrm{4}} +{Ct}^{\mathrm{3}} +{Dt}^{\mathrm{2}} +{Et}+{F}=\mathrm{0} \\ $$$${wont}\:{even}\:{assume}\:{A}=\mathrm{1},\:{or}\:{B}=\mathrm{0}. \\ $$$${but}\:{if}\:{A}+{C}+{E}={B}+{D}+{F}\: \\ $$$${then}\:{my}\:{formula}\:{dont}\:{work} \\ $$$${but}\:{then}\:{obviously}\:{t}=−\mathrm{1}\:{is}\:{a}\:{root}! \\ $$

Question Number 66381 Answers: 0 Comments: 0

Question Number 66302 Answers: 0 Comments: 3

solved the general quintic, despite whatever proof that it cant be solved in a simple way!

$${solved}\:{the}\:{general}\:{quintic}, \\ $$$${despite}\:{whatever}\:{proof}\:{that}\:{it} \\ $$$${cant}\:{be}\:{solved}\:{in}\:{a}\:{simple}\:{way}! \\ $$

Question Number 66290 Answers: 1 Comments: 0

(x^2 )^(1/(√3)) + x^(√3) − 392 = 0

$$\:\sqrt[{\sqrt{\mathrm{3}}}]{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:+\:\boldsymbol{\mathrm{x}}^{\sqrt{\mathrm{3}}} \:−\:\mathrm{392}\:=\:\mathrm{0} \\ $$

Question Number 66199 Answers: 0 Comments: 3

calculate cos(79)=?

$${calculate} \\ $$$${cos}\left(\mathrm{79}\right)=? \\ $$

Question Number 66197 Answers: 3 Comments: 3

If x+(1/x)=1,prove that: x^n +x^(n−2) +x^(n−4) =0

$$\mathrm{If}\:\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{1},\mathrm{prove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{n}} +\mathrm{x}^{\mathrm{n}−\mathrm{2}} +\mathrm{x}^{\mathrm{n}−\mathrm{4}} =\mathrm{0} \\ $$

Question Number 66126 Answers: 1 Comments: 0

Is there any formula to find sum of 1 + n^2 + n^4 + n^6 + n^8 + ... + n^(2k) + ... where n,k ∈ Z^+

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{find}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{1}\:+\:{n}^{\mathrm{2}} \:+\:{n}^{\mathrm{4}} \:+\:{n}^{\mathrm{6}} \:+\:{n}^{\mathrm{8}} \:+\:...\:+\:{n}^{\mathrm{2}{k}} \:+\:... \\ $$$$\mathrm{where}\:{n},{k}\:\in\:\mathbb{Z}^{+} \: \\ $$

Question Number 66004 Answers: 1 Comments: 0

x^x^x^⋰ =3 find the value of x

$${x}^{{x}^{{x}^{\iddots} } } =\mathrm{3} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x} \\ $$

Question Number 68063 Answers: 1 Comments: 0

Question Number 65981 Answers: 1 Comments: 0

Question Number 65911 Answers: 0 Comments: 1

Question Number 65866 Answers: 0 Comments: 5

Question Number 65859 Answers: 1 Comments: 0

x^4 +5x^2 +20x+104=0 solve for x.

$${x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}{x}+\mathrm{104}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$

Question Number 65853 Answers: 1 Comments: 0

x^4 −23x^2 +18x+40=0 solve for x.

$${x}^{\mathrm{4}} −\mathrm{23}{x}^{\mathrm{2}} +\mathrm{18}{x}+\mathrm{40}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$

Question Number 65830 Answers: 1 Comments: 3

Question Number 65745 Answers: 0 Comments: 1

Question Number 65743 Answers: 0 Comments: 1

Question Number 65683 Answers: 0 Comments: 0

Question Number 65680 Answers: 1 Comments: 0

Question Number 65724 Answers: 2 Comments: 0

Question Number 65723 Answers: 1 Comments: 0

Question Number 65569 Answers: 1 Comments: 1

reposting this: x^4 +(1−2a)x^2 −2ax+1=0 solve for a>0∧x>0 a^5 −((23)/8)x^4 −((49)/4)x^3 +((27)/8)x^2 +3x+(9/8)=0 ⇒ 3 real solutions a_1 <a_2 <a_3 we cannot solve exactly (?) so it′s not possible to get all the below solutions exactly, only rough approximations for the following α,β,γ,δ ∈R the possible natures of the zeros: (1) a<a_1 ∨a_2 <a<a_3 ⇒ x=α∨x=β∨x=γ±δi (2) a>a_3 ⇒ x=α∨x=β∨x=γ∨x=δ (3) a_1 <a<a_2 ⇒ x=α±βi∨x=γ±δi (4) a=a_3 ⇒ x=α=β∨x=γ∨x=δ (5) a=a_1 ∨a=a_2 ⇒ x=α=β∨x=γ±δi

$$\mathrm{reposting}\:\mathrm{this}: \\ $$$${x}^{\mathrm{4}} +\left(\mathrm{1}−\mathrm{2}{a}\right){x}^{\mathrm{2}} −\mathrm{2}{ax}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{a}>\mathrm{0}\wedge{x}>\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{5}} −\frac{\mathrm{23}}{\mathrm{8}}{x}^{\mathrm{4}} −\frac{\mathrm{49}}{\mathrm{4}}{x}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{8}}{x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{8}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:{a}_{\mathrm{1}} <{a}_{\mathrm{2}} <{a}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{exactly}\:\left(?\right)\:\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{not} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{possible}\:\mathrm{to}\:\mathrm{get}\:\mathrm{all}\:\mathrm{the}\:\mathrm{below}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{exactly},\:\mathrm{only}\:\mathrm{rough}\:\mathrm{approximations} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{following}\:\alpha,\beta,\gamma,\delta\:\in\mathbb{R} \\ $$$$\mathrm{the}\:\mathrm{possible}\:\mathrm{natures}\:\mathrm{of}\:\mathrm{the}\:\mathrm{zeros}: \\ $$$$\left(\mathrm{1}\right)\:\:{a}<{a}_{\mathrm{1}} \vee{a}_{\mathrm{2}} <{a}<{a}_{\mathrm{3}} \:\Rightarrow\:{x}=\alpha\vee{x}=\beta\vee{x}=\gamma\pm\delta\mathrm{i} \\ $$$$\left(\mathrm{2}\right)\:\:{a}>{a}_{\mathrm{3}} \:\Rightarrow\:{x}=\alpha\vee{x}=\beta\vee{x}=\gamma\vee{x}=\delta \\ $$$$\left(\mathrm{3}\right)\:\:{a}_{\mathrm{1}} <{a}<{a}_{\mathrm{2}} \:\Rightarrow\:{x}=\alpha\pm\beta\mathrm{i}\vee{x}=\gamma\pm\delta\mathrm{i} \\ $$$$\left(\mathrm{4}\right)\:\:{a}={a}_{\mathrm{3}} \:\Rightarrow\:{x}=\alpha=\beta\vee{x}=\gamma\vee{x}=\delta \\ $$$$\left(\mathrm{5}\right)\:\:{a}={a}_{\mathrm{1}} \vee{a}={a}_{\mathrm{2}} \:\Rightarrow\:{x}=\alpha=\beta\vee{x}=\gamma\pm\delta\mathrm{i} \\ $$

Question Number 65495 Answers: 1 Comments: 0

{ (((a/b)+(b/(a+b))=(√3))),(((b/a)+(a/(a+b))=(√2))) :} [a,b∈R]

$$\begin{cases}{\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}+\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}=\sqrt{\mathrm{3}}}\\{\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}=\sqrt{\mathrm{2}}}\end{cases}\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}\right] \\ $$

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