Question and Answers Forum

All Questions   Topic List

AlgebraQuestion and Answers: Page 307

Question Number 61591    Answers: 1   Comments: 8

solve for z∈C (z)^(1/2) =−1 (z)^(1/3) =−1 (z)^(1/4) =−1

$$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$\sqrt[{\mathrm{2}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{z}}=−\mathrm{1} \\ $$

Question Number 61526    Answers: 0   Comments: 0

Solve for n: D/A×{1−((P×((((1+i)^n ×i)/((1+i)^n −1))))/((P×((((1+i)^r ×i)/((1+i)^r −1))))−(R/i)×[((1/n)+i)×((((1+i)^r ×i)/((1+i)^r −1)))−((1/n)+i)×((((1+i)^n ×i)/((1+i)^n −1)))]))}−1=0

$${Solve}\:{for}\:{n}:\:{D}/{A}×\left\{\mathrm{1}−\frac{{P}×\left(\frac{\left(\mathrm{1}+{i}\right)^{{n}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{n}} −\mathrm{1}}\right)}{\left({P}×\left(\frac{\left(\mathrm{1}+{i}\right)^{{r}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{r}} −\mathrm{1}}\right)\right)−\frac{{R}}{{i}}×\left[\left(\frac{\mathrm{1}}{{n}}+{i}\right)×\left(\frac{\left(\mathrm{1}+{i}\right)^{{r}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{r}} −\mathrm{1}}\right)−\left(\frac{\mathrm{1}}{{n}}+{i}\right)×\left(\frac{\left(\mathrm{1}+{i}\right)^{{n}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{n}} −\mathrm{1}}\right)\right]}\right\}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$

Question Number 61495    Answers: 1   Comments: 7

Question Number 61490    Answers: 0   Comments: 10

(√(a−(√(a+x))))+(√(a+(√(a−x))))=2x this is the solution Sir Aifour and me found trivial solution a=x=0 a, x ∈R x=((√2)/8)(r+(√(r^2 +4)))(√(2(4a−1)−r^2 −r(√(r^2 +4)))) with r=−2(√((4a−3)/3))sin ((1/3)arcsin ((3(√3))/((4a−3)^(3/2) ))) no solution for a<a_0 with a_0 ≈1.509830340886

$$\sqrt{{a}−\sqrt{{a}+{x}}}+\sqrt{{a}+\sqrt{{a}−{x}}}=\mathrm{2}{x} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{Sir}\:\mathrm{Aifour}\:\mathrm{and}\:\mathrm{me}\:\mathrm{found} \\ $$$$ \\ $$$$\mathrm{trivial}\:\mathrm{solution}\:{a}={x}=\mathrm{0} \\ $$$$ \\ $$$${a},\:{x}\:\in\mathbb{R} \\ $$$$ \\ $$$${x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left({r}+\sqrt{{r}^{\mathrm{2}} +\mathrm{4}}\right)\sqrt{\mathrm{2}\left(\mathrm{4}{a}−\mathrm{1}\right)−{r}^{\mathrm{2}} −{r}\sqrt{{r}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\mathrm{with} \\ $$$${r}=−\mathrm{2}\sqrt{\frac{\mathrm{4}{a}−\mathrm{3}}{\mathrm{3}}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\left(\mathrm{4}{a}−\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right) \\ $$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a}<{a}_{\mathrm{0}} \:\mathrm{with}\:{a}_{\mathrm{0}} \approx\mathrm{1}.\mathrm{509830340886} \\ $$

Question Number 61470    Answers: 1   Comments: 0

Is there any other solution besides {x=a,y=b} or {x=b,y=a} of the following system of equations x+y=a+b ∧ x^7 +y^7 =a^7 +b^7 ?

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{besides} \\ $$$$\left\{\mathrm{x}=\mathrm{a},\mathrm{y}=\mathrm{b}\right\}\:\mathrm{or}\:\left\{\mathrm{x}=\mathrm{b},\mathrm{y}=\mathrm{a}\right\}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\:\:\:\:\mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b}\:\:\wedge\:\mathrm{x}^{\mathrm{7}} +\mathrm{y}^{\mathrm{7}} =\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} \:\:? \\ $$$$ \\ $$

Question Number 61343    Answers: 1   Comments: 0

Question Number 61320    Answers: 1   Comments: 0

solve for x: x^4 −x=12

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\boldsymbol{\mathrm{x}}=\mathrm{12} \\ $$

Question Number 61272    Answers: 0   Comments: 0

Question Number 61273    Answers: 2   Comments: 2

Suppose α ,β,γ,δ are real numbers such that α+β+γ+δ = α^7 +β^7 +γ^7 +δ^7 =0 Prove that α(α+β)(α+γ)(α+δ)=0

$${Suppose}\:\alpha\:,\beta,\gamma,\delta\:{are}\:{real}\:{numbers} \\ $$$${such}\:{that}\:\alpha+\beta+\gamma+\delta\:=\:\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0} \\ $$$${Prove}\:{that}\:\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\delta\right)=\mathrm{0} \\ $$

Question Number 61269    Answers: 3   Comments: 0

Let p(x) be a quadratic polynomial such that for distinct α and β , p(α) = α and p(β) =β prove that α and β are roots of p[p(x)]−x=0 Find the remaining roots .

$${Let}\:{p}\left({x}\right)\:{be}\:{a}\:{quadratic}\:{polynomial}\:{such} \\ $$$${that}\:{for}\:{distinct}\:\alpha\:{and}\:\beta\:, \\ $$$${p}\left(\alpha\right)\:=\:\alpha\:{and}\:{p}\left(\beta\right)\:=\beta \\ $$$${prove}\:{that}\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\:\:{p}\left[{p}\left({x}\right)\right]−{x}=\mathrm{0}\: \\ $$$${Find}\:{the}\:{remaining}\:{roots}\:. \\ $$

Question Number 61268    Answers: 0   Comments: 0

Let a,b,c,d,e ≥ −1 and a+b+c+d+e=5 Find the maximum and minimum value of S =(a+b)(b+c)(c+d)(d+e)(e+a)

$${Let}\:{a},{b},{c},{d},{e}\:\geqslant\:−\mathrm{1}\:{and}\:{a}+{b}+{c}+{d}+{e}=\mathrm{5} \\ $$$${Find}\:{the}\:{maximum}\:{and}\:{minimum} \\ $$$${value}\:{of}\:{S}\:=\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{d}\right)\left({d}+{e}\right)\left({e}+{a}\right) \\ $$

Question Number 61267    Answers: 0   Comments: 0

Question Number 61211    Answers: 1   Comments: 3

Solve for x in terms of a (√(a−(√(a+x )))) + (√(a+(√(a−x)))) = 2x Please sir i request you to solve this question =_=

$${Solve}\:{for}\:{x}\:{in}\:{terms}\:{of}\:{a}\: \\ $$$$\sqrt{{a}−\sqrt{{a}+{x}\:}}\:+\:\:\sqrt{{a}+\sqrt{{a}−{x}}}\:=\:\mathrm{2}{x} \\ $$$${Please}\:{sir}\:{i}\:{request}\:{you}\:{to}\:{solve}\:{this}\: \\ $$$${question}\:=\_= \\ $$

Question Number 61162    Answers: 1   Comments: 3

if sin(x) = ((x − 20)/(20)) , find x

$$\mathrm{if}\:\:\:\:\:\mathrm{sin}\left(\mathrm{x}\right)\:\:=\:\:\frac{\mathrm{x}\:−\:\mathrm{20}}{\mathrm{20}}\:\:,\:\:\:\mathrm{find}\:\:\mathrm{x} \\ $$

Question Number 61140    Answers: 1   Comments: 1

can we find an exact solution? t^6 +4t^4 −12t^3 +24t^2 −24t+8=0

$$\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}? \\ $$$${t}^{\mathrm{6}} +\mathrm{4}{t}^{\mathrm{4}} −\mathrm{12}{t}^{\mathrm{3}} +\mathrm{24}{t}^{\mathrm{2}} −\mathrm{24}{t}+\mathrm{8}=\mathrm{0} \\ $$

Question Number 61137    Answers: 1   Comments: 0

What is the sum of first 3n term of an AP , if the sunm of first n term is 2n and sum of first 2n term is 5n

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{3n}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:,\:\mathrm{if}\:\mathrm{the}\:\mathrm{sunm}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{term}\:\mathrm{is} \\ $$$$\mathrm{2n}\:\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{2n}\:\mathrm{term}\:\mathrm{is}\:\:\mathrm{5n} \\ $$

Question Number 61117    Answers: 2   Comments: 0

The 2nd, 4th and 8th term of an AP are the consecutive term of a GP. If the sum of the 3rd and 4th term of the AP is 20. Find the sum of the first four terms of the AP.

$$\mathrm{The}\:\mathrm{2nd},\:\mathrm{4th}\:\mathrm{and}\:\mathrm{8th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{are}\:\mathrm{the}\:\mathrm{consecutive}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{GP}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3rd}\:\mathrm{and}\:\mathrm{4th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{AP}\:\mathrm{is}\:\mathrm{20}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{AP}. \\ $$

Question Number 61111    Answers: 1   Comments: 2

Please what does the 2 on the C mean. C_1 ^2 + 2 C_2 ^2 + 3 C_3 ^2 + ... + n C_n ^2 = (((2n − 1)!)/([(n − 1)!]^2 )) Does the 2 on C mean square ?? I mean: (C_1 )^2 + 2(C_2 )^2 + 3(C_3 )^2 + ... + n (C_n )^2 which is also ( ^n C_1 )^2 + 2( ^n C_2 )^2 + 3( ^n C_3 )^2 + ... + n ( ^n C_n )^2 I just want to know what the 2 on C represent . Thanks. C_1 ^2 + 2 C_2 ^2 + 3 C_3 ^2 + ... + n C_n ^2 = (((2n − 1)!)/([(n − 1)!]^2 ))

$$\mathrm{Please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{the}\:\mathrm{C}\:\mathrm{mean}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{2}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{3}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\:\:\:=\:\:\:\frac{\left(\mathrm{2n}\:−\:\mathrm{1}\right)!}{\left[\left(\mathrm{n}\:−\:\mathrm{1}\right)!\right]^{\mathrm{2}} } \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{C}\:\mathrm{mean}\:\mathrm{square}\:?? \\ $$$$\:\:\:\:\mathrm{I}\:\mathrm{mean}:\:\:\:\:\:\:\left(\mathrm{C}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\:\mathrm{2}\left(\mathrm{C}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{3}\left(\mathrm{C}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\left(\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{2}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{also} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\:\mathrm{2}\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{3}\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{what}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{C}\:\mathrm{represent}\:.\:\:\mathrm{Thanks}. \\ $$$$\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{2}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{3}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\:\:\:=\:\:\:\frac{\left(\mathrm{2n}\:−\:\mathrm{1}\right)!}{\left[\left(\mathrm{n}\:−\:\mathrm{1}\right)!\right]^{\mathrm{2}} } \\ $$

Question Number 60980    Answers: 1   Comments: 2

Solve for x, y, z x(y + z) = 33 ..... (i) y(z + x) = 35 ..... (ii) z(x + y) = 14 ..... (iii)

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x},\:\mathrm{y},\:\mathrm{z} \\ $$$$\:\:\:\:\:\mathrm{x}\left(\mathrm{y}\:+\:\mathrm{z}\right)\:=\:\mathrm{33}\:\:\:\:\:\:\:\:.....\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\mathrm{y}\left(\mathrm{z}\:+\:\mathrm{x}\right)\:=\:\mathrm{35}\:\:\:\:\:\:\:\:.....\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\mathrm{z}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:\mathrm{14}\:\:\:\:\:\:\:\:.....\:\left(\mathrm{iii}\right) \\ $$

Question Number 60946    Answers: 3   Comments: 4

Find x: x^x = 2x

$$\mathrm{Find}\:\mathrm{x}:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}} \:\:=\:\:\mathrm{2x} \\ $$

Question Number 60905    Answers: 3   Comments: 0

if x+(1/x)=(√3).find x^(24) +x^(18) +x^6 +1

$${if}\:{x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{3}}.{find} \\ $$$${x}^{\mathrm{24}} +{x}^{\mathrm{18}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$

Question Number 60910    Answers: 1   Comments: 7

Question Number 60854    Answers: 2   Comments: 4

(x^4 −3x^2 +2x+1)/(x−1)

$$\left({x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)/\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$

Question Number 60817    Answers: 4   Comments: 2

V=(4/3)𝛑R^3 prove

$$\boldsymbol{\mathrm{V}}=\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{R}}^{\mathrm{3}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$

Question Number 60816    Answers: 1   Comments: 0

S=4𝛑R^2 prove

$$\boldsymbol{\mathrm{S}}=\mathrm{4}\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$

Question Number 60745    Answers: 4   Comments: 5

  Pg 302      Pg 303      Pg 304      Pg 305      Pg 306      Pg 307      Pg 308      Pg 309      Pg 310      Pg 311   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com