Question and Answers Forum

AlgebraQuestion and Answers: Page 302

Question Number 55108 Answers: 0 Comments: 0

Question Number 55088 Answers: 0 Comments: 0

Question Number 55076 Answers: 1 Comments: 0

Known polynom P(z)=a_0 z^n +a_1 z^(n−1) +…+a_(n ) With explain real number. If z_0 =3−4i form root is from polynom, then one other root defonitely appeared is..

$$\mathrm{Known}\:\mathrm{polynom} \\ $$$$\mathrm{P}\left({z}\right)={a}_{\mathrm{0}} {z}^{{n}} +{a}_{\mathrm{1}} {z}^{{n}−\mathrm{1}} +\ldots+{a}_{{n}\:} \mathrm{With} \\ $$$$\mathrm{explain}\:\mathrm{real}\:\mathrm{number}.\:\mathrm{If}\:{z}_{\mathrm{0}} =\mathrm{3}−\mathrm{4}{i} \\ $$$$\mathrm{form}\:\mathrm{root}\:\mathrm{is}\:\mathrm{from}\:\mathrm{polynom}, \\ $$$$\mathrm{then}\:\mathrm{one}\:\mathrm{other}\:\mathrm{root}\:\mathrm{defonitely}\:\mathrm{appeared} \\ $$$$\mathrm{is}.. \\ $$

Question Number 55070 Answers: 0 Comments: 1

Factorised the polynom z^4 +1 be polynom with lower degree, but have real coefficient

$$\mathrm{Factorised}\:\mathrm{the}\:\mathrm{polynom}\:{z}^{\mathrm{4}} +\mathrm{1}\: \\ $$$$\mathrm{be}\:\mathrm{polynom}\:\mathrm{with}\:\mathrm{lower}\:\mathrm{degree}, \\ $$$$\mathrm{but}\:\mathrm{have}\:\mathrm{real}\:\mathrm{coefficient} \\ $$

Question Number 55069 Answers: 1 Comments: 3

Known analytic function f(z)=((2(z−2))/(z(z−4))) and written as f(z)=Σ_(n=0) ^(∝) a_n (z−1)^n The value of a_(100) is...

$$\mathrm{Known}\:\mathrm{analytic}\:\mathrm{function} \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}\left({z}−\mathrm{2}\right)}{{z}\left({z}−\mathrm{4}\right)} \\ $$$$\mathrm{and}\:\mathrm{written}\:\mathrm{as}\:{f}\left({z}\right)=\underset{{n}=\mathrm{0}} {\overset{\propto} {\Sigma}}\:{a}_{{n}} \left({z}−\mathrm{1}\right)^{{n}} \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}_{\mathrm{100}} \:\mathrm{is}... \\ $$

Question Number 55066 Answers: 0 Comments: 0

Find radius convergence for series 1−z^2 +z^4 −z^6 +...

$$\mathrm{Find}\:\mathrm{radius}\:\mathrm{convergence}\:\mathrm{for}\:\mathrm{series} \\ $$$$\mathrm{1}−{z}^{\mathrm{2}} +{z}^{\mathrm{4}} −{z}^{\mathrm{6}} +... \\ $$

Question Number 55039 Answers: 1 Comments: 3

α and β,are 2 roots of eq: ax^2 +bx+c=0 with conditions: { ((α^2 =β+b)),((β^2 =α+a)) :} find: c in terms of: a and b.

$$\alpha\:{and}\:\beta,{are}\:\mathrm{2}\:{roots}\:{of}\:{eq}: \\ $$$$\:\:\:\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{with}\:{conditions}: \\ $$$$\:\:\:\:\begin{cases}{\alpha^{\mathrm{2}} =\beta+{b}}\\{\beta^{\mathrm{2}} =\alpha+{a}}\end{cases} \\ $$$${find}:\:\:\boldsymbol{{c}}\:{in}\:{terms}\:{of}:\:\boldsymbol{{a}}\:\:{and}\:\:\boldsymbol{{b}}. \\ $$

Question Number 55011 Answers: 0 Comments: 1

Σ_(k=o) ^(n−1) (1/(2−x^k )) find out the summetion

$$\underset{{k}={o}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}−{x}^{{k}} }\:\:\:\:{find}\:{out}\:{the}\:{summetion} \\ $$$$ \\ $$

Question Number 55014 Answers: 3 Comments: 0

what is the value of t that makes x^2 +10x+t a perfect square?

$${what}\:{is}\:{the}\:{value}\:{of}\:{t}\:{that}\:{makes}\: \\ $$$${x}^{\mathrm{2}} +\mathrm{10}{x}+{t}\:{a}\:{perfect}\:{square}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 54983 Answers: 1 Comments: 2

Find shortest distance from origin to the cubic y=x^3 −10x^2 +27x−18.

$${Find}\:{shortest}\:{distance}\:{from}\:{origin} \\ $$$${to}\:{the}\:{cubic}\:\:\:{y}={x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{27}{x}−\mathrm{18}. \\ $$

Question Number 54948 Answers: 0 Comments: 1

α,β are the roots and prove that α^n +β^n =2[cos nΠ/2]

$$\alpha,\beta\:{are}\:{the}\:{roots}\:{and}\:{prove}\:{that}\:\alpha^{{n}} +\beta^{{n}} =\mathrm{2}\left[\mathrm{cos}\:{n}\Pi/\mathrm{2}\right] \\ $$

Question Number 54923 Answers: 0 Comments: 9

Is tan 1° rational or irrational? Give your proof.

$${Is}\:\mathrm{tan}\:\mathrm{1}°\:{rational}\:{or}\:{irrational}? \\ $$$${Give}\:{your}\:{proof}. \\ $$

Question Number 54857 Answers: 1 Comments: 0

Question Number 54808 Answers: 1 Comments: 2

let p(x)=(1+x^2 )(1+x^4 )...(1+x^2^n ) with n integr natural 1) find a simple form of p(x) 2) find roots of p(x)and decompose p(x) inside C[x] 3)calculate ∫_0 ^1 p(x)dx 4) decompose the fraction F(x)=(1/(p(x))) .

$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)...\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$${with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{roots}\:{of}\:{p}\left({x}\right){and}\:{decompose} \\ $$$${p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{p}\left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{decompose}\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)}\:. \\ $$

Question Number 54787 Answers: 0 Comments: 0

if: (a/((√b)+(√c)))=(b/((√c)+(√d)))=(c/((√d)+(√a)))=(d/((√a)+(√b))) ⇒(a/d)=?

$$\:{if}:\:\:\frac{{a}}{\sqrt{{b}}+\sqrt{{c}}}=\frac{{b}}{\sqrt{{c}}+\sqrt{{d}}}=\frac{{c}}{\sqrt{{d}}+\sqrt{{a}}}=\frac{{d}}{\sqrt{{a}}+\sqrt{{b}}} \\ $$$$\Rightarrow\frac{{a}}{{d}}=? \\ $$

Question Number 54786 Answers: 0 Comments: 5

solve for:a,b,c,d∈R a^2 =b+(√c) b^2 =c+(√d) c^2 =d+(√a) d^2 =a+(√b)

$${solve}\:{for}:{a},{b},{c},{d}\in\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} ={b}+\sqrt{{c}} \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} ={c}+\sqrt{{d}} \\ $$$$\:\:\:\:\:\:{c}^{\mathrm{2}} ={d}+\sqrt{{a}} \\ $$$$\:\:\:\:\:\:{d}^{\mathrm{2}} ={a}+\sqrt{{b}} \\ $$

Question Number 54775 Answers: 0 Comments: 2

found something interesting, it was published by Tschirnhaus in 1683 we can reduce x^3 +ax^2 +bx+c=0 (1) to y^3 +py+q=0 (2) and further to z^3 =t (1) is the well known linear substitution y=x+(a/3) → x=y−(a/3) ⇒ y^3 −((a^2 −3b)/3)y+((2a^3 −9ab+27c)/(27))=0 p=−((a^2 −3b)/3) and q=((2a^3 −9ab+27c)/(27)) ⇒ y^3 +py+q=0 (2) quadratic substitution z=y^2 +αy+β → y^2 +αy+(β−z)=0 we could solve this for y and then plug in above... (Tschirnhaus did) but there′s an easier way: we calculate the determinant of the Sylvester Matrix we have (a) 1y^3 +0y^2 +py+q=0 (b) 0y^3 +y^2 +αy+(β−z)=0 the matrix is [(1,0,p,q,0),(0,1,0,p,q),(0,1,α,(β−z),0),(0,0,1,α,(β−z)),(1,α,(β−z),0,0) ] the determinant is −z^3 +(3β−2p)z^2 −(pα^2 +3qα+3β^2 −4pβ+p^2 )z −(qα^3 −pα^2 β−3qαβ+pqα−β^3 +2pβ^2 −p^2 β−q^2 )p we want the square and the linear terms to disappear so we set their constants zero to get α and β ⇒ β=((2p)/3); α=−((3q)/(2p))±((√(12p^3 +81q^2 ))/(6p)) this leads to z^3 =((8p^3 )/(27))+((27q^4 )/(2p^3 ))+4q^2 ±((q(√(3(4p^3 +27q^2 )^3 )))/(18p^3 )) Tschirnhaus thought he could solve polynomes of any degree with this method but it′s getting harder to solve because you need a cubic substitution to eliminate 3 constants and so on...

$$\mathrm{found}\:\mathrm{something}\:\mathrm{interesting},\:\mathrm{it}\:\mathrm{was}\:\mathrm{published} \\ $$$$\mathrm{by}\:\mathrm{Tschirnhaus}\:\mathrm{in}\:\mathrm{1683} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{reduce} \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{to} \\ $$$${y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{further}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} ={t} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{well}\:\mathrm{known}\:\mathrm{linear}\:\mathrm{substitution} \\ $$$${y}={x}+\frac{{a}}{\mathrm{3}}\:\rightarrow\:{x}={y}−\frac{{a}}{\mathrm{3}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} −\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}{y}+\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}}=\mathrm{0} \\ $$$${p}=−\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}\:\mathrm{and}\:{q}=\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{quadratic}\:\mathrm{substitution} \\ $$$${z}={y}^{\mathrm{2}} +\alpha{y}+\beta\:\rightarrow\:{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{y}\:\mathrm{and}\:\mathrm{then}\:\mathrm{plug}\:\mathrm{in} \\ $$$$\mathrm{above}...\:\left(\mathrm{Tschirnhaus}\:\mathrm{did}\right)\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{an} \\ $$$$\mathrm{easier}\:\mathrm{way}:\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{Sylvester}\:\mathrm{Matrix} \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\left({a}\right)\:\mathrm{1}{y}^{\mathrm{3}} +\mathrm{0}{y}^{\mathrm{2}} +{py}+{q}=\mathrm{0} \\ $$$$\left({b}\right)\:\mathrm{0}{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{matrix}\:\mathrm{is} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}\\{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}\\{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\mathrm{the}\:\mathrm{determinant}\:\mathrm{is} \\ $$$$−{z}^{\mathrm{3}} \\ $$$$\:\:+\left(\mathrm{3}\beta−\mathrm{2}{p}\right){z}^{\mathrm{2}} \\ $$$$\:\:−\left({p}\alpha^{\mathrm{2}} +\mathrm{3}{q}\alpha+\mathrm{3}\beta^{\mathrm{2}} −\mathrm{4}{p}\beta+{p}^{\mathrm{2}} \right){z} \\ $$$$\:\:−\left({q}\alpha^{\mathrm{3}} −{p}\alpha^{\mathrm{2}} \beta−\mathrm{3}{q}\alpha\beta+{pq}\alpha−\beta^{\mathrm{3}} +\mathrm{2}{p}\beta^{\mathrm{2}} −{p}^{\mathrm{2}} \beta−{q}^{\mathrm{2}} \right){p} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{the}\:\mathrm{square}\:\mathrm{and}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{terms} \\ $$$$\mathrm{to}\:\mathrm{disappear}\:\mathrm{so}\:\mathrm{we}\:\mathrm{set}\:\mathrm{their}\:\mathrm{constants}\:\mathrm{zero} \\ $$$$\mathrm{to}\:\mathrm{get}\:\alpha\:\mathrm{and}\:\beta \\ $$$$\Rightarrow\:\beta=\frac{\mathrm{2}{p}}{\mathrm{3}};\:\alpha=−\frac{\mathrm{3}{q}}{\mathrm{2}{p}}\pm\frac{\sqrt{\mathrm{12}{p}^{\mathrm{3}} +\mathrm{81}{q}^{\mathrm{2}} }}{\mathrm{6}{p}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} =\frac{\mathrm{8}{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{27}{q}^{\mathrm{4}} }{\mathrm{2}{p}^{\mathrm{3}} }+\mathrm{4}{q}^{\mathrm{2}} \pm\frac{{q}\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{18}{p}^{\mathrm{3}} } \\ $$$$\mathrm{Tschirnhaus}\:\mathrm{thought}\:\mathrm{he}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{polynomes} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{degree}\:\mathrm{with}\:\mathrm{this}\:\mathrm{method}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\ $$$$\mathrm{harder}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{because}\:\mathrm{you}\:\mathrm{need}\:\mathrm{a}\:\mathrm{cubic} \\ $$$$\mathrm{substitution}\:\mathrm{to}\:\mathrm{eliminate}\:\mathrm{3}\:\mathrm{constants}\:\mathrm{and} \\ $$$$\mathrm{so}\:\mathrm{on}... \\ $$

Question Number 54737 Answers: 0 Comments: 0