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AlgebraQuestion and Answers: Page 307

Question Number 63485    Answers: 1   Comments: 0

f(x−3)+f(x)=2x−3 F(2)=0. F(−2)=?

$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{3}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{F}}\left(\mathrm{2}\right)=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{F}}\left(−\mathrm{2}\right)=? \\ $$

Question Number 63474    Answers: 1   Comments: 0

let P(x)=x^2 +(1/2)x+b and Q(x)=x^2 +cx+d be to polynomials with real coefficient such that P(x) Q(x)=Q(P(x)) find all the real roots of P(Q(x))=0

$${let}\:{P}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+{b} \\ $$$$ \\ $$$${and}\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d} \\ $$$$ \\ $$$${be}\:{to}\:{polynomials}\:{with}\:{real}\:{coefficient}\:{such}\:{that} \\ $$$$ \\ $$$${P}\left({x}\right)\:{Q}\left({x}\right)={Q}\left({P}\left({x}\right)\right) \\ $$$$ \\ $$$${find}\:{all}\:{the}\:{real}\:{roots}\:{of}\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0} \\ $$

Question Number 63449    Answers: 0   Comments: 2

Do you remember that π≈3.1415926535897932⋱ 38462643383279502884⋱ 197...

$${Do}\:{you}\:{remember}\:{that} \\ $$$$\pi\approx\mathrm{3}.\mathrm{1415926535897932}\ddots \\ $$$$\mathrm{38462643383279502884}\ddots \\ $$$$\mathrm{197}... \\ $$

Question Number 63614    Answers: 0   Comments: 0

solve the equation with using (FPI_Fixed Point Iteration) x−tan(x)=0 , in[4,5]

$${solve}\:{the}\:{equation}\:{with}\:{using}\:\left({FPI\_Fixed}\:{Point}\:{Iteration}\right) \\ $$$$ \\ $$$${x}−{tan}\left({x}\right)=\mathrm{0}\:,\:{in}\left[\mathrm{4},\mathrm{5}\right] \\ $$

Question Number 63383    Answers: 1   Comments: 1

solve this equation in all part of complex number: (√((x^9 −3x^2 +1)(x−6)+4))=(x^9 −3x^2 +1)(x−6)−16

$${solve}\:{this}\:{equation}\:{in}\:{all}\: \\ $$$${part}\:{of}\:{complex}\:{number}: \\ $$$$\sqrt{\left({x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{6}\right)+\mathrm{4}}=\left({x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{6}\right)−\mathrm{16} \\ $$

Question Number 63381    Answers: 0   Comments: 2

solve for both x and n in equation: x^n =216 in all part of integer A {_(n=3) ^(x=6) B{_(n=4) ^(x=5) C{_(n=5) ^(x=4) D {_(n=6) ^(x=3)

$${solve}\:{for}\:{both}\:{x}\:{and}\:{n} \\ $$$${in}\:{equation}:\:{x}^{{n}} =\mathrm{216}\:{in}\:{all} \\ $$$${part}\:{of}\:{integer} \\ $$$$\mathscr{A}\:\underset{{n}=\mathrm{3}} {\overset{{x}=\mathrm{6}} {\left\{}}\right. \\ $$$$\mathscr{B}\underset{{n}=\mathrm{4}} {\overset{{x}=\mathrm{5}} {\left\{}}\right. \\ $$$$\mathscr{C}\underset{{n}=\mathrm{5}} {\overset{{x}=\mathrm{4}} {\left\{}}\right. \\ $$$$\mathscr{D}\:\underset{{n}=\mathrm{6}} {\overset{{x}=\mathrm{3}} {\left\{}}\right. \\ $$

Question Number 63373    Answers: 0   Comments: 2

just found this on the web I thought it might help in some cases where quartics appear i.e. Sir Aifour′s geometric questions. sometimes we know the nature of the roots, but how to use this information? ax^4 +bx^3 +cx^2 +dx+e=0 1. divide by a 2. x=z−(b/(4a)) this leads to the reduced z^4 +pz^2 +qz+r=0 now we find the nature of the roots: T_1 =16p^4 r−4p^3 q^2 −128p^2 r^2 +144pq^2 r−27q^4 +256r^3 T_2 =p^2 +12r T_3 =−p^2 +4r T_1 <0 ⇒ 2 distinct real and 2 conjugated complex roots T_1 >0∧(p<0∧T_3 <0) ⇒ 4 distinct real roots T_1 >0∧(p>0∨T_3 >0) ⇒ 2 pairs of conjugated complex roots T_1 =0∧(p<0∧T_3 <0∧T_2 ≠0) ⇒ 1 real double and 2 real simple roots T_1 =0∧(T_3 >0∨(p>0∧(T_3 ≠0∨q≠0))) ⇒ 1 real double and 2 conjugated complex roots T_1 =0∧(T_2 =0∧T_3 ≠0) ⇒ 1 real triple and 1 real simple roots T_1 =0∧(T_3 =0∧p<0) ⇒ 2 real double roots T_1 =0∧(T_3 =0∧p>0∧q=0) ⇒ 2 conjugated complex double roots T_1 =0∧T_2 =0 ⇒ all roots are equal

$$\mathrm{just}\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web} \\ $$$$\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{might}\:\mathrm{help}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{quartics}\:\mathrm{appear}\:\mathrm{i}.\mathrm{e}.\:\mathrm{Sir}\:\mathrm{Aifour}'\mathrm{s}\:\mathrm{geometric} \\ $$$$\mathrm{questions}.\:\mathrm{sometimes}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots},\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{this}\:\mathrm{information}? \\ $$$$ \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0} \\ $$$$\mathrm{1}.\:\mathrm{divide}\:\mathrm{by}\:{a} \\ $$$$\mathrm{2}.\:{x}={z}−\frac{{b}}{\mathrm{4}{a}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{reduced} \\ $$$$ \\ $$$${z}^{\mathrm{4}} +{pz}^{\mathrm{2}} +{qz}+{r}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}: \\ $$$${T}_{\mathrm{1}} =\mathrm{16}{p}^{\mathrm{4}} {r}−\mathrm{4}{p}^{\mathrm{3}} {q}^{\mathrm{2}} −\mathrm{128}{p}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{144}{pq}^{\mathrm{2}} {r}−\mathrm{27}{q}^{\mathrm{4}} +\mathrm{256}{r}^{\mathrm{3}} \\ $$$${T}_{\mathrm{2}} ={p}^{\mathrm{2}} +\mathrm{12}{r} \\ $$$${T}_{\mathrm{3}} =−{p}^{\mathrm{2}} +\mathrm{4}{r} \\ $$$${T}_{\mathrm{1}} <\mathrm{0}\:\Rightarrow\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\right)\:\Rightarrow\:\mathrm{4}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}>\mathrm{0}\vee{T}_{\mathrm{3}} >\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\wedge{T}_{\mathrm{2}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} >\mathrm{0}\vee\left({p}>\mathrm{0}\wedge\left({T}_{\mathrm{3}} \neq\mathrm{0}\vee{q}\neq\mathrm{0}\right)\right)\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{2}} =\mathrm{0}\wedge{T}_{\mathrm{3}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{triple}\:\mathrm{and}\:\mathrm{1}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}<\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{real}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}>\mathrm{0}\wedge{q}=\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge{T}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\mathrm{all}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{equal} \\ $$

Question Number 63247    Answers: 0   Comments: 1

Prove that (√(abc)) + (√((1−a)(1−b)(1−c))) ≤ 1 for 0 ≤ a,b,c ≤ 1

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\sqrt{{abc}}\:+\:\sqrt{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)}\:\leqslant\:\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{0}\:\leqslant\:{a},{b},{c}\:\leqslant\:\mathrm{1} \\ $$

Question Number 63645    Answers: 0   Comments: 4

n integr natural prove that 5 divide n^5 −n

$${n}\:{integr}\:{natural}\:{prove}\:{that}\:\mathrm{5}\:{divide}\:{n}^{\mathrm{5}} −{n} \\ $$

Question Number 63291    Answers: 0   Comments: 3

find some of all real x such that ((4x^2 +15x+17)/(x^2 +4x+12)) = ((5x^2 +16x+18)/(2x^2 +5x+13))

$${find}\:{some}\:{of}\:{all}\:{real}\:{x}\:{such}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{17}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{12}}\:=\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{13}} \\ $$

Question Number 63289    Answers: 0   Comments: 0

Question Number 63175    Answers: 0   Comments: 2

solve for x x^x^x = 16 x = 2, but how to use Lambert W function

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:=\:\:\mathrm{16} \\ $$$$\mathrm{x}\:=\:\mathrm{2},\:\:\:\:\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Lambert}\:\mathrm{W}\:\mathrm{function} \\ $$

Question Number 63162    Answers: 1   Comments: 3

Find the set of values of x which satisfy the inequalities (2/(x−1))≤(1/x) and x^2 −∣3x∣+2<0

$${Find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{x}\:{which}\:{satisfy}\:{the}\:{inequalities}\: \\ $$$$\frac{\mathrm{2}}{{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{{x}}\:\:{and}\:\:{x}^{\mathrm{2}} −\mid\mathrm{3}{x}\mid+\mathrm{2}<\mathrm{0} \\ $$

Question Number 63095    Answers: 1   Comments: 0

Question Number 63090    Answers: 0   Comments: 0

s=(√(a^2 +(a^2 −d)^2 ))+(√((b−a)^2 +(b^2 −a^2 )^2 )) +(√(b^2 +(c−b^2 )^2 ))+c−d p= a(a^2 −d)+(a+b)(b^2 −a^2 ) +b(c−b^2 ) Find a,b,c, or d in terms of s if p is maximum. Assume a,b,c,d ≥0 .

$${s}=\sqrt{{a}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{d}\right)^{\mathrm{2}} }+\sqrt{\left({b}−{a}\right)^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt{{b}^{\mathrm{2}} +\left({c}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} }+{c}−{d} \\ $$$$\:{p}=\:{a}\left({a}^{\mathrm{2}} −{d}\right)+\left({a}+{b}\right)\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{b}\left({c}−{b}^{\mathrm{2}} \right) \\ $$$${Find}\:{a},{b},{c},\:{or}\:{d}\:\:{in}\:{terms}\:{of}\:{s} \\ $$$${if}\:\:{p}\:{is}\:{maximum}.\: \\ $$$${Assume}\:\:\:\:{a},{b},{c},{d}\:\geqslant\mathrm{0}\:. \\ $$

Question Number 63076    Answers: 0   Comments: 0

show that f:A→B is bijection then f(A_1 ^c )=[f(A_1 )]^c

$${show}\:{that}\:{f}:{A}\rightarrow{B}\:{is}\:{bijection}\:{then}\:{f}\left({A}_{\mathrm{1}} ^{{c}} \right)=\left[{f}\left({A}_{\mathrm{1}} \right)\right]^{{c}} \\ $$

Question Number 63021    Answers: 2   Comments: 3

solve this equation x^y =y^x x,y∈R.

$${solve}\:{this}\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{y}} ={y}^{{x}} \\ $$$$ \\ $$$$ \\ $$$${x},{y}\in\mathbb{R}. \\ $$

Question Number 62998    Answers: 0   Comments: 12

Solve for x: 5^x +6x=7

$${Solve}\:{for}\:{x}:\:\:\mathrm{5}^{\boldsymbol{{x}}} +\mathrm{6}\boldsymbol{{x}}=\mathrm{7} \\ $$

Question Number 62945    Answers: 1   Comments: 0

Find the greatest coefficient in the expansion of (6 − 4x)^(−3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\:\:\left(\mathrm{6}\:−\:\mathrm{4x}\right)^{−\mathrm{3}} \\ $$

Question Number 62942    Answers: 1   Comments: 10

Make r the subject of the formular: S = ((a(r^n − 1))/(r − 1))

$$\mathrm{Make}\:\:\mathrm{r}\:\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formular}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}} \:−\:\mathrm{1}\right)}{\mathrm{r}\:−\:\mathrm{1}} \\ $$

Question Number 62895    Answers: 1   Comments: 2

Question Number 62844    Answers: 1   Comments: 0

Let p(x) = ax^2 + bx + c be such that p(x) takes real values for real values of x and non−real values for non−real values of x . Prove that a = 0 and find all possible values of c.

$${Let}\:{p}\left({x}\right)\:=\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:\:{be}\:{such}\:{that}\:{p}\left({x}\right)\:{takes}\:{real}\:{values} \\ $$$${for}\:{real}\:{values}\:{of}\:{x}\:{and}\:{non}−{real}\:{values}\:{for}\:{non}−{real} \\ $$$${values}\:{of}\:{x}\:.\:{Prove}\:{that}\:{a}\:=\:\mathrm{0}\:{and}\:{find}\:{all} \\ $$$${possible}\:{values}\:{of}\:{c}. \\ $$

Question Number 62798    Answers: 0   Comments: 0

Calculate tg(20°)+4sin(20°)+1

$$\boldsymbol{\mathrm{Calculate}} \\ $$$$\boldsymbol{\mathrm{tg}}\left(\mathrm{20}°\right)+\mathrm{4}\boldsymbol{\mathrm{sin}}\left(\mathrm{20}°\right)+\mathrm{1} \\ $$

Question Number 62676    Answers: 1   Comments: 1

Question Number 62672    Answers: 1   Comments: 9

Question Number 62669    Answers: 2   Comments: 0

calculate the value of Σ_(n=0) ^(1947) (1/(2^n +(√2^(1947) )))

$${calculate}\:{the}\:{value}\:{of}\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{1947}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} +\sqrt{\mathrm{2}^{\mathrm{1947}} }} \\ $$

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