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AlgebraQuestion and Answers: Page 347

Question Number 26227    Answers: 1   Comments: 0

if x^2 +(1/x^2 )=98 find x^3 +(1/x^3 )

$$\mathrm{if}\:\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{98}\:\mathrm{find}\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} } \\ $$

Question Number 26224    Answers: 1   Comments: 1

find the value of x−(1/x).when x^4 +(1/x^4 )=332

$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}.\mathrm{when}\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }=\mathrm{332} \\ $$

Question Number 26218    Answers: 0   Comments: 2

∫_2 ^4 sinθ dθ

$$\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\mathrm{sin}\theta\:{d}\theta \\ $$

Question Number 26207    Answers: 1   Comments: 0

I think of a two digit number.The sum of the digits is 9. When the number is reversed and subtracted from the original, the result is 45. Find the original number

$$\mathrm{I}\:\mathrm{think}\:\mathrm{of}\:\mathrm{a}\:\mathrm{two}\:\mathrm{digit}\:\mathrm{number}.\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{9}. \\ $$$$\mathrm{When}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{reversed}\:\mathrm{and}\:\mathrm{subtracted}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{original},\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{45}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{original}\:\mathrm{number} \\ $$

Question Number 26153    Answers: 1   Comments: 0

ratio of income of two persons is 9 is to 7.ratio of their expenses is 4 is to 3 .every person saves rupees 200. find income of each.

$${ratio}\:{of}\:{income}\:{of}\:{two}\:{persons}\:{is} \\ $$$$\mathrm{9}\:{is}\:{to}\:\mathrm{7}.{ratio}\:{of}\:{their}\:{expenses} \\ $$$${is}\:\mathrm{4}\:{is}\:{to}\:\mathrm{3}\:.{every}\:{person}\:{saves}\: \\ $$$${rupees}\:\mathrm{200}.\:{find}\:{income}\:{of}\:{each}. \\ $$

Question Number 26147    Answers: 1   Comments: 0

There are 5 more girls than boys in a class. If 2 boys join the class, the ratio of girls to boys will be 5:4. Find the number of of girls in the class.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{5}\:\mathrm{more}\:\mathrm{girls}\:\mathrm{than}\:\mathrm{boys}\:\mathrm{in}\:\mathrm{a}\:\mathrm{class}.\:\mathrm{If}\:\mathrm{2}\:\mathrm{boys}\:\mathrm{join} \\ $$$$\mathrm{the}\:\mathrm{class},\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{girls}\:\mathrm{to}\:\mathrm{boys}\:\mathrm{will}\:\mathrm{be}\:\mathrm{5}:\mathrm{4}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{of}\:\mathrm{girls}\:\mathrm{in}\:\mathrm{the}\:\mathrm{class}. \\ $$

Question Number 26143    Answers: 2   Comments: 1

2000^(3000) vs 3000^(2000) who is stronger ?

$$\mathrm{2000}^{\mathrm{3000}} \:\:\boldsymbol{{vs}}\:\mathrm{3000}^{\mathrm{2000}} \\ $$$$ \\ $$$$\:\boldsymbol{{who}}\:\boldsymbol{{is}}\:\boldsymbol{{stronger}}\:? \\ $$

Question Number 26133    Answers: 1   Comments: 1

find the value of (C_n ^(0 ) )^2 +(C_n ^1 )^2 +(C_n ^2 )^2 +...(C_n ^n )^2 .

$${find}\:{the}\:{value}\:{of}\:\:\left({C}_{{n}} ^{\mathrm{0}\:\:} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{1}} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{2}} \right)^{\mathrm{2}} \:+...\left({C}_{{n}} ^{{n}} \right)^{\mathrm{2}} . \\ $$

Question Number 26117    Answers: 1   Comments: 0

(1/6)(√((3log1728)/(1+(1/2)log36+(1/3)log8))) simplify the question above

$$\frac{\mathrm{1}}{\mathrm{6}}\sqrt{\frac{\mathrm{3}{log}\mathrm{1728}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{36}+\frac{\mathrm{1}}{\mathrm{3}}{log}\mathrm{8}}} \\ $$$${simplify}\:{the}\:{question}\:{above} \\ $$

Question Number 26115    Answers: 1   Comments: 0

x^2 +(1/x^2 )=3 thrn find the valu of( x−(1/x))^2

$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{3}\:\mathrm{thrn}\:\mathrm{find}\:\mathrm{the}\:\mathrm{valu}\:\mathrm{of}\left(\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \\ $$

Question Number 26067    Answers: 2   Comments: 0

Find the value of ((2 + 3^2 )/(1! + 2! + 3! + 4!)) + ((3 + 4^2 )/(2! + 3! + 4! + 5!)) + ... + ((2013 + 2014^2 )/(2012! + 2013! + 2014! + 2015!))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{2}\:+\:\mathrm{3}^{\mathrm{2}} }{\mathrm{1}!\:+\:\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!}\:+\:\frac{\mathrm{3}\:+\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!\:+\:\mathrm{5}!}\:+\:...\:+\:\frac{\mathrm{2013}\:+\:\mathrm{2014}^{\mathrm{2}} }{\mathrm{2012}!\:+\:\mathrm{2013}!\:+\:\mathrm{2014}!\:+\:\mathrm{2015}!} \\ $$

Question Number 26053    Answers: 1   Comments: 0

If x^2 + 9x + 2 = 0 and x^2 + kx + 5 = 0 have a common root, show that 2k^2 + 63k − 414 = 0 , hence find the value of k such that k > 9.3

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{9x}\:+\:\mathrm{2}\:=\:\mathrm{0}\:\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{kx}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{2k}^{\mathrm{2}} \:+\:\mathrm{63k}\:−\:\mathrm{414}\:=\:\mathrm{0}\:,\:\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{k}\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{k}\:>\:\mathrm{9}.\mathrm{3} \\ $$

Question Number 26042    Answers: 0   Comments: 2

ax^2 −bx=0

$${ax}^{\mathrm{2}} −{bx}=\mathrm{0} \\ $$

Question Number 25971    Answers: 1   Comments: 2

In finding the equations of the bisectors of the angles between two lines a_1 x+b_1 y+c_1 =0 and a_2 x+b_2 y+c_2 =0, why we observe a_1 a_2 +b_1 b_2 >0 or <0 for obtuse and acute angle bisectors?

$${In}\:{finding}\:{the}\:{equations}\:{of}\:{the} \\ $$$${bisectors}\:{of}\:{the}\:{angles}\:{between}\:{two} \\ $$$${lines}\:{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0}\:{and}\:{a}_{\mathrm{2}} {x}+{b}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} =\mathrm{0}, \\ $$$${why}\:{we}\:{observe}\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}} >\mathrm{0}\:{or}\:<\mathrm{0} \\ $$$${for}\:{obtuse}\:{and}\:{acute}\:{angle}\:{bisectors}? \\ $$

Question Number 25961    Answers: 0   Comments: 0

8cos^4 x−8cos^2 x+1=0 solution:8cos^2 x(cos^2 x−1)+1=0⇒ −8cos^2 xsin^2 x=−1⇒ sin^2 2x=(1/2)⇒sinx=+_− ((√2)/2)⇒ { ((2x=2kπ+(π/4))),((2x=2kπ+((3π)/4))) :} { ((2x=2kπ−(π/4))),((2x=2kπ+((5π)/4))) :} and so we have { ((x=kπ+(π/8))),((x=kπ+((3π)/8))) :} { ((x=kπ−(π/8))),((x=kπ+((5π)/8))) :} why x=((kπ)/4)+(π/8)? can we solve by another way?

$$\mathrm{8cos}^{\mathrm{4}} \mathrm{x}−\mathrm{8cos}^{\mathrm{2}} \mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solution}:\mathrm{8cos}^{\mathrm{2}} \mathrm{x}\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$−\mathrm{8cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}=−\mathrm{1}\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{2x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sinx}=\underset{−} {+}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi+\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi−\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{5}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi+\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi−\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{5}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\mathrm{why}\:\mathrm{x}=\frac{\mathrm{k}\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}? \\ $$$$ \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{another}\:\mathrm{way}? \\ $$

Question Number 25937    Answers: 1   Comments: 0

For a certain amount of work,Ade takes 6hours less than Bode.if they work together it takes them 13hours 20 minutes.How long will it take Bode alone to complete the work?

$${For}\:{a}\:{certain}\:{amount}\:{of}\:{work},{Ade}\:{takes} \\ $$$$\mathrm{6}{hours}\:{less}\:{than}\:{Bode}.{if}\:{they}\:{work}\:{together} \\ $$$${it}\:{takes}\:{them}\:\mathrm{13}{hours}\:\mathrm{20}\:{minutes}.{How} \\ $$$${long}\:{will}\:{it}\:{take}\:{Bode}\:{alone}\:{to}\:{complete} \\ $$$${the}\:{work}? \\ $$

Question Number 25853    Answers: 0   Comments: 2

simplify: 3^x +4^x =5^x anybody to solve this

$${simplify}:\:\mathrm{3}^{{x}} +\mathrm{4}^{{x}} \:=\mathrm{5}^{{x}} \\ $$$${anybody}\:{to}\:{solve}\:{this} \\ $$

Question Number 25823    Answers: 0   Comments: 1

z_1 =4−3i z_2 =9+2i Find (a) z_1 oz_2 (b) z_1 ×z_2 (c) angle between z_1 and z_2 .

$${z}_{\mathrm{1}} =\mathrm{4}−\mathrm{3}{i} \\ $$$${z}_{\mathrm{2}} =\mathrm{9}+\mathrm{2}{i} \\ $$$${Find}\:\:\left({a}\right)\:{z}_{\mathrm{1}} {oz}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}\right)\:{z}_{\mathrm{1}} ×{z}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({c}\right)\:{angle}\:{between}\:{z}_{\mathrm{1}} {and}\:{z}_{\mathrm{2}} . \\ $$

Question Number 25822    Answers: 0   Comments: 0

answer to the question of p(X)= (1+iX)^n −(1−iX)^n key of solutionafter resolving p(X)=0 the roots of p(X) are x_k =tan(kπ/n) with k in [[0.n−1]] so p(X)= ∝Π_(k=0) ^(k=n−1) (X−x_k^ ) let searsh ∝ by using binome formula p(X)= 2iΣ_(p=0) ^ (−1)^(p ) C_n ^(2p+1) X^(2p+1) so ∝= 2i(−1)^([n−1/2]) C_n ^ case1 n=2N p(X)=∝ Π_(k=0) ^(k=2N−1) (X−tan(kπ/2N)) and ∝=4in(−1)^(N−1) case2 n=2N+1 p(X)=∝Π_(k=0) ^(k=2N) (X−tan(kπ/2N+1) and ∝=2i(−1)^N

$${answer}\:{to}\:{the}\:{question}\:{of}\:{p}\left({X}\right)=\:\left(\mathrm{1}+{iX}\right)^{{n}} −\left(\mathrm{1}−{iX}\right)^{{n}} \\ $$$${key}\:{of}\:{solutionafter}\:{resolving}\:{p}\left({X}\right)=\mathrm{0}\:\:{the}\:{roots}\:{of}\:{p}\left({X}\right) \\ $$$${are}\:\:{x}_{{k}} ={tan}\left({k}\pi/{n}\right)\:{with}\:{k}\:\:{in}\:\left[\left[\mathrm{0}.{n}−\mathrm{1}\right]\right]\:{so} \\ $$$${p}\left({X}\right)=\:\propto\prod_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \left({X}−{x}_{{k}^{} } \right)\:{let}\:{searsh}\:\propto\:{by}\:{using}\:{binome}\:{formula} \\ $$$${p}\left({X}\right)=\:\mathrm{2}{i}\sum_{{p}=\mathrm{0}} ^{} \left(−\mathrm{1}\right)^{{p}\:} {C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} {X}^{\mathrm{2}{p}+\mathrm{1}} {so} \\ $$$$\propto=\:\mathrm{2}{i}\left(−\mathrm{1}\right)^{\left[{n}−\mathrm{1}/\mathrm{2}\right]} \:{C}_{{n}} ^{} \\ $$$${case}\mathrm{1}\:{n}=\mathrm{2}{N}\:\:\:{p}\left({X}\right)=\propto\:\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}−\mathrm{1}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}\right)\right) \\ $$$${and}\:\propto=\mathrm{4}{in}\left(−\mathrm{1}\right)^{{N}−\mathrm{1}} \\ $$$${case}\mathrm{2}\:\:{n}=\mathrm{2}{N}+\mathrm{1}\:\:\:\:{p}\left({X}\right)=\propto\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}+\mathrm{1}\right)\right. \\ $$$${and}\:\propto=\mathrm{2}{i}\left(−\mathrm{1}\right)^{{N}} \\ $$

Question Number 25787    Answers: 0   Comments: 2

2^x =x^(2 ) ,hence x=2 . prove

$$\mathrm{2}^{{x}} ={x}^{\mathrm{2}\:} ,{hence}\:{x}=\mathrm{2}\:.\:{prove} \\ $$

Question Number 25709    Answers: 0   Comments: 1

Q. 25444 (another solution) z^ +1 =iz^2 +∣z∣^2 let z=x+iy , ⇒ x−iy+1=i(x^2 −y^2 )−2xy+x^2 +y^2 or (x−y)^2 =x+1 ,and (y^2 −x^2 )=y let y−x=u and y+x=v ⇒ 2u^2 =v−u+2 ...(i) 2uv=u+v ....(ii) ⇒ v=(u/(2u−1)) , substituting in (i) 2u^2 +u=(u/(2u−1))+2 u=1 satisfies this , so ⇒ (2u+3)(u−1)=(u/(2u−1))−1 ⇒ (2u+3)(u−1)=(((1−u))/(2u−1)) ⇒ u=1 or (2u+3)(2u−1)+1=0 or 4u^2 +4u−2=0 or 2u^2 +2u−1=0 ⇒ u=((−2±(√(4+8)))/4) =((−1±(√3))/2) for u_1 =1 , v_1 =(u_1 /(2u_1 −1))=1 so x_1 =((v_1 −u_1 )/2)=0 and y_1 =((v_1 +u_1 )/2) =1 hence z_1 =i for u_2 =(((√3)−1)/2) , v_2 =(((√3)−1)/(2((√3)−1−1))) =−((((√3)−1)(2+(√3)))/2) =−((((√3)+1))/2) x_2 =((v_2 −u_2 )/2)=−((√3)/2) y_2 =((v_2 +u_2 )/2) =−(1/2) hence z_2 =−(1/2)((√3)+i) for u_3 =−((((√3)+1))/2) v_3 =(u_3 /(2u_3 −1)) =((((√3)+1))/(2((√3)+2))) =((((√3)+1)(2−(√3)))/2) =(((√3)−1)/2) x_3 =((v_3 −u_3 )/2)=((√3)/2) and y_3 =((v_3 +u_3 )/2)=−(1/2) ⇒ z_3 =(1/2)((√3)−i) So to summarize, z_1 =i , z_2 =−(1/2)((√3)+i), and z_3 =(1/2)((√3)−i) .

$${Q}.\:\mathrm{25444}\:\:\left({another}\:{solution}\right) \\ $$$$\:\:\bar {{z}}+\mathrm{1}\:={iz}^{\mathrm{2}} +\mid{z}\mid^{\mathrm{2}} \\ $$$${let}\:\:{z}={x}+{iy}\:,\:\Rightarrow \\ $$$${x}−{iy}+\mathrm{1}={i}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)−\mathrm{2}{xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${or}\:\:\left({x}−{y}\right)^{\mathrm{2}} ={x}+\mathrm{1}\:,{and}\: \\ $$$$\:\:\:\:\:\:\:\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)={y} \\ $$$${let}\:\:\:{y}−{x}={u}\:\:{and}\:\:{y}+{x}={v} \\ $$$$\Rightarrow\:\:\mathrm{2}{u}^{\mathrm{2}} ={v}−{u}+\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{uv}={u}+{v}\:\:\:\:\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\:{v}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}\:\:,\:{substituting}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +{u}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}+\mathrm{2} \\ $$$${u}=\mathrm{1}\:{satisfies}\:{this}\:,\:{so} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}−\mathrm{1} \\ $$$$\Rightarrow\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{\left(\mathrm{1}−{u}\right)}{\mathrm{2}{u}−\mathrm{1}} \\ $$$$\Rightarrow\:\:{u}=\mathrm{1}\:{or}\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{4}{u}^{\mathrm{2}} +\mathrm{4}{u}−\mathrm{2}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{u}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:{u}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{8}}}{\mathrm{4}}\:=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${for}\:{u}_{\mathrm{1}} =\mathrm{1}\:\:,\:{v}_{\mathrm{1}} =\frac{{u}_{\mathrm{1}} }{\mathrm{2}{u}_{\mathrm{1}} −\mathrm{1}}=\mathrm{1} \\ $$$${so}\:\:{x}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} −{u}_{\mathrm{1}} }{\mathrm{2}}=\mathrm{0}\:\:\:{and}\: \\ $$$$\:\:\:\:\:{y}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} +{u}_{\mathrm{1}} }{\mathrm{2}}\:=\mathrm{1} \\ $$$${hence}\:\:{z}_{\mathrm{1}} ={i} \\ $$$${for}\:{u}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\:,\:\:{v}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}−\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:{x}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} −{u}_{\mathrm{2}} }{\mathrm{2}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:{y}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} +{u}_{\mathrm{2}} }{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${hence}\:\:\:{z}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+{i}\right) \\ $$$${for}\:\:{u}_{\mathrm{3}} =−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:{v}_{\mathrm{3}} =\frac{{u}_{\mathrm{3}} }{\mathrm{2}{u}_{\mathrm{3}} −\mathrm{1}}\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} −{u}_{\mathrm{3}} }{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and} \\ $$$$\:{y}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} +{u}_{\mathrm{3}} }{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{z}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−{i}\right) \\ $$$${So}\:\:{to}\:{summarize}, \\ $$$$\boldsymbol{{z}}_{\mathrm{1}} =\boldsymbol{{i}}\:\:,\:\:\boldsymbol{{z}}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+\boldsymbol{{i}}\right),\:{and} \\ $$$$\:\boldsymbol{{z}}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\boldsymbol{{i}}\right)\:. \\ $$

Question Number 25680    Answers: 0   Comments: 0

let p(X) = (1 +iX )^(n) − ( 1 − iX )^(n) if p(X ) =∝Σ ( X −xk) find xk and α

$${let}\:{p}\left({X}\right)\:=\:\left(\mathrm{1}\:+{iX}\:\overset{{n}} {\right)}\:−\:\left(\:\mathrm{1}\:−\:{iX}\:\overset{{n}} {\right)}\:\:{if}\:\:{p}\left({X}\:\right)\:=\propto\Sigma\:\left(\:{X}\:−{xk}\right)\:\:{find}\:{xk}\:{and}\:\alpha \\ $$

Question Number 25638    Answers: 0   Comments: 1

Question Number 25619    Answers: 1   Comments: 0

Question Number 25589    Answers: 0   Comments: 3

possible or not ? a^a +b^b > a^b +b^a with below conditions: case 1) a>b>1 case 2) 0<b<a<1

$$\boldsymbol{{possible}}\:\boldsymbol{{or}}\:\boldsymbol{{not}}\:? \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\boldsymbol{{a}}} +\boldsymbol{{b}}^{\boldsymbol{{b}}} >\:\boldsymbol{{a}}^{\boldsymbol{{b}}} +\boldsymbol{{b}}^{\boldsymbol{{a}}} \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{below}}\:\boldsymbol{{conditions}}: \\ $$$$\left.\:\boldsymbol{{case}}\:\mathrm{1}\right)\:\:\boldsymbol{{a}}>\boldsymbol{{b}}>\mathrm{1} \\ $$$$\left.\boldsymbol{{case}}\:\mathrm{2}\right)\:\:\:\:\mathrm{0}<\boldsymbol{{b}}<\boldsymbol{{a}}<\mathrm{1} \\ $$

Question Number 25536    Answers: 1   Comments: 0

If x = 2sec(3t) and y = 4tan(3t), Show that y = 4(x^2 − 4)

$$\mathrm{If}\:\:\:\mathrm{x}\:=\:\mathrm{2sec}\left(\mathrm{3t}\right)\:\:\mathrm{and}\:\:\mathrm{y}\:=\:\mathrm{4tan}\left(\mathrm{3t}\right),\:\:\:\mathrm{Show}\:\mathrm{that}\:\:\:\:\mathrm{y}\:=\:\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4}\right) \\ $$

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