Menu Close

Question-121585

Tinku Tara 0 Comments
Pin




Question Number 121585 by Algoritm last updated on 09/Nov/20
Commented by Algoritm last updated on 09/Nov/20
answer: x=2(√2)−3
$$\boldsymbol{\mathrm{answer}}:\:\boldsymbol{\mathrm{x}}=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3} \\ $$
Commented by MJS_new last updated on 09/Nov/20
x=re^(iθ) if we stay strictly real x∈R ⇒ θ=0∨θ=π usually we define x∈R: x^(1/3) =∣x∣^(1/3) sign (x) but if we′re in C we use the root x^(1/3) =r^(1/3) e^(i(θ/3)) which is ∉R for x∈R^− so we have to decide. if we use the 1^(st) definition (strictly real) we get (x^3 +7x^2 +16x+5)^3 =(−3x^2 −7x+5)(1−2x)^3 x^9 +21x^8 +195x^7 +1030x^6 +3306x^5 +6571x^4 +7637x^3 +4266x^2 +1237x+120=0 (x+3)(x^2 +6x+1)(x^6 +12x^5 +68x^4 +187x^3 +295x^2 +159x+40)=0 x_1 =−3−2(√2) x_2 =−3 x_3 =−3+2(√2) if we use the 2^(nd) definition we get x_1 =−3+2(√2) x_(2, 3) ≈−3.99799±3.26066i
$${x}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{stay}\:\mathrm{strictly}\:\mathrm{real}\:{x}\in\mathbb{R}\:\Rightarrow\:\theta=\mathrm{0}\vee\theta=\pi \\ $$$$\mathrm{usually}\:\mathrm{we}\:\mathrm{define}\:{x}\in\mathbb{R}:\:{x}^{\mathrm{1}/\mathrm{3}} =\mid{x}\mid^{\mathrm{1}/\mathrm{3}} \mathrm{sign}\:\left({x}\right) \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{we}'\mathrm{re}\:\mathrm{in}\:\mathbb{C}\:\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{root} \\ $$$${x}^{\mathrm{1}/\mathrm{3}} ={r}^{\mathrm{1}/\mathrm{3}} \mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{3}}} \:\mathrm{which}\:\mathrm{is}\:\notin\mathbb{R}\:\mathrm{for}\:{x}\in\mathbb{R}^{−} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{decide}.\:\mathrm{if}\:\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{definition} \\ $$$$\left(\mathrm{strictly}\:\mathrm{real}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\left({x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{5}\right)^{\mathrm{3}} =\left(−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{5}\right)\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{9}} +\mathrm{21}{x}^{\mathrm{8}} +\mathrm{195}{x}^{\mathrm{7}} +\mathrm{1030}{x}^{\mathrm{6}} +\mathrm{3306}{x}^{\mathrm{5}} +\mathrm{6571}{x}^{\mathrm{4}} +\mathrm{7637}{x}^{\mathrm{3}} +\mathrm{4266}{x}^{\mathrm{2}} +\mathrm{1237}{x}+\mathrm{120}=\mathrm{0} \\ $$$$\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{1}\right)\left({x}^{\mathrm{6}} +\mathrm{12}{x}^{\mathrm{5}} +\mathrm{68}{x}^{\mathrm{4}} +\mathrm{187}{x}^{\mathrm{3}} +\mathrm{295}{x}^{\mathrm{2}} +\mathrm{159}{x}+\mathrm{40}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =−\mathrm{3} \\ $$$${x}_{\mathrm{3}} =−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{definition}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}_{\mathrm{1}} =−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} \approx−\mathrm{3}.\mathrm{99799}\pm\mathrm{3}.\mathrm{26066i} \\ $$
Commented by Dwaipayan Shikari last updated on 09/Nov/20
Sir, how did you factorize it?
$${Sir},\:{how}\:{did}\:{you}\:{factorize}\:{it}? \\ $$
Commented by MJS_new last updated on 09/Nov/20
−3 was easy to find (try factors of the constant). the rest usually depends on good luck. you can approximate the real solutions and try to see if they “are something” or you must use software in this case −3+2(√2) was given, so I tried it and then very often the conjugate is also a solution for solutions that cannot be written as a±(√b) you have to use software. or spend lots of time...
$$−\mathrm{3}\:\mathrm{was}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{find}\:\left(\mathrm{try}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{constant}\right).\:\mathrm{the}\:\mathrm{rest}\:\mathrm{usually}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{good} \\ $$$$\mathrm{luck}.\:\mathrm{you}\:\mathrm{can}\:\mathrm{approximate}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\mathrm{and}\:\mathrm{try}\:\mathrm{to}\:\mathrm{see}\:\mathrm{if}\:\mathrm{they}\:“\mathrm{are}\:\mathrm{something}''\:\mathrm{or} \\ $$$$\mathrm{you}\:\mathrm{must}\:\mathrm{use}\:\mathrm{software} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{was}\:\mathrm{given},\:\mathrm{so}\:\mathrm{I}\:\mathrm{tried} \\ $$$$\mathrm{it}\:\mathrm{and}\:\mathrm{then}\:\mathrm{very}\:\mathrm{often}\:\mathrm{the}\:\mathrm{conjugate}\:\mathrm{is}\:\mathrm{also} \\ $$$$\mathrm{a}\:\mathrm{solution} \\ $$$$\mathrm{for}\:\mathrm{solutions}\:\mathrm{that}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\: \\ $$$${a}\pm\sqrt{{b}}\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{use}\:\mathrm{software}.\:\mathrm{or}\:\mathrm{spend}\:\mathrm{lots} \\ $$$$\mathrm{of}\:\mathrm{time}… \\ $$
Commented by Dwaipayan Shikari last updated on 09/Nov/20
Suppose a given x^5 −x^3 +x−2=0 And we are asked to factorize it , then which method we should apply sir?
$${Suppose}\:{a}\:{given}\:{x}^{\mathrm{5}} −{x}^{\mathrm{3}} +{x}−\mathrm{2}=\mathrm{0}\: \\ $$$${And}\:{we}\:{are}\:{asked}\:{to}\:{factorize}\:{it}\:,\:{then}\:{which}\:{method}\:{we}\:{should} \\ $$$${apply}\:{sir}? \\ $$
Commented by MJS_new last updated on 09/Nov/20
first, try ±2, ±1 ⇒ no solution then find the number of real zeros. plot it or go on like this: y′=5x^4 −3x^2 +1=0 this has no real solutions ⇒ y has no extremes ⇒ as it′s a (2n+1)^(th) degree polynomial it must have exactly 1 real solution if we are lucky it can be written as y=(x^2 +ax+b)(x^3 +αx^2 +βx+γ) which gives (1) a+α=0 (2) aα+b+β+1=0 (3) aβ+bα+γ=0 (4) aγ+bβ−1=0 (5) bγ+2=0 solving this is not easy but possible we get a “nice” solution a=β=−1 b=α=1 γ=−2 but in most cases this won′t lead to exact values for these factors. try x^5 +2x^3 −3x+5
$$\mathrm{first},\:\mathrm{try}\:\pm\mathrm{2},\:\pm\mathrm{1}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{zeros}.\:\mathrm{plot}\:\mathrm{it}\:\mathrm{or} \\ $$$$\mathrm{go}\:\mathrm{on}\:\mathrm{like}\:\mathrm{this}: \\ $$$${y}'=\mathrm{5}{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{solutions}\:\Rightarrow\:{y}\:\mathrm{has}\:\mathrm{no}\:\mathrm{extremes} \\ $$$$\Rightarrow\:\mathrm{as}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{polynomial}\:\mathrm{it} \\ $$$$\mathrm{must}\:\mathrm{have}\:\mathrm{exactly}\:\mathrm{1}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{are}\:\mathrm{lucky}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as} \\ $$$${y}=\left({x}^{\mathrm{2}} +{ax}+{b}\right)\left({x}^{\mathrm{3}} +\alpha{x}^{\mathrm{2}} +\beta{x}+\gamma\right) \\ $$$$\mathrm{which}\:\mathrm{gives} \\ $$$$\left(\mathrm{1}\right)\:{a}+\alpha=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{a}\alpha+{b}+\beta+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:{a}\beta+{b}\alpha+\gamma=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:{a}\gamma+{b}\beta−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{5}\right)\:{b}\gamma+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{solving}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{easy}\:\mathrm{but}\:\mathrm{possible} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:“\mathrm{nice}''\:\mathrm{solution} \\ $$$${a}=\beta=−\mathrm{1} \\ $$$${b}=\alpha=\mathrm{1} \\ $$$$\gamma=−\mathrm{2} \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{most}\:\mathrm{cases}\:\mathrm{this}\:\mathrm{won}'\mathrm{t}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{exact} \\ $$$$\mathrm{values}\:\mathrm{for}\:\mathrm{these}\:\mathrm{factors}. \\ $$$$\mathrm{try} \\ $$$${x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5} \\ $$
Commented by Dwaipayan Shikari last updated on 10/Nov/20
Thanking you
$${Thanking}\:{you}\: \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *