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25

If $z$ is a root of a real polynomial, say $p(z) =\sum_{j=0}^n r_jz^j = 0$, then $\overline{z}$ is also a root of $p$ as $p(\overline{z}) =\sum_{j=0}^n r_j\overline{z}^j = \sum_{j=0}^n r_j\overline{z^j}= \sum_{j=0}^n \overline{r_jz^j} = \overline{p(z)} = 0$. Thus, non-real roots of real polynomials always come in pairs and their number is thus even. There ...


22

Let $p(x) = x^3+x^2-2x-1$, we have $$p(t + t^{-1}) = t^3 + t^2 + t + 1 + t^{-1} + t^{-2} + t^{-3} = \frac{t^7-1}{t^3(t-1)}$$ The RHS has roots of the form $t = e^{\pm \frac{2k\pi}{7}i}$ ( coming from the $t^7 - 1$ factor in numerator ) for $k = 1,2,3$. So $p(x)$ has roots of the form $$e^{\frac{2k\pi}{7} i} + e^{-\frac{2k\pi}{7} i} = 2\cos\left(\frac{2 k\...


10

Who told a quartic polynomial can't have 3 real roots.? It can have provided the coefficients are complex... If coefficients have to be real then you must see that to cancel the $i$ of one root there must be one more $i$ in another root.!


8

Consider the equation $$\cos4\theta=\cos3\theta$$ whose roots are $$\theta=n\cdot\frac{2\pi}{7}$$ Representing this as a polynomial in $c=\cos\theta$, we have $$8c^4-4c^3-8c^2+3c+1=0$$ $$\Rightarrow (c-1)(8c^3+4c^2-4c-1)=0$$ Now write $x=2c$ and we see that the polynomial equation $$x^3+x^2-2x-1=0$$ has roots as stated in your question. Note that $$2\cos\...


7

For $(a,b,c,d)=(1,2,3,1)$ we have $ad=1$ and $bc=6$, but nevertheless the polynomial $$ x^4+ax^3+bx^2+cx+d=x^4 + x^3 + 2x^2 + 3x + 1=(x+1)(x^3+2x+1) $$ has a rational root. Did I misunderstand something ? Edit: Suppose that $$ x^4+ax^3+bx^2+cx+d=(x-a_1)(x-a_2)(x-a_3)(x-a_4) $$ where we may assume that $a,b,c,d$ and all $a_i$ are integers. Then we obtain, ...


7

HINT: Let $x=b\cos A$ $$\implies b^3\cos^3A-3(b\cos A)=1$$ As $\cos3A=4\cos^3A-3\cos A,$ $$\dfrac43=\dfrac{b^3}{3b}\implies b^2=4\text{ as }b\ne0$$ Let $b=2$ Consequently, $$2\cos3A=1\iff\cos3A=\dfrac12=\cos\dfrac\pi3$$ $$3A=2n\pi\pm\dfrac\pi3=\dfrac\pi3(6n\pm1)$$ where $n$ is any integer $$A=\dfrac\pi9(6n\pm1)$$ where $n\equiv0,\pm1\pmod3$ What if ...


6

Set $x=t+t^{-1}$. Then the equation becomes $$ t^3+3t+3t^{-1}+t^{-3}+t^2+2+t^{-2}-2t-2t^{-1}-1=0 $$ and, multiplying by $t^3$, $$ t^6+t^5+t^4+t^3+t^2+t+1=0 $$ and it should be now clear what the solutions are. For each root there's another one giving the same solution in $x$.


5

$2x^4+x^3-11x^2+x+2=0$ Note that the coefficients: $2,1,-11,1,2$ are symmetrical. $2(x^4+1)+(x^3+x)-11x^2=0$ $2(x^4+4x^3+6x^2+4x+1)-7(x^3+x)-23x^2=0$ $2(x^4+4x^3+6x^2+4x+1)-7(x^3+2x^2+x)-9x^2=0$ $2(x+1)^4-7(x+1)^2x-9x^2=0$ $2\left(\frac{(x+1)^2}{x}\right)^2-7\left(\frac{(x+1)^2}{x}\right)-9=0$


5

$$2x^4+x^3-11x^2+x+2 = (x-2)(2x^3+5x^2-x-1)=(x-2)(2x-1)(x^2+3x+1)$$ Now you can solve it easily EDIT: You can use the rational root theorem to get the possible rational roots


4

The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots are produced in a similar manner, ...


4

As $\alpha\beta=\dfrac71,$ let $y=\dfrac{\alpha\beta+1}\alpha=\dfrac{7+1}\alpha\iff\alpha=\dfrac8y$ Put the value of $\alpha$ in $$x^2-6x+7=0$$ and rearrange.


4

If $\alpha,\beta$ are the roots of $x^2-6x+7=0$, then all quadratic equations with roots $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$ are $$a\left(x-\left(\alpha+\frac{1}{\beta}\right)\right)\left(x-\left(\beta+\frac{1}{\alpha}\right)\right)=0,$$ where $a\in\mathbb R$, $a\neq 0$ (saying "the equation" is wrong, because there are infinitely many of ...


4

We may take $\xi$ as a primitive $19$-th root of unity, $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then check that the elementary symmetric functions of $$ a = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17} $$ $$ b = \xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15} $$ $$ c = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18} $$ simplify giving the coefficients of $x^...


3

Hint: $$\sum_{k=0}^{n-1} a r^k = a \frac{1-r^n}{1-r}.$$ Substitute $n=16$ and $r=1-a$.


3

For such a case, there is the trigonometric method


3

Use the number of digits. For your example $\sqrt[11]{34567887654323456780987634567895}$ I would count that there are $32$ digits, so the radicand can be bounded as follows: $$\left(10^2\right)^{11} = 10^{22} < 34567887654323456780987634567895 < 10^{33}=\left(10^3\right)^{11}$$ That means $$ 100 < \sqrt[11]{34567887654323456780987634567895} &...


3

Extend to complex plane, you can use Rouche's theorem to conclude that zeros of polynomial $a_nz^n+...+a_0$ (where $a_n\neq 0$ and $a_0\neq 0$) lies in $A(0;r,R)=\{r<|z|<R\}$ where $R=1+\frac{\max\{|a_j|:0\leq j\leq n-1\}}{|a_n|}$, $r=(1+\frac{\max\{|a_j|:1\leq j\leq n\}}{|a_0|})^{-1}$. See Problem 6.


3

Between any two real roots of a polynomial there is a root of its derivative. Between any two real roots of the derivative there is a root of the second derivative. So your second derivative should have all real roots. Take another look at it.


3

In A Note on Trigonometric Algebraic Numbers by D. H. Lehmer (and also here with a different notation) we find that $$ z^{-d}\Phi_n(z) = \psi_n(z+z^{-1}) $$ where $\Phi_n$ is the $n$-th cyclotomic polynomial and $d=\frac{\phi(n)}{2}$ is the degree of $\psi_n$, which is half the degree of $\Phi_n$. Lehmer proves that $\psi_n$ is irreducible. The roots of $\...


3

By the rational root theorem, there are $6$ possibilities to check: $\pm\frac12,\pm1,$ or $\pm2$. The easiest values to test are $\pm1$ and neither works. The next easiest value to test is $2$, which is a solution. In addition, using Kenny Lau's observation from his comment, the equation can be re-written as $$2x^2+x-11+\frac1x+\frac2{x^2}=0$$ Since $x=...


2

Up to division by a constant, $$p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$$ Call $M=\max\{1,\lvert a_0\rvert,\lvert a_1\rvert,\cdots,\lvert a_{n-1}\rvert\}$ For $\lvert x\rvert\ge 1$, by inverse triangular identity you know that $$\lvert p(x)\rvert\ge \lvert x\rvert^n-nM\lvert x\rvert^{n-1}=\lvert x\rvert^{n-1}(\lvert x\rvert -nM)$$ So, for $n\ge2$, $p(x)\ne 0$ ...


2

Hint: $\alpha$ is a root of $f(x)$ if and only if $(x-\alpha)$ divides $f(x)$. The complex numbers $\mathbb{C}$ are algebraically closed (which is what your teacher probably meant). This mean that a polynomial $p(x)$ of degree $n$with coefficients in $\mathbb{C}$ always "factors completely" as \begin{equation} p(x)=a(x-\alpha_{1})\dots(x-\alpha_{n}), \end{...


2

Let $x=t+t^{-1}$ in $y=x^3-3x-1$. So we have $$\left(t+\frac {1}{t}\right)^{3}-3\left(t+\frac {1}{t}\right)-1=0$$ Expanding that and simplifying, we get $\frac {t^{6}-t^{3}+1}{t^{3}}$ and multiplying both the numerator and the denominator by $(t^3+1)$ we get $\frac {t^{9}+1}{t^{3}(t^{3}+1)}$. We let $t=\cos(\alpha)+i\sin(\alpha)$ and we get $\cos(9\alpha)+...


2

Given a complex number z, a polynomial of degree two having $z$ and $\bar z$ as roots is $(x-z)(x-\bar z) = x^2 -(z + \bar z)x +z\bar z$. $z +\bar z$ is real and $z\bar z$ is real too. Thus to obtain a polynomial of degree two with conjugate complex roots and complex coefficients you only have trivial examples like $3i(x^2+1)$ or the one you mentioned ...


2

Notice that $x=c$ is a root of $f(x)$ iff $x=c-z$ is a root of $f(x+z)$. So the roots of $f(x+z)$ are just the points $c-z$ where $c$ can be any of the vertices of our parallelogram. But if $c_1$ is a vertex of the parallelogram and $c_2$ is the opposite vertex, then $z=(c_1+c_2)/2$. Thus $c_2-z=-(c_1-z)$. That is, if $\alpha=c_1-z$ is a root of $f(x+z)$,...


2

$(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta=0\\ (\alpha + \beta) = 6\\ \alpha\beta = 7$ $(x-\alpha - \frac 1\beta)(x-\beta - \frac 1\alpha) = x^2 - (\alpha + \beta + \frac 1\alpha + \frac 1\beta)x + (\alpha\beta + 2+ \frac {1}{\alpha\beta})=0\\ x^2 - (\alpha + \beta + \frac{(\alpha + \beta)}{\alpha\beta})x + (\alpha\beta + 2+ \frac {1}{\...


2

$$(x-(\alpha + 1/\beta))(x - (\beta + 1/\alpha)) = x^2 - (\alpha + \beta + 1/\alpha + 1/\beta) x + (\alpha + 1/\beta)(\beta + 1/\alpha $$ You know $\alpha + \beta = 6$ and $\alpha \beta = 7$. $$ \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha \beta} = \dfrac{6}{7}$$ $$\left( \alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\...


2

By the translation $x\to x-\frac13$, you cancel the square term. $$(x-\frac13)^3+(x-\frac13)^2-6(x-\frac13)-7=x^3-\frac{19x}3-\frac{133}{27}.$$ Then multiplying by $27$ and replacing $x\to3x$, you get $$x^3-57x-133=0.$$ Next, setting $x=2\sqrt{19}\cos(\theta)$ (with the aim to let $4\cos^3(\theta)-3\cos(\theta)=\cos(3\theta)$ appear), $$152\sqrt{19}\cos^...


2

The question is $\dfrac{20\sqrt3-23}3?\dfrac{\sqrt6+12}3$ Multiply both sides by $3$, then add $23$: $20\sqrt3$ $?$ $\sqrt6+35$ Now $\sqrt3<1.8$, as $3<3.24=1.8^2$ and $\sqrt6>1$ So $20\sqrt3<36=1+35<\sqrt6+35$ $\therefore\dfrac{20\sqrt3-23}3<\dfrac{\sqrt6+12}3$


1

First note that the problem is equivalent to proving that for every $\epsilon>0$, $\exists N$ such that for $\forall n\geq N$,$f_n(\dfrac{1}{z})=P_n(z)$ has all zeroes outside $B(0,\epsilon)$. Assume on the contrary the existence of such an $\epsilon.$ Then for each $P_n$, $\exists z_n \in B(0,\epsilon)$ such that $P_n(z_n)=0.$ Observe that $P_n \to f$, ...



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