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15

Given : $ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}} {1+\sqrt{2-2x}} $ Let $\alpha = x+3 $ ; $\beta = 2x+2$ ; $\left(\beta - \alpha\right) = x-1$ $\implies 1+ \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = x + \dfrac{\sqrt{\beta}} {1+\sqrt{4-\beta}} $ $\implies \alpha + \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = \beta + ...


15

The Rational Root Test shows that the only possible rational solutions are $\pm 1$. Substituting gives that $x = -1$ is one (but $x = 1$ is not), so polynomial long division gives $p(x) = -(x + 1) q(x)$ for some quintic $q$. Substituting $x = -1$ gives that $-1$ is not a root of $q$, so if $q$ factors over $\Bbb Q$, it does so into an irreducible quadratic ...


9

Consider this system of polynomial equations: $$\begin{align} a+b+c &= 2 \\ a^3 &= 7x + 1 \\ b^3 &= 8 + x - x^2 \\ c^3 &= x^2 - 8 x - 1 \end{align}$$ Using the method of resultants (via the oh-so-convenient Resultant[] function in Mathematica), we can eliminate $a$, $b$, $c$ to get $$x^3\;(x-1)^6\;(x+1)^3\;(x - 9 )^3 \;=\; 0$$ Therefore, the ...


8

Why does this equation has $4$ roots? The short answer is it is an accident. For similar problem like $$(7x-6)^{1/3}+(8+x-x^2)^{1/3}+(x^2-8x-1)^{1/3}=1 $$ You can get $5$ instead of $4$ solutions. Define $a,b,c$ like Blue's answer. $$\begin{cases} a^3 &= 7x + 1 \\ b^3 &= 8 + x - x^2 \\ c^3 &= x^2 - 8 x - 1 \end{cases} $$ The key of the whole ...


7

If we set $$ p_a(x) = x^{a+2}-x^{a+1}-1 $$ we may easily see that $p_a(x)$ is negative on $[0,1]$, increasing and convex on $[1,+\infty)$, so the largest real root is in a right neighbourhood of $x=1$. We may also notice that: $$ p_a\left(1+\frac{\log(a+1)}{a+1}\right) = \frac{\log(a+1)}{a+1}\left(1+\frac{\log(a+1)}{a+1}\right)^{a+1}-1>0 $$ by ...


7

Firstly, we need to find what values of $x$ are acceptable. Solving the inequalities: $$\begin{array}[t]{rl} x^2-x+2 &\ge 0\\ x^2+x &\ge 0 \\ -x^2+x+2 &\ge 0\\ -x^2-x+4 &\ge 0 \end{array} $$ yields $$\left.\begin{array}{l} x \in \mathbb R\\ x \in (-\infty,-1] \cup [0,+\infty)\\ x \in [-1,2] \\ x\in \left[ ...


7

Hint: Between any two roots of $f$ there is a root of $f'$


7

Quoting from Locating the zeros of partial sums of $\exp(z)$ with Riemann-Hilbert methods: We denote by $p_{n}(z) := 1 + z + \cdots + \frac{z^{n}}{n!}$ the partial sums of the exponential series. The problem to describe the asymptotic distribution of the zeros of $p_n$ was posed and solved in the classical paper of Szegő [11]. He proved that the ...


6

One can check that $1$ and $-1$ are both roots of this polynomial. After factoring out $x-1$ and $x+1$, you're left with a quadratic polynomial. As a general rule, the rational root theorem is a good place to start for questions like these.


5

Let $a,b,c,d$ be the roots of $x^4+x^3-1=0$. By Vieta's formula, $$a+b+c+d=-1\quad\Rightarrow\quad c+d=-1-(a+b)\tag1$$ $$abcd=-1\quad\Rightarrow \quad cd=-\frac{1}{ab}\tag2$$ Since we have $$a^4+a^3=1\quad\text{and}\quad b^4+b^3=1$$ we can have $$1=(a^4+a^3)(b^4+b^3)$$ $$(ab)^4+(ab)^3(a+b+1)=1,$$ i.e. $$a+b=\frac{1-(ab)^4}{(ab)^3}-1\tag3$$ Similarly, ...


5

This really is similar in some way to your previous problem. $$\dfrac{\sqrt{x+1}}{2+\sqrt{2-x}} - \dfrac{\sqrt{x^2-x+2}}{2+\sqrt{-x^2+x+1}} = x^3-x^2-x+1$$ Firstly, for all the terms to be valid, we do domain checking. Solving $x+1\ge0$, $2-x\ge0$, $x^2-x+2\ge0$ and $-x^2+x+1\ge0$, we arrive at $x\in[-0.618, 1.618]$ (actually from the last inequality only) ...


5

We have $$x^2+7x+12-(x+1) \sqrt{x+2} - (x+6)\sqrt{x+7}=0$$ Multiplying the both sides by $3$ gives $$3x^2+21x+36-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0\tag1$$ Now since $$\begin{align}&3x^2+21x+36\\&=x^2+5x+4+x^2+13x+42+x^2+3x-10\\&=(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)\end{align}$$ we have, from $(1)$, $$(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)-3(x+1) ...


5

If a degree 4 polynomial has 4 real roots, then it must have at least 3 local extremes, so its derivative must have 3 real roots; by repeating this argument, its second derivative must have 2 real roots. But in fact, the second derivative of this function is $$ 12((x-1)^2+1), $$ which is clearly positive everywhere. Edit (thank you almagest): This argument ...


4

Let: $(7x+1)^{1/3} = a,(x^2-8x-1)^{1/3} = b, (8+x-x^2)^{1/3} = c.$ You have: $$\begin{cases} a+b+c = 2 \\ a^3+b^3+c^3 = 8 \end{cases} \Rightarrow a^3+b^3+c^3 = (a+b+c)^3$$ $$ \Leftrightarrow a^3+b^3 = (a+b+c)^3-c^3 = (a+b)[(a+b+c)^2+c(a+b+c)+c^2)]$$ $$ \Leftrightarrow a+b = 0 \text{ or }a^2-ab+b^2 = (a+b+c)^2+c(a+b+c)+c^2 $$ The first case is easy: $a = -b ...


4

Let $X=4x+3$. Then, $$3(2x^2-X)\sqrt{X}=-2x(2x^2+X)$$ Squaring the both sides gives $$9(2x^2-X)^2X=4x^2(2x^2+X)^2,$$ i.e. $$4x^4X-4x^2X^2+9X^3-16x^6+16x^4X-36x^2X^2=0$$ $$X(4x^4-4x^2X+9X^2)-4x^2(4x^4-4x^2X+9X^2)=0$$ $$(X-4x^2)(4x^4-4x^2X+9X^2)=0$$ since $9X^2=X^2+8X^2$ $$(X-4x^2)(4x^4-4x^2X+X^2+8X^2)=0$$ to have $$(X-4x^2)((2x^2-X)^2+8X^2)=0$$ finally, ...


4

The first part is obvious. Since $\alpha,\beta, P(\alpha)$ and $P(\beta)$ are all integers, $\mid \beta - \alpha \mid$ divides $\mid P(\beta) - P(\alpha) \mid = 2$. For the second part, notice that $(P(x))^2-1=(P(x)+1)(P(x)-1)$. So, all the roots of those two polynomials would be a root of this polynomial. There are $2d$ such (not necessarily distinct) ...


4

Let $P_k(x)=P(P(...(x))$ with $k$ iterations of $P$. Suppose (for contradiction) that $P(x)$ has a non-positive root $s$ and let $r$ be a positive root of $P(x)$ so that $P(x)=(x-r)(x-s)f(x)$. Then $P_2(x)= (P(x)-r)(P(x)-s)g(x)$. Note that $P(x)$ must have arbitarily large negative values or arbit. large positive values to the left of $s$; thus $P(x)=r$ or ...


4

I think the best way to look at this issue is to write the equation as \begin{equation} \epsilon x^5 = 1-x. \tag{1} \end{equation} When $\epsilon=1$, the situation is as shown in the picture below. Since $x^5$ is monotonically increasing (and $1-x$ is strictly monotonically decreasing), it's clear that $(1)$ has one real solution, where the two curves ...


3

There is no nice formula to get the roots of $P+Q$ from the roots of $P$ and of $Q$. For example, the roots of $x^5$ and $2x+1$ are easy to find, but the sum of these polynomials is $x^5 + 2 x + 1$, an irreducible quintic whose roots can't be expressed in radicals.


3

For $z^{3}+w z^{2}+\bar{w} z+1=0$, there's a paper discussed. Still need to check for your case.


3

Hint: $$f'(x)=101x^{100}-100^2x^{99}=x^{99}\left(101x-100^2\right)$$ has two solutions. $x_{\max}=0, x_{\min}=\frac{100^2}{101}$ Then $$x_1<x_{\max}=0<x_2<x_{\min}<x_3$$ Then $x_2,x_2>0$


3

Lemma: If $\alpha,\beta,\gamma$ are the roots of an irreducible polynomial over $\mathbb{Q}$ with degree $3$, $\alpha\beta$ is a root of: $$ q(x)=(x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma)\in\mathbb{Q}[x]. $$ Proof: The coefficients of $q(x)$ are symmetric polynomials in $\alpha,\beta,\gamma$, hence $q(x)\in\mathbb{Q}[x]$. The same argument also ...


3

Here is a way. Suppose that we want to visualize the roots of the cubic equation $$ z^3+z+1=0 $$ Write $z=x+iy$ and expand: \begin{align} (x+iy)^3+(x+iy)+1&=0\\ x^3+3ix^2y-3xy^2-iy^3+x+iy+1&=0\\ \end{align} Taking real and imaginary parts, we get \begin{align} x^3-3xy^2+x+1&=0\\ 3x^2y-y^3+y&=0 \end{align} Plotting the solution sets of ...


3

Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly). Module[{roots}, roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x]; Transpose[{ N[x /. roots], FullSimplify[Element[x, Reals] /. roots], x /. ...


3

$1$ is a root of $p(x)$ iff $p(1) = 0$. If $p,q$ are two such polynomials $(p+q)(1) = p(1) + q(1) = 0$ and for $\lambda \in \mathbb{C}, \lambda p(1) = \lambda \cdot 0 = 0$. (Note you need to bound $\deg(p+q)$ above also but that is straightforward) Then since your set is non-empty it is a subspace. The set of all polynomials over $\mathbb{C}$ where $1$ is ...


3

A polynomial is separable (has distinct roots) if it shares no zeroes with its formal derivative. If $ P(X) - w $ is a polynomial of degree $ n $ for $ w $ a constant, then its formal derivative $P'(X) $ is a polynomial of degree at most $ n-1 $ and does not depend on $ w $. Let the distinct zeroes of the formal derivative be $ z_1, z_2, \ldots, z_r $ where ...


3

There is a different approach that I have found many times useful: the conversion of a single equation into a system of 2 cartesian equations, here: $$x^{a+2}-x^{a+1}-1=0 \ \ (E) \ \ \ \ \Longleftrightarrow \ \ \ \ x \ \text{is solution to} \ \begin{cases}y=f_1(x)=x^{a+1}\\y=f_2(x)=\dfrac{1}{x-1}\end{cases}$$ (for a certain $y$). This kind of unfolding ...


3

I will show, by elementary means that the root is beteween $1+\dfrac1{\sqrt{a+1}}$ and $1+\dfrac1{a+1}$. Since the root of $x^{a+2}-x^{a+1}-1 $ is close to $1$, let $x = 1+y$. Then we want $1 =(1+y)^{a+2}-(1+y)^{a+1} =(1+y)^{a+1}(1+y-1) =y(1+y)^{a+1} $. Since $(1+y)^{a+1} \gt 1+y(a+1) $, $1 \gt y(1+y(a+1)) \gt y^2(a+1) $ so $y \lt \dfrac1{\sqrt{a+1}} $. ...


2

Hint Show that the evaluation map $\Bbb C [x] \to \Bbb C$ defined by $1 \mapsto p(1)$ is a linear map. By definition its kernel is the set of polynomials for which $1$ is a root, and the kernel of any linear map is a (linear) subspace.


2

The multiplicative group of $F$ is cyclic of order $255=3 \cdot 5 \cdot 17$, a multiple of $15$. Therefore, $x^{15}-1$ has exactly $15$ roots in $F$ and $f$ can be equal to $x^{15}-1$. $x^{63}=1$ and $x^{255}=1$ imply $x^3=1$ because $3=\gcd(63,255)$. Therefore, $x^{63}-1$ has exactly $3$ roots in $F$ and so $f$ cannot be equal to $x^{63}-1$ because ...



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