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11

Here is an easier way. If your polynomial $P_k(x) = 2x^3-9x^2+12x-k$ has a double real root $r$, then $P_k'(r) = 0$. For such a root, you get $P_k'(r) = 6r^2-18r+12 = 0$, i.e. $r=1$ or $r=2$. Then you solve $P_k(r)=0$ with respect to $k$. More precisely : If $r=1$, one needs $P_k(1)=5-k=0$ i.e. $k=5$. If $r=2$, one has $P_k(2)=4-k=0$ i.e. $k=4$.

6

Suppose $x=\frac{m}{n}$ is rational for $n,m\in\Bbb N$. It's clear that $2^x=\sqrt[n]{a}$ and $3^x\sqrt[n]{b}$ are non-integers, where $a=2^m,b=3^m$, hence are irrational (it's an elementary result that $n$-th root of an integer is either an integer or is irrational). By positive answer to this question it follows that a sum of irrational roots of positive ...

5

It is better, I think, to try with $x^4-2x^2-8\sqrt{15}\space x+69$ where $x=2T$. By some tedious calculation of undetermined coefficients we have $x^4-2x^2-8\sqrt{15}\space x+69=[x^2-2\sqrt {5}\space x +(9-2\sqrt 3)]\cdot[x^2+2\sqrt {5}\space x +(9+2\sqrt 3)]$ Hence $x_{1,2}=\sqrt 5\pm\sqrt{4-2\sqrt 3}\space i$ and $x_{3,4}=-\sqrt 5\pm \sqrt{4+2\sqrt ... 5 Hint: the function$f$defined by$f(x) = ax + b - \sin x$is continuous, and has limits$-\infty$and$+\infty$respectively at$-\infty$and$+\infty$(if$a > 0$; the other way if$a < 0$). Apply the IVT on$f$. 5 The square root of a nonnegative real number$x$is, by definition, the positive real number$a$such that$a^2 = x$. As a consequence, you have the property $$\sqrt{x^2} = |x|$$ which reduces your question to: what real numbers satisfy $$|x| = -x$$ If you know how the absolute value works (or: check its definition), this shouldn't be hard anymore. 5 Edit: After reading the comment of lhf to the original question, I figured I might point out that what's going on here is simply inverse iteration. How might we compute the roots of$f^n$, given its very high degree? Well, computing the roots of$fisn't hard, say we get $$z_{1},z_{2},z_{3},z_{4},z_{5}.$$ Now, each of those points has five more ... 5 This kind of equation does not show analytical solution and only numerical methods would be able to solve the problem. What is nice if that we know the range which contains the bounds $$1 < x < \frac{\log (3)}{\log (2)}$$ However, since the function is quite stiff (it varied very fast), it should be better to consider the function ... 5 \begin{aligned} x^4+2x^3+2x^2-2x-3 &=(x^4+2x^2-3)+(2x^3-2x) \\ &=(x^2+3)(x^2-1)+2x(x^2-1) \\ &=(x^2-1)(x^2+3+2x) \\ &=(x^2-1)(x^2+2x+3) \end{aligned} 4 Note that $$\dfrac1{\sqrt{k}+\sqrt{k+3}} = \dfrac{\sqrt{k+3}-\sqrt{k}}3$$ Hence, $$\sum_{k=1}^N \dfrac1{\sqrt{k}+\sqrt{k+3}} = \sum_{k=1}^N \dfrac{\sqrt{k+3}-\sqrt{k}}3 = \dfrac13 \left(\sqrt{N+3}+\sqrt{N+2}+\sqrt{N+1}-\sqrt3 -\sqrt2-1\right)$$ Similarly, we can obtain an expression for\displaystyle \sum_{k=1}^N \dfrac1{\sqrt{k+m}+\sqrt{k+n}}$4 First note that$\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)$and also note that$-\frac{b}{a}=\alpha+\beta$and$\frac{c}{a}=\alpha\beta(do you see why?) We can make $$\alpha^2++2\alpha\beta+\beta^2=(\alpha+\beta)^2=\frac{b^2}{a^2}$$ so our final outcome will be \begin{align} ... 4 There is a way to build an (approximate) explicit formula, using Euler's method of substitution (I think Euler was the first to use it anyway). First, we need to transform our equation. Let's take3^x=y$and$\alpha=\frac{log(2)}{log(3)}$with$log$being the natural logarithm. Then we have: $$y^\alpha+y-6=0$$ $$y=6-y^\alpha$$ $$... 4 From Vieta's Formulas, we can easily get a_1+a_2+a_3=-19 and a_1a_2+a_2a_3+a_3a_1=12. Also, a_1^2+a_2^2+a_3^2=(a_1+a_2+a_3)^2-2(a_1a_2+a_2a_3+a_3a_1)=361-2\times12=337. For the three roots, we have x^3+19x^2+12x+3=0 and x^4+19x^3+12x^2+3x=0. We add this expression for a_1, a_2, a_3 to get:$$\sum a_i^3+19\sum a_i^2+12\sum a_i+9=0$$and ... 4$$x^3-3=2\sqrt{x+2}x^6-6x^3+9=4x+8x^6-6x^3-4x+1=0$$At this point we get a sixth degree polynomial, and you can estimate the roots accordingly. Wolfram Alpha gives the positive solution as x \approx 1.90874157798\dots, but cannot find a closed form for the points (this can happen while solving any polynomial of degree 5 of higher, by a proof by ... 3 We can exploit the homogeneity of the polynomial and (on the set where y \neq 0) denote u := \frac{x}{y}, where we see that the equation is equivalent to$$u^4 - 2 u^3 + 3 u - 2 = 0 .$$By the Rational Root Theorem, the only possible rational roots (in u) are \pm 1, \pm 2, and checking shows that among these only +1 is. Polynomial long division of ... 3 Note that we define \sqrt{x^2} = |x| so the only real number that satisfies your equation is 9. 3 Use Viete formulas:$$\alpha\beta = c/a\alpha + \beta = - b/a$$Therefore$$\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha^2\beta - 3\alpha\beta^2 = (-b/a)^3 + 3bc/a^2$$3 First of all, you can do the same sort of trick to get a closed form for this as we use for any other linear recurrence. The roots of U^2-xU -y=0 are u_1,u_2=\frac{x\pm\sqrt{x^2+4y}}{2}. And a_n = bu_1^n + cu_2^n for some b,c\in R=\mathbb Q(x,y)[\sqrt{x^2+4y}]. Solving for b and c, we get b=-c and b(u_1-u_2)=b\sqrt{x^2+4y}=1. So$$a_n = ... 2 You have$z=-1+i$, so$|z|=\sqrt{2}$. Note$\tan^{-1}(\tfrac{1}{-1})=-\frac{\pi}{4}$, so you need$\pi-\frac{\pi}{4}=\tfrac{3\pi}{4}$as$z$is in the second quadrant. Now using$z=\sqrt{2}e^{i\left(2k\pi+\tfrac{3\pi}{4}\right)}$you need the cube roots of this. To this end let$w^3=z, then \begin{align} w_k=\left(\sqrt{2}e^{i ... 2 Using Galois theory, this result is trivial. LetK = \mathbb{Q}(\alpha_1, \dots, \alpha_n)$. Then$K$is the splitting field of$f$, hence is a Galois extension of$\mathbb{Q}$. Now if$c = \sum 1/\alpha_i^2$, then$c$is fixed under any automorphism$\sigma$of$K$, since$\sigma$must permute the roots of$f$. Therefore$c$belongs to the fixed field of ... 2 Remember, if$\alpha_j$are roots of$f(x)$, then$\frac{1}{\alpha_j}$are roots of$x^n f\left(\frac{1}{x}\right)$. A proof of$\sum_{j=1}^n \frac{1}{\alpha_j^2} \in \mathbb{Q}$is not that different from a proof that$\sum_{j=1}^n \alpha_j^2 \in \mathbb{Q}$. Following the same spirit of the proof of the latter case, we have: $$\sum_{j=1}^n ... 2 Well$$(x + 4.56)(x + 0.44) = x^2 + 5x + 2.0064 \neq x^2 + 5x + 2 = g(x), $$so no, anything using that factorization couldn't tell you anything about g(x). Your Q(x) looks OK, but your R(x) certainly doesn't satisfy f(x) = Q(x)g(x) + R(x). Now, once you use g(x) to get the correct R(x) you will have an expression of the form$$ R(x) = f(x) - ... 2 Hint Let$\alpha $a root of$f$. Then $$f(f(\alpha ))=f(0)=d,$$ If$\beta $is a root of$f(f(x))$, then $$f(f(\beta ))=f(\beta )^2+af(\beta )+d\cos(f(\beta ))=0.$$ 2 The derivative$3x^2+2ax+b$has also two real, negative roots. Then $$4a^2-12b\ge0$$ I hope this helps. 2 The Galois group of the splitting field of, for example,$x^5 - x^4 + 1$, is isomorphic to$S_5$, which is not solvable, and hence the roots of that polynomial are not solvable in radicals. Hence, there is no general formula for the roots of a real polynomial of the form$A x^n + B x^{n - 1} + C$in terms of radicals. (See the Wikipedia article on the ... 2 Rearranging, we have that$x^{13} = -1.$An obvious solution is$x = -1.$To find the others, we take a look at the complex plane. The$13$roots can be shown to be spaced evenly around the unit circle. We already know one solution, so the others follow as such:$\boxed{x = -1, i\sin(\frac{\pi}{13}) - \cos(\frac{\pi}{13}), i\sin(\frac{2\pi}{13}) - ...

2

$$x^4+1=0\Longleftrightarrow$$ $$x^4=-1\Longleftrightarrow$$ $$x^4=|-1|e^{\arg(-1)i}\Longleftrightarrow$$ $$x^4=e^{\pi i}\Longleftrightarrow$$ $$x=\left(e^{\left(2\pi k+\pi\right)i}\right)^{\frac{1}{4}}\Longleftrightarrow$$ $$x=e^{\frac{1}{4}\left(2\pi k+\pi\right)i}$$ With $k\in\mathbb{Z}$ and $k:0-3$ So, the solutions are: ...

2

$a,b,c$ and $d$ are distinct integers. So $r-a,r-b,r-c$ and $r-d$ are all distinct integers. Again, $9=3^2$ Hence the only possible factorisation: $$(r-a)(r-b)(r-c)(r-d)=3\cdot 3\cdot 1 \cdot 1$$ or $$(a-r)(r-b)(c-r)(r-d)=3\cdot 3\cdot 1 \cdot 1$$ So we have $a=r+3$, $b=r-3$, $c=r+1$, $d=r-1$. Thus $\color{blue}{a+b+c+d-4r} = \color{red}{0}$

2

HINT Now $$e^{-4x} = 5x^2$$ becomes $$5x^2 \approx 8x^2-4x+1$$ which is a quadratic you can solve. The roots should approximate the rotts of the transcendental equation you started with.

2

For $\;x\;$ close enough to zero we have the good approximation you gave, so $$8x^2-4x+1=5x^2\iff3x^2-4x+1=0\implies x=\begin{cases}1\\{}\\\frac13\end{cases}$$ Take $\;\frac13\;$ as a good approx. to the wanted zero.

1

$$0=0.001 + \frac{-0.0018 x+0.009 x^2}{\left(\sqrt{0.04 - x^2}\right)^3}$$ $$0=1 + \frac{-1.8 x+9 x^2}{\left(\sqrt{0.04 - x^2}\right)^3}$$ $$1 =\frac{1.8 x-9 x^2}{\left(\sqrt{0.04 - x^2}\right)^3}$$ $$\left(\sqrt{0.04 - x^2}\right)^3=1.8 x-9 x^2$$ $$\left(0.04 - x^2 \right) \left(\sqrt{0.04 - x^2}\right)=1.8 x-9 x^2$$  (0.04 - x^2 )^2 (0.04 - ...

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