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since $$\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$ let $$P(x,x^2),A(3,2),B(0,1)$$ so $$|PA|-|PB|\le |AB|=\sqrt{10}$$ if and only is $A,P,B$ on a line.
The answer is $x=\operatorname{li}^{-1}(100) \approx 488.871909852807532.\;$ My implementation computes $\operatorname{li}^{-1}(x)$ as the zero of the function $f(z)=\operatorname{li}(z)-x$ using Halley iterations $$z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)}\left[1 - \frac{f(z_n)}{f'(z_n)}\frac{f''(z_n)}{2f'(z_n)}\right]^{-1}\\ =z_n-\Delta_n \ln(z_n) \left[1 + ... 7 A multiple root is a common root of both f(x) and f^{\prime}(x).$$f^{\prime}(x)=5x^4-1$$and if you are interested only in real roots (i get this impression that you are) then$$x=\frac{1}{\sqrt[4]{5}}$$and further we must have$$f\left(\frac{1}{\sqrt[4]{5}}\right)=0$$this gives$$\alpha=\frac{4}{5\sqrt[4]{5}}$$6 If$$z = e^z$$then$$-ze^{-z} = -1$$so$$-z = W(-1)$$and thus$$z = - W(-1),$$where W is any branch of the Lambert W function. 5 The fundamental theorem of algebra states that an n^{th} degree polynomial has exactly n roots, provided you count them with their multiplicity. This is equivalent to saying that any polynomial has a root. Because if you divide a polynomial of the n^{th} degree by the monomial x-r (where r is a root), you get an (n-1)^{th} polynomial, which has ... 4 Yes, what you have done is a very good start for the full root calculation. You now need to solve \alpha^3=1 and \alpha^3=8. The roots of the second equation are twice the roots of the first. To solve \alpha^3=1, note that \alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1). 4 Hint : Rewrite your equation as (z+1)^6-z^6=0 and since z=0 isn't a solution, show that \frac{z+1}{z} is a 6-th root of unity. 4 Use Vieta's formulas and the following identity:$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)$$4 The easiest description of the quadratic extension is that it has the same 25 elements as yours, with addition defined in the obvious way, and x^2 being defined as -x-2, that is, 4x+3. Then x is a solution of your quadratic equation. Once we have defined x^2, the product ax+b)(cx+d) can be defined in the "natural" way: Multiply as usual, and ... 3 The answer to your question is no for the trivial reason you already observed: suppose you have a polynomial f(x) for which all roots are unsolvable, then xf(x) has the solvable root 0 but all others are unsolvable. The situation changes if you assume that f to be irreducible: then if one root can be expressed by radicals, the same is true for all ... 3 There are two problems involved here: One, is a general local root finding problem, for which many method's (such as Newton's) can be used. The second, is the problem of finding all (at most three) solutions. One method I could suggest is the "Homotopy Continuation Method". To use it, first find all three solutions for some a> e^e\sim15.15. Now, vary ... 3 Hint: (\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha \beta (\alpha + \beta) You're basically there with \displaystyle\alpha^3 + \beta^3 = -\frac1{27} + \frac{3\alpha}2 + \frac{3\beta}2 = -\frac1{27} + \frac32(\alpha + \beta)=\ldots (Note that it's \frac{1}{27} not \frac19) And \displaystyle \frac1{\alpha^3} + \frac1{\beta^3} = \frac{\alpha^3 + ... 2 This is a very general question and there is no definite answer. It depends on polynomials you have. But, if you want to get an insight, these methods are usually discussed in textbooks for numerical/computational mathematics. There is always a section called non-linear equations, where they present Interval Bisection, Secant and Newton method and study ... 2 See the following paper: Charles Edward Siewert and Ernest Edmund Burniston, Exact analytical solutions of ze^{z} = a, Journal of Mathematical Analysis and Applications 43 #3 (September 1973), 626-632. Author's Abstract: By means of the theory of complex variables, the solutions of ze^{z} = a, where a is in general complex, are established ... 2 As shown elsewhere, z\sin z=1, with z\in C, has only real solutions. So we are ultimately left with solving x\sin x=1, with x\in R, which, unlike the solution to your other question, does not possess a closed form, not even in terms of the special Lambert W function. However, what we can say for certain, is that it has exactly two solutions in each ... 2 The method which you mentioned is helpful. In fact with \alpha=x^3 we have$$\alpha^2-9\alpha+8=0$$so the roots are$$\alpha_1=1\quad \alpha_2=8$$now$$x^3=1\iff x=\exp\left(\frac{2ik\pi}{3}\right),\quad k=0,1,2$$and$$x^3=8\iff x=2\exp\left(\frac{2ik\pi}{3}\right),\quad k=0,1,2$$2 What M. Strochyk did was solve 1-x^5 = 0 for real and complex roots using De Moivre's Theorem. First, we want to solve:$$x^5 = 1$$We first convert the "1" into complex form, which is 1+0i. Then we convert 1+0i into polar form, and we need to find r and \theta. r = \sqrt{1^2+0^2} = 1 and \theta = \arctan(\frac{0}{1}) = 0, \pi, 2\pi,...  So ... 2 Hint Why not considering$$F(x,\epsilon)=(1 + \epsilon)x^3 - 2ax^2 + (a - 3\epsilon)x + 2\epsilon = 0$$and use implicit differentiation to obtain the value of \frac{dx}{d\epsilon} and compute its value for \epsilon=0 and then use$$x \simeq x_0+\frac{dx}{d\epsilon} \Delta \epsilon$$For sure, you must do the above for each of the roots$$x_1=0$$... 2 If you write a polynomial p(x)=(x-r_1)^{n_1}\ldots (x-r_k)^{n_k} where we assume each r_i\ne r_j when i\ne j and n_i\ge 1 for 1\le i\le k. Then if we take derivatives, the product rule gives us$$p'(x)=\sum_{i=1}^k n_i(x-r_i)^{n_i-1}\prod_{i\ne j}(x-r_j)^{n_j}$$If n_m=1 for some 1\le m\le k, then we get that (x-r_m) divides each term ... 2 Hint:$$2^{x}=16=2^{4}$${}{}{}{}{}{}{}{} 2$$2^{x}=16Take logarithm from both sides. \begin{align} \require{cancel} x\ln 2&=4\ln2\\ x\color{red}{\cancel{\color{black}{\ln 2}}}&=4\color{red}{\cancel{\color{black}{\ln 2}}}\\ x&=4\\ \end{align} 1 For \left|z \right|=1 we have that \left| 3z^{100} \right| = 3 and \left|e^z \right| = \left| e^{ \mathfrak{R}z } \right| \leq e<3. Thus we have that \left| 3z^{100} \right|>\left| e^z \right| on the unit circle. Therefore, using Roche's theorem we obtain that 3z^{100} and 3z^{100} - e^z have the same number of zeros in the unit disc. ... 1 As Daniel pointed out, since p(x) will have a double root, p'(x) must have the same root as well. Also, by using Descartes rules of signs,p(x) = x^5 -x +\alphap(-x) = -x^5 +x +\alpha$$Therefore, p(x) has either 2 or 0 positive roots, 1 negative root, and either 2 or 4 complex root. Since we are assumed there is a positive root, p(x) will ... 1 Somebody posted a hint answer which made solving it easy, but they quickly deleted it. Here's a solution using that answer. When the discriminant of a polynomial is zero, atleast two roots coincide. The discriminant of x^5-x+\alpha is 3125\alpha^4-256 solving for \alpha$$3125\alpha^4-256=03125\alpha^4=256\alpha^4=\frac{256}{3125}$$... 1 Given the roots of a polynomial, call them r_1, r_2, r_3, \ldots r_{201}, then the coefficients are given by the elementary symmetric functions on the roots. If you are not familiar with these, read this article on Wikipedia So, a_{2k+1} = \dfrac{e_{201-2k -1}(r_1, r_2, \ldots r_{201})}{a_{201}} This gives a "general" solution. Unless he/she gave you ... 1 Multiplicity usually means "the number of times it is repeated" (http://en.wikipedia.org/wiki/Multiplicity_%28mathematics%29), but without knowing what exactly your -\frac{1}{3} is supposed to be, it's impossible to figure out what is repeated. 1 Let me answer by showing how a homogenous polynomial of degree n in the projective line has n solutions. A typical such polynomial is$$a_nx^n+a_{n-1}x^{n-1}y+\cdots +a_0y^n$$let us assume that a_{m}\neq 0 is the first nonzero coefficient then the equation factors$$a_m y^{n-m}(x-\beta_1 y) \cdots (x-\beta_m y)$$now for y=1 we have the finite ... 1 We want all polynomials of degree n to have n roots (or equivalently, n linear factors). But polynomials may have repeated roots. E.g. the polynomial$$f(x)=x^2-4x+4=(x-2)^2$$has only one root, namely 2. But we say it has multiplicity 2 since its linear factor (x-2) occurs twice in its factorization. Another example: the polynomial ... 1 Solving x^3=1 gives x=1 and the other roots satisfy x^2+x+1=0. For these you may use the quadratic formula x=\frac{-1\pm \sqrt{-3}}{2}. Solving x^3=8 gives x=2 and the other roots satisfy x^2+2x+4=0. Use the quadratic formula x=2\frac{-1\pm \sqrt{-3}}{2}=-1\pm \sqrt{-3}. 1 If you know about the derivative, this can be done directly. Let f(x) = 13 x^{13} - e^{-x} - 1; then$$f'(x) = 169 x^{12} + e^{-x} which is always positive; hence $f$ is strictly increasing, and there is at most $1$ real solution. Finally, since $f(0) < 0$ and $f(1) > 0$, the real solution is between $0$ and $1$ (Wolfram gives about $.844$). ...