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9

HINT: For real $x,\dfrac{2^x+2^{-x}}2\ge\sqrt{2^x\cdot 2^{-x}}=1$ and for real $y,\cos y\le1$


8

The terminology varies a bit between people and fields, but what I would say is that $x^n=0$ has one root (namely $0$) of multiplicity $n$. If we explicitly say that we count that root "with multiplicity", of course, there are $n$ of it.


5

$x^3+2x^2-3=x^2(x-1)+3(x-1)(x+1)=(x-1)(x^2+3x+3)$. So, the roots of the original equation are $x=1$ and the two roots of $x^2+3x+3$. You can find the roots of $x^2+3x+3$ by using the quadratic formula.


4

Note by the Rational Root Theorem that $1$ is a root of the cubic. Proceeding by polynomial long division gives the factorization $$(x-1)(x^2+3x+3)$$ You can then use the quadratic formula and find the roots of the second factor.


4

You won't get a closed-form solution, but numerical methods can be used.


4

There is no positive root, or any root at all, of $x^3-x^3-1=0$, since, as you noted, $x^3-x^3-1=-1$ for all $x$, and there is no $x$ that satisfies $-1=0$. This is definitely a typo. The positive root of $$x^3-x^2-1=0$$ is exactly $$\frac{1}{3}\left(\sqrt[3]{\frac{29+9\sqrt{31/3}}{2}}+\sqrt[3]{\frac{29-9\sqrt{31/3}}{2}}+1\right)$$ so that is almost ...


4

We are given: $$f(x) = 2^{-x}+e^x+2 \cos x-6$$ We are asked to use Newton's Method to find the roots with accuracy $10^{-5}$ in the interval $(1,2)$. A plot of the function over the range shows: There is a root at $x \approx 1.83$, so lets verify that using Newton's Method. The Newton iteration is given by: $$x_{n+1} = x_n- \dfrac{f(x_n)}{f'(x_n)} ...


4

I think you mean inverse rather than opposite. The inverse of square is squareroot which could be expressed as:$$f(x)=x^2$$$$\therefore f^{-1}(x)=\sqrt{x}=x^{\frac{1}{2}}$$In a similar vein the inverse of cubing a number would be expressed as:$$f(x)=x^3$$$$\therefore f^{-1}(x)=\sqrt[3]{x}=x^{\frac{1}{3}}$$And in general:$$f(x)=x^n$$$$\therefore ...


4

You can factor the polynomial as follows: $$x^2(x-1)-4(x-1)=0,$$$$(x^2-4)(x-1)=0.$$ The first factor can be factored further using difference of two squares, and then you easily get the three roots.


4

Any nonzero polynomial can be written as $$ a_0+a_1x+\dots+a_nx^n $$ with $a_n\ne0$. This $n$ is called the degree of the polynomial. The zero polynomial is usually assigned degree $-\infty$ (or no degree at all). If we multiply two nonzero polynomials, the degree of the product is the sum of the degrees of the factors. Theorem. If $r$ is a root of the ...


3

The Rational root theorem can be used here to find all roots of this polynomial


3

Hint $\ 6f(t) = (6t)^3\! -19(6t) + 30 = x^3\! - 19x + 30 =: g(x)\,$ for $\,x = 6t.$ By the Rational Root Test the only possible rational roots of $g$ are integer factors of $30.\,$ Remark $\ $ This is a generalization of the high-school "AC method" and it works generally to reduce finding rational roots to finding integer roots. See here for much further ...


3

Here's a very simple construction: let $R$ be the ring $$ \mathbf{Z}[x_0, x_1, x_2, \ldots] / \langle x_0^2 + 1, x_1^2 + 1, x_2^2 + 1, \ldots \rangle $$ Then every $x_i$ is a root of the polynomial $t^2 + 1$ over $R$. A more trimmed down example is the ring $$ \mathbf{Z}[x,y] / \langle x^2, xy, y^2 \rangle $$ This is the ring of all polynomials of the ...


3

Have you tried using the rational root theorem?


3

This kind of equations which mix polynomial and trigonometric terms do not, in general, show solutions which have a closed form expression and almost only numerical methods would solve the problem. Newton method is among the simplest root-finding methods. Starting from a reasonable guess $x_0$, the method will update it according to ...


3

The discriminant is $$\Delta=(4a-2)^2-4(a-1)(4a+1)$$ a) For which values of $a$ is $\Delta$ non-negative? b) For which values of $a$ is $\displaystyle \frac{2-4a-\sqrt{\Delta}}{a-1}>0$?


3

You're almost entirely correct. Certainly, the polynomial will be some constant multiple of $f(x) = (x+3)(x-4)(x-8)$. It seems you've made one tiny error, however. Multiplying this out, the coefficient on $x^2$ should be $-9$, and not $-7$. From there, you just solve $-9a = -18$, which unfortunately you were trying to do with the wrong coefficient.


3

You cannot reconstruct a quadratic equation from its roots, only a monic quadratic equation. More precisely, if $q(x)=0$ is a quadratic equation with roots $x_1$ and $x_2$, then $q(x)=a(x-x_1)(x-x_2)$ for some $a\ne 0$. The roots do not determine $a$.


2

As we see that $x=1$ is a root of the equation, we therefore factor it into $(x-1)(x^2+3x+3)$. Now, with the quadratic formula, the roots are: $$\begin{align} x=&\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =&\frac{-3\pm \sqrt{9-12}}{2}\\ =&\frac{-3\pm i\sqrt 3}{2} \end{align}$$


2

As said in comments and answers, there is no analytical solutions fo such an equation and numerical methods should be used. Probably, the simplest root-finding method is Newton which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)} {f'(x_n)}$$ This could be applied to the orginal equation ...


2

The cubics that have the desired roots have the shape $k(x+3)(x-4)(x-8)$. Note that the sum of the roots of $(x+3)(x-4)(x-8)$ is $9$, so the coefficient of $x^2$ is $-9$. We want it to be $-18$, so we must take $k=2$.


2

You don't have all integer coefficients of your middle terms in the expanded polynomial. Multiplying through by three will get rid of fractional coefficients and give you the possibility of 3 being a root. The rational root theorem requires all integer coefficients.


2

Galois theory tells us that the only constructible roots are the $2^n$'th roots. These can be constructed by repeating the square root construction $n$ times. More precisely, if you start with the rational numbers, and allow only straightedge and compass operations on them, you generate the field of constructible numbers, which is characterized by the ...


2

To follow the hint. The hint is suggesting to use the argument principle. Argument principle (Geometric interpretation): If $f$ is analytic in $\Omega$ (simply connected to make it simple), $\gamma$ is a simple closed path in $\Omega$ such that $f$ has no zeros on $\gamma$, then the number of times $f(\gamma)$ winds around the origin is equal to the ...


2

Roots of polynomials are an unordered set, so you can only do transformations of coefficients that are invariant to the permutation of roots. You can, for instance, negate all roots by appropriately changing the signs of coefficients, but you cannot even specify what you require if you want to only negate a subset of roots, unless you already have the roots. ...


2

First, note that $\gamma$ must be an integer for the l.h.s. to be a polynomial. And no, it is not in general possible to solve explicitly for the roots, at least not in terms of radicals. For example: Taking $a = 1$, $\gamma = 5$ gives the polynomial $$2x^5 + x^4 - 2;$$ its Galois group is isomorphic to $S_5$, which is not solvable, and hence its roots ...


2

The motivation of using mod 8 in the first step is that if $x$ is even $x = 2k$, then $x^3 = 8k^2$ so $x^3$ is zero mod 8; and then you can see that you are solving for $y^2 = -1 \mod 8$ which looks like it might have no solution (as is in fact the case). All the author was trying to do there is to show that $x$ can't be even. The motivation for the mod 4 ...


2

Let $f(x) = \begin{cases} 0 & x \le 0 \\ x & x > 0 \end{cases}$. $f$ is continuous and is $0$ for $x \le 0$, which has cardinality $\mathfrak c$. Even if we specify our function to be $C^{\infty}$ (infinitely differentiable), we can still have uncountably many zeroes. Consider: $$g(x) = \begin{cases} 0 & x \le 0 \\ e^{-1/x^2} & x > ...


2

I am not sure I answer exactly the question as posted; so, forgive me if what I write is out of topic. If you have two curves defined by $y_1(x)$ and $y_2(x)$, the square of the distance between the two curves is given $$\Phi(x_1,x_2)=\big(x_1-x_2\big)^2+\big(y_1(x_1)-y_2(x_2)\big)^2$$ and you want to minimize this function with respect to $x_1$ and $x_2$. ...


2

hint: it separates out into $(x^3-x^2)-(4x-4)$. Try out rational roots if you want how to do it without, uh, "seeing" it, though.



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