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32

Transform the equation. Since all the roots are symmetric, say $$y=\frac {1+x}{1-x}\implies x(y+1)=y-1\implies x=\frac {y-1}{y+1}$$ Substitute this expression in place of $x$ in the original equation and simplify. $$\require{cancel}\begin{align}f(y)&=\biggl(\frac {y-1}{y+1}\biggr)^3-\frac ...


12

Since the polynomial $x^3-x-1$ is irreducible over$\def\Q{\Bbb Q}~\Q$ by the rational root test, one approach would be to identify the element $\frac{1+a}{1-a}$ where $a$ is the image of $x$ in the field $K=\Q[x]/(x^3-x-1)$, and to compute its minimum polynomial over$~\Q$; since the Galois group of the splitting field of $x^3-x-1$ permutes its roots ...


12

$$(x^2-4)(x^2-2x)=2$$ $$\Rightarrow x^4-2x^3-4x^2+8x-2=0$$ $$\Rightarrow (x^2-x-1)^2-3(x-1)^2=0$$ $$\Rightarrow (x^2-x-1+\sqrt 3\ (x-1))(x^2-x-1-\sqrt 3\ (x-1))=0$$ $$\Rightarrow x^2+(\sqrt 3-1)x-1-\sqrt 3=0\ \ \text{or}\ \ x^2-(\sqrt 3+1)x-1+\sqrt 3=0$$ $$\Rightarrow x=\frac{-\sqrt 3+1\pm\sqrt{8+2\sqrt 3}}{2},\frac{\sqrt 3+1\pm\sqrt{8-2\sqrt 3}}{2}.$$


8

Adapting the argument from this answer by Noam Elkies. If $p(x)$ has only real zeros, then so does its reciprocal $$ x^np(\frac1x)=x^n+x^{n-1}+x^{n-2}+\cdots+a_{n-1}x+a_n. $$ But by Vieta relations the sum of the roots of this polynomial is $-1$ and the sum of their pairwise products is $1$. Therefore the sum of the squares of the roots is ...


8

Hint: $$ z^n-z = z \, (z^{n-1}-1) $$ and the $n-1$ roots of unity are different from each other (and different from $0$).


6

Let the three roots be $x_1,x_2,x_3$. Method $1$: Let $f(x) = ax^2+bx+c$. Since $x_1$ and $x_2$ are roots, this means $f(x) = (x-x_1)(x-x_2)g(x)$. Since $f(x)$ has degree $2$, this forces $g(x)$ to be a constant say $k$. Further, we have $f(x_3) = 0$. This means $k(x_3-x_1)(x_3-x_2) = 0$. Since $x_3 \neq x_1$ and $x_3 \neq x_2$, this forces $k$ to be zero. ...


5

Interesting question. What kind of class was this? We can approximate this in the continuous case by letting $(a, b, c)$ reside in the open unit cube in the positive octant. Given any value of $b$ in the interval $(0, 1)$, the allowable values of $(a, c)$ are those in the region of the first-quadrant unit square "inside" the curve $4ac = b^2$. The area ...


4

Here's an expression for $\xi$ in terms of radicals, according to Maple: $$1/10\,\sqrt {20\, \left( -1/4-1/4\,\sqrt {5}-1/4\,\sqrt {-10+2\,\sqrt {5}} \right) \sqrt [5]{-{\frac {-3125\,i\sqrt {5}\sqrt {2}+12500\, \sqrt {5}\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}+15625\,i\sqrt {2}- 25000\,\sqrt {{\frac {\sqrt {5}}{\sqrt {5}-1}}}}{2048\, \left( \sqrt { 5}-1 ...


4

The only way I can think of doing it, short of doing a numerical simulation, may be approximating it with a triple integral, i.e. approximating this probability by the probability that the quadratic has solutions if a, b, and c can take any real values between 1 and 100: ...


4

Let $z=\frac{k\pi}{j}$. If $4\cos^2z$ is the root of $x^3−7x^2+14x−7=0$, then $2\cos z$ is the root of $x^6−7x^4+14x^2−7=0$. Let's plug $x=2\cos z=2\frac{e^{iz}+e^{-iz}}{2}$ in and see what happens. The expression comes out to be $e^{-6 i z} \left(-e^{2 i z}+e^{4 i z}-e^{6 i z}+e^{8 i z}-e^{10 i z}+e^{12 i z}+1\right)=0$. Let's change the variables again, ...


4

The polynomial has a root $x=2$ as can be seen; this can be verified by replacing x by 2 in the polynomial. By division, the polynomial can be seen to factor into: $(x-2)(x^2+px+5)$. For the polynomial to have only 2 real roots, $x^2+px+5$ must have only one root. This happens if its discriminant $p^2-20=0$. The two values for $p$ are then $2 \sqrt5$ and $-2 ...


4

Assume that $f(x)=x$ has two solutions $x\neq y$ so that $f(x) = x$ and $f(y) = y$. But then for some $\xi \in \left(x,\,y\right)$ the Mean Value Theorem gives $$f'(\xi)=\frac{f(x)-f(y)}{x-y}=\frac{x-y}{x-y}=1,$$ a contradiction. I'm assuming that you came across this question in Baby Rudin with the requirement that $f$ is differentiable. Both Rolle's and ...


4

This might not be rigorous, but note that one has $$(x+a)^x=x^{x+a}\Rightarrow x\ln(x+a)=(x+a)\ln x\Rightarrow \frac{\ln(x+a)}{x+a}=\frac{\ln x}{x}.$$ Then, let $f(x)=\frac{\ln x}{x}$. One has $f'(x)=0\iff x=e$, and considering the graph of $y=f(x)$ should give you the answer.


4

The extreme point is a maximum or minimum, in the quadratic function case it will be global maximum of minimum. If you know calculus, this is easy to understand: at the extreme point, the first derivative at that point is $0$. Differentiate $ax^2+bx+c$ you get $2ax+b=0$, thus $x = -\frac{b}{2a}$ If you don't know calculus, then this is the way you may ...


4

Let $f(x)=x^4-7x^3+1$. Note that $f(x)$ is a continuous function. Since $$f(0)=0^4-7\cdot 0^3+1=1\gt 0$$ and $$f(1)=1^4-7\cdot 1^3+1=-5\lt 0,$$ there exists a real number $\alpha$ such that $$f(\alpha)=0\ \ \text{and}\ \ 0\lt\alpha\lt 1$$ by the intermediate value theorem.


3

We do have fractional roots but, I write it like - $$x^{\frac{a}{b}}=C \to x=C^{\frac{b}{a}}$$ But anyways there are a lot of ways you can represent math symbols and I can't find anything wrong in your representation too.


3

Let $\lambda_1,\ldots,\lambda_n$ be the roots of $p(x)$. We have coefficient of $x^0 = 1 \implies \prod_{i=1}^n \lambda_i = (-1)^n$ coefficient of $x^1 = 1 \implies \sum_{j=1}^n \prod_{i=1,\ne j}^n \lambda_i = (-1)^{n-1}$ coefficient of $x^2 = 1 \implies \sum_{1\le j < k \le n}\prod_{i=1,\ne j, k}^n \lambda_i = (-1)^{n-2}$ Combine these, we have ...


3

The polynomial $f(z)$ of degree $n>0$ has $n$ distinct roots if and only if it has no common root with the derived polynomial $f'(z)$. If we assume this result, we have to check $f(z)=z^n-z$ and $f'(z)=nz^{n-1}-1$. Let $f'(\alpha)=0$, so $\alpha^{n-1}=\frac{1}{n}$. Since $$ f(\alpha)=\alpha(\alpha^{n-1}-1)=\alpha\left(\frac{1}{n}-1\right)\ne0 $$ ...


3

You could use the fact that the product of the two roots is $-a$. If they have different signs, their product has to be negative. That means $-a<0$. Also two distinct roots means the discriminant is greater than $0$. This gives you $a^2+4a>0$. Combining the two should give you the answer.


3

The equation is: $$e^{\frac 1x}=x$$ Raise everything to the $x$ power: $$e=x^x$$ Now using super-square root in terms of Lambert's function leads us to: $$x=\sqrt e_s =e^{W(1)}=\frac 1{W(1)}$$


3

The problem is that $$ (5\sqrt 5)^3 = 5\sqrt 5 \cdot 5\sqrt 5 \cdot 5\sqrt 5 $$ and $$ (5\sqrt 5\cdot 5\sqrt 5)^2 = 5\sqrt 5 \cdot 5\sqrt 5 \cdot 5\sqrt 5\cdot 5\sqrt 5 $$ so you have one $5\sqrt 5$ too much. EDIT: In case you want your answer as $5^n$, note that we have $$ (5\sqrt 5)^3 = 5\sqrt 5 \cdot 5\sqrt 5 \cdot 5\sqrt 5 = 5 \cdot 5^\frac12\cdot5\cdot ...


3

Suppose $n\cdot 2^n=160$. Since $160=2^5\cdot 5$, you know that $5$ divides $n$, so $n=5m$. Then $$ 5m\cdot2^{5m}=160 $$ becomes $m\cdot 2^{5m}=32$ and so $m=1$. However, the general solution of $x\cdot 2^x=a$, for arbitrary $a$, cannot be determined “explicitly”, without using “higher level” functions such as Lambert's $W$.


3

You can modify your equation as such $$ \ln(2)n e^{\ln(2)n} = 160\ln(2) $$ So that $$ n = \frac{1}{\ln(2)} W(160\ln(2))=5$$ Where $W$ is the Lambert W function. I do not know any identity on W that would lead the result 5 without numerically computing W (which is usually done using root-finding).


3

We first consider the case $n$ odd, since the solution is simpler. Let $n = 2m + 1$. Choose $a_i: 1 \le i \le m$ nonzero distinct rational values such that the sum of squares is itself a square: $$a_1^2 + a_2^2 + \ldots + a_{m}^2 = A^2$$ (One way to generate such a list would be to iteratively use Pythagorean triples, ie something like $a_1 = 3, a_2 = 4, ...


3

We have $\sin(z) = \sin(x+iy) = \frac{e^{ix-y}-e^{-ix+y}}{2i}$ $\frac{e^{ix}e^{-y}-e^{-ix}e^{y}}{2i}$ So $|\sin(z)| = \left|\frac{e^{ix}e^{-y}-e^{-ix}e^{y}}{2i}\right|$ $=\frac{\left|e^{ix}e^{-y}-e^{-ix}e^{y}\right|}{2}$ $\geq \left|\frac{\left|e^{ix}e^{-y}\right|-\left|e^{-ix}e^{y}\right|}{2}\right|$ Because $|e^{ix}| = 1$, we can simplify, $= ...


3

It's exactly as you say. If $g$ has two zeroes, then by Rolle there is a point where $g'=0$. This of course depends on $f$ being differentiable.


3

The answer to this question depends on exactly what you mean by expressed in radicals. In the sense which is usually meant in Galois theory courses, $\cos \pi/25$ is expressible in radicals, but in a dumb sense: We have $\cos (\pi/25) = \frac{1^{1/50}+1^{-1/50}}{2}$ for one of the $50$-th roots of $1$. You may argue that this is a useless expression, but it ...


2

Well, you can write $${7\over 432}+ i\sqrt{49\over6912}={7\sqrt{7}\over216}\,e^{i\arctan\left(3\sqrt{3}\right)}=\left({\sqrt{7}\over6}\right)^3\,e^{i\arctan\left(3\sqrt{3}\right)}\\ {7\over 432}- i\sqrt{49\over6912}={7\sqrt{7}\over216}\,e^{-i\arctan\left(3\sqrt{3}\right)}=\left({\sqrt{7}\over6}\right)^3\,e^{-i\arctan\left(3\sqrt{3}\right)}$$ And go from ...


2

To get a real solution, the discriminant $b^2-4ac\geq0$, so $b\geq 2 \sqrt{ac}$, and $\frac{b^2}{4}\geq ac$. All of our variables are real so there is no need to consider negatives. Values of $a$ and $b$ can range from $1$ to $n$. The probability of each separate combination is $\frac{1}{n}*\frac{1}{n}$. Considering each combination of $a$ and $b$ ...


2

The analytical method for the irreducible case compares $$ y^3−\frac{7}{12}y=\frac{7}{216} $$ with $y=r\cosϕ$, $$ y^3=r^3\frac{(e^{iϕ}+e^{-iϕ})^3}{8}=\frac{r^3}4(\cos(3ϕ)+3\cosϕ) \\ \iff\\ y^3-\frac34 r^2y=\frac{r^3}4\cos(3ϕ) $$ to read off $r=\frac{\sqrt{7}}3$ and $$ \frac{7\sqrt{7}}{4·3^3}\cos(3ϕ)=\frac{7}{2^3·3^3} \iff ...



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