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28

$x^2$ and $x^4$ are not negative, so $$1+x^2+x^4\ge1$$ for any real number $x$.


16

Consider the identity $(x^4-4x-1)^2=x^8-8x^5-2x^4+16x^2+8x+1$ Differentiating both sides w.r.t. $x$, we get, $x^7-5x^4-x^3+4x+1=(x^4-4x-1)(x^3-1)$ Now, the equation becomes, $(x^4-4x-1)(x^3-1)=0$ $\implies x=1,\omega, \omega^2$ (where $\omega$ is a non real cube root of unity) or $x^4-4x-1=0$ $\implies x^{4}+2x^{2}+1=2x^{2}+4x+2$ ...


8

More specific; if you have a function with $$k * x^2$$ where k is a real number, and this term is only added or subtracted to the function, not multiplied with any other term, does the function have maximum 2 roots? No, take as an example $\sin x+0.01 x^2$, it has $6$ roots. Reducing the coefficient of the $x^2$ term will significantly increase the ...


7

If $r$ and $s$ are solutions to your equation, then we consider $r^2$ and $s^2$: $$r^2+s^2=2(a+b)$$ and $$r^2s^2=(a-b)^2$$ These simultaneous equations can be solved fairly easily, giving us two similar formulas, with a $\pm$, for $r^2$ and $s^2$. Taking square roots gives your four solutions, with another $\pm$. All this shows why the solutions are so ...


7

Hint: The value $x=0$ can be eliminated immediately. So it is valid to multiply both sides of the equation by $x^n$. Doing this and moving all terms to the left side gives you $$(x^n)^2-k(x^n)+1 = 0$$ which is a quadratic in $x^n$. Apply the quadratic formula to get $$x^n = \tfrac{k}{2} \pm \sqrt{\left(\tfrac{k}{2}\right)^2 - 1}$$ You should be able to ...


7

Consider $x = \frac{10}{12}$: $$\begin{align*} f\left(\frac{10}{12}\right) &= a\left(\frac{10}{12}\right)^2+b\left(\frac{10}{12}\right)+c\\ &= \frac{10}{144}\left(10a+12b+\frac{144}{10}c\right)\\ &= \frac{10}{144}\left(10a+12b+15c-\frac{6}{10}c\right)\\ &= -\frac{6}{144}c \end{align*}$$ And consider $x=0$: $$f(0) = 0^2a + 0b + c = c$$ If ...


6

since $$(x^2-a-b)^2=4ab\Longrightarrow x^2=a+b\pm 2\sqrt{ab}=(\pm\sqrt{a}\pm \sqrt{b})^2$$


6

let $$x=\sqrt[3]{\cos{\dfrac{2\pi}{7}}},y=\sqrt[3]{\cos{\dfrac{4\pi}{7}}},z=\sqrt[3]{\cos{\dfrac{6\pi}{7}}},$$ then we have $$\begin{cases} x^3+y^3+z^3=-\dfrac{1}{2}\\ (xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\ (xyz)^3=\dfrac{1}{8} \end{cases}$$ use this identity $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$ so $$\begin{cases} ...


5

Let $S(n)$ be the statement that " $P_n(x)$ has no real root for $n$ even, and has exactly one real roots for $n$ odd" You can check directly that $S(1)$ and $S(2)$ are true. Assume that $S(k)$ is true. Consider $k+1$-case: if $k$ is even, the induction hypothesis says that $P_k(x)$ has no real roots. As $P_k(0) = 1$, we have $P_k(x) >0$ for all ...


5

By the rational root theorem, if there is a rational root $p/q$ in lowest terms, then $p$ divides $a_{0}$ (i.e. $1$ is divisible by $p$, so what can $p$ be?). We can also say that $q$ divides $a_{3}=1$. What must $q$ be? Make sure you check whether the solution you get works!


5

Easy! Since $x^4 \ge 0$ and $x^2 \ge 0$ for all $x \in \Bbb R$, we have $x^4 + x^2 \ge 0$, whence $x^4 + x^2 + 1 > 0$, $\forall x \in \Bbb R$. No zeroes in $\Bbb R$, no solution! QED!!! Of course, if we allow $x \in \Bbb C$, we have a different story altogether, which is told in a different place. Hope this helps. Cheers! And as ever, Fiat Lux!!! ...


5

Elementary solution We exploit the insight that the product of all of the roots (and hence the product of the moduli of the roots) is $1$: Since $p$ has odd degree and real coefficients, it has at least one real root. But substituting gives that $p(-1) = 4$ and $p(1) = 6$, so $p$ has a real root that in particular does not have unit modulus. Thus $p$ must ...


5

The main thing to note is that you mustn't directly square the terms. In order that the inequality makes sense, you need $x^2-1\ge0$ Recall that $\sqrt{a}\ge0$ (when $a\ge0$) by definition, so an inequality $\sqrt{a}>b$ where $b<0$ is automatically true as soon as $a\ge0$. If $a\ge0$ and $b\ge0$, the inequality $a>b$ is equivalent to $a^2>b^2$. ...


4

I'm not sure the name of it but a simple proof would be: Since $a$ is a root of $f(x)$ we can write $f(x) = (x-a)g(x)$ for some polynomial $g(x)$. Then taking the derivative we get $$f'(x) = g(x) + (x-a)g'(x)$$ Now plugging in $a$ we get: $$0 =f'(a) = g(a) + (a-a)g'(a) = g(a)$$ So $a$ is a root of $g$ and we can write $g(x) = (x-a)h(x)$. Hence $f(x) = ...


4

I would solve the equation. Take $y=x^2$. Then the equation is $y^2+y+1=0$. This is quadratic and easy to solve.


4

This is a quadratic equation problem in disguise. Since we have two occurrences of $x^n$, let's make this slightly easier on ourselves by setting $y=x^n$, then we want to solve $$y + \frac{1}{y} = k.$$ As written, it's a little difficult to solve so the most reasonable thing to do is multiply everything by $y$ so that we get only get positive powers of $y$ ...


4

I am assuming that by $\sin \theta_1 z^3$ our OP Jackie means $z^3 \sin \theta_1$ and so forth; with this understanding, we have the given equation $z^3 \sin \theta_1 + z^2 \sin \theta_2 + z \sin \theta_3 + \sin \theta_4 = 3; \tag{1}$ taking absolute values and using the triangle inequality (several times) yields $3 = \vert 3 \vert \le \vert z^3 \sin ...


3

Calculate the determinant:$$4(a+b)^2-4(a-b)^2 = 16ab$$ Then find the roots as following: $$x^2 = \frac{2(a+b)+\sqrt{16ab}}{2},\frac{2(a+b)-\sqrt{16ab}}{2}$$ $$x^2 = a+b+2\sqrt {ab}, a+b-2\sqrt {ab}$$ $$x = \pm(\sqrt a + \sqrt b), \pm(\sqrt a - \sqrt b)$$


3

If $g$ is a primitive root, $S:\{s, 1\le s\le p-1\}, T:\{g^r\pmod p, 1\le r\le p-1\}$ will be the same set $$\implies\sum_{s=1}^{p-1}s^k\equiv\sum_{r=1}^{p-1}(g^r)^k\equiv \sum_{r=1}^{p-1}(g^k)^r\pmod p$$ If $(p-1)\mid k, g^k\equiv1\pmod p $ $$ \sum_{r=1}^{p-1}(g^k)^r\equiv \sum_{r=1}^{p-1}1^r\pmod p\equiv p-1\equiv-1$$ Else $g^k\not\equiv1\pmod ...


3

Hint: $\sin x + \cos x = 0$ if and only if $\sin x = -\cos x$ if and only if $\tan x = -1$.


3

Here is a kind of cheaty proof: $$f(-5) = 1351 \\ f(-1) = -1 \\ f(1) = 7 \\ f(4) = -71 \\ f(7) = 463$$ So there are zeroes in the intervals $(-5,-1), (-1,1), (1,4)$ and $(4,7)$, because polynomials are continuous functions.


3

Yes. And no. A second order polynomial $p:\mathbb R\to\mathbb R$, i.e. a real function defined as $$p(x) = ax^2+bx+c$$ where $a,b,c$ are real numbers, has a maximum of $2$ roots as it can have either $1$ root ($x^2$), $2$ roots ($x^2-1$) or $0$ roots ($x^2+1$). A second order polynomial $p:\mathbb C\to\mathbb C$, i.e. a complex function defined as ...


3

you will get $$(x-1)(x^2+x+1)(x^4-4x-1)=0$$ and you can solve your problem


3

Let's define $$ p_n(x) = \sum_{k=0}^n 2^{k(n-k)} x^k. $$ Since all its coefficients are positive, all roots of $p_n(x)$ must be negative. I'll start with some heuristics to try to explain the substitutions we'll make before actually proving the result. We observe that $p_n(x/2^n)$ converges compactly to an entire function as $n \to \infty$, so presuming ...


3

If $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ then $(x-\sqrt{a})^2=(\sqrt{b}+\sqrt{c})^2$, that is, $\xi=\sqrt{\alpha}+\sqrt{\beta}$ with $\alpha=4ax^2$, $\beta=4bc$ and $\xi=x^2+a-b-c$. The result in your previous question yields $\xi^4-2(\alpha+\beta)\xi^2+(\alpha-\beta)^2=0$, that is, $$(x^2+a-b-c)^4-8(ax^2+bc)(x^2+a-b-c)^2+16(ax^2-bc)^2=0.$$ This polynomial is ...


3

If $a > 0$, you can use the Intermediate Value Theorem. If $a < 0$, let $v$ be the minimum value of $x^a + x$ on $(0,\infty)$. Then $x^a + x - b$ has no positive real roots if $b < v$, one if $b = v$ and two if $b > v$. If $a$ is complex, $x^a + x - b$ is real only for a discrete set of positive real $x$'s, so there will usually be no ...


3

Here is another approach. Rolle's Theorem (a special case of the Mean Value Theorem) states that for a function $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, $f(a) = f(b)$ implies the existence of a $c \in (a,b)$ such that $f(c) = 0$. We can then say that if a polynomial has $k$ real roots, its derivative must have $k-1$ real roots where each ...


3

(In this answer i switched $x$ with $-x$) The polynomial $((x^2-14)^2-14)^2-x-14$ factors into the two cubic with cyclic galois groups (I already did the work in Exploring 3-cycle points for quadratic iterations) $(x^3 + 4x^2 - 11x - 43)(x^3 - 3x^2- 18x + 55)$, whose discriminants are $49^2$ and $63^2$. Both cubic have all real roots, so we can assume ...


3

The function $g(x)=\cos x-kx$ is the sum of two strictly decreasing functions in $[0,\pi/2]$, so it can not have more than one zero. On the other hand, $g(0)=1>0$ and $g(\pi/2)=-k\pi/2<0$. By Bolzano's theorem, it must have one zero.


3

For this to even make sense, $x$ must be in $(-\infty,-1]\cup[1,\infty)$. If it's in $[1,\infty)$, then $$\sqrt{x^2-1}>x\iff x^2-1>x^2,$$ which is impossible. If, on the other hand, $x$ is in $(-\infty,-1]$, it's trivially true as the left side can't be negative.



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