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## Hot answers tagged roots

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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