The Roots are an American hip hop band, formed in 1987 by Tariq "Black Thought" Trotter and Ahmir "Questlove" Thompson in Philadelphia, Pennsylvania, United States. The Roots serve as the house band on NBC's The Tonight Show Starring Jimmy Fallon, having served in the same role on Late Night with Jimmy Fallon from 2009 to 2014.
The Roots are known for a jazzy and eclectic approach to hip-hop featuring live musical instruments and the group's work has consistently been met with critical acclaim. ThoughtCo ranked the band #7 on its list of the 25 Best Hip-Hop Groups of All-Time, calling them "Hip-hop's first legitimate band."In addition to the band's music, several members of the Roots are involved in side projects, including record production, acting, and regularly serving as guests on other musicians’ albums and live shows.
##y+3=3\sqrt{y+7}##
Square both sides:
##\Rightarrow y^2+6y+9=9y+63##
##\Rightarrow y^2-3y-54=0##
##\Rightarrow (y-9)(y+6)=0##
##y=9, -6##
But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?
The problem and solution are posted... no. 8
I may need insight on common difference ...
In my lines i have,
Let the roots be ##(b), (b-1)## and ##(b+1)##.
Then,
##x^3-3bx^2+3cx-d = a(x-b(x-b+1)(x-b-1)##
##x^3-3bx^2+3cx-d= a(x^3-3bx^2+3b^2x-x-b^3+b)##
##a=1##.
Let...
Say we have the following conditions:
For an any degree polynomial with integer coefficients, the root of the polynomial is n. There should be infinite polynomials that satisfy this condition. What is the general way to generate one of the polynomial?
In my approach i have the roots of the equation being ##x=a## and ##x=b##.
There are two assumptions,
In the first assumption,
##a=\dfrac{1}{2}b##
##2a=b##
then,
##4=k(-a)^2(-2a)##
##4=-2ka^3##
##⇒ -2=ka^3##
Now since ##2a=b## then ##a=1, b=2⇒k=-2##.
our equation becomes...
This is part of a longer exercise I struggled with. I checked the solutions manual, and there was a bit where they performed the following steps:
$$x^3=3a^2x-2a^3 \\$$
$$(x-a)^2(x+2a)=0$$
And then concluded that the roots were ##a## and ##-2a##, which is clear. What I can't work out is how...
Going through this, am still checking but will post all the same; which method did they apply to find the roots of the attachment below.
My thinking;
Let
##p+qi##
be the cube root of
##x^3-6x+2=0##
then,
##\sqrt{x(x^2-6)}=i\sqrt{2}##
##(p^2-q^2+2pqi)(p+qi)= x^3-6x+2##
We know that...
When I look at the left hand side of the equation in above question then I can see that the highest degree of x would be 6 after the denominators are eliminated.
I know that a polynomial of degree n will have n roots, but this one is not a pure polynomial since there is also a trigonometric...
Kindly see attached...I just want to understand why for the case; ##(-1+i)^\frac {1}{3}## they divided by ##3## when working out the angles...
Am assuming they used;
##(\cos x + i \sin x)^n = \cos nx + i \sin nx## and here, we require ##n## to be positive integers...unless I am not getting...
In a lesson on square roots this came up (Root) 27 simplifies too 3(root)3 ok. when I work that out it's
= 5.196... or if I say 3squard (root)3 this works out to 15.588.... What am I missing?
https://www.technologyreview.com/2019/12/06/131673/a-new-way-to-make-quadratic-equations-easy/
An interesting article about solving ax2 + bx + c = 0 = (x-R)(x-S), where R and S are the roots.
## x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} ##
In my classes, we were never 'spoon fed' any formula, but...
I am wondering if someone can look over my proof, and point out any mistakes I might have made.There is no value of m such that
x^3 - 3x + m = 0
has two distinct roots on the interval 0 <= x <= 1.
Proof.
Let f(x) = x^3 - 3x + m. Suppose, to the contrary, that there is a value of m such that f...
This is a textbook question and i have no solution. My attempt:
We know that ##\cosh x = \dfrac{e^x + e^{-x}}{2}##
and ##\cosh u = \dfrac{{x^2 + 1}}{2x}## it therefore follows that;
##e^{2u} = x^2##
##⇒u = \dfrac {2\ln...
Hey all,
I am having trouble solving the following equation for C
$$A(-\sqrt{C^2+4F_{+}}-C) = B(\sqrt{C^2+4F_{-}}+C)$$
I don't know how to get ride of the square roots on both sides.
Any help would be appreciated, thanks!
The characteristic equation ## m^3 -6m^2 + 12m -8 = 0## has just one single, I mean all three are equal, root ##m=2##. So, one of the particular solution is ##y_1 = e^{2x}##. How can we find the other two? The technique ##y_2 = u(x) e^{2x}## doesn't seem to work, and even if it were to work how...
I need to find the zeros of this function where d,L,v are constants.
After several calculations I faced this equation.
I tried everything I know, but I can't solve this. Maybe I'm missing something or I must made a mistake earlier in the problem.
Thus, I would like to know if it is possible to...
Let ##a,\,b,\,c## and ##d## be any four real numbers but not all equal to zero.
Prove that the roots of the polynomial ##f(x)=x^6+ax^3+bx^2+cx+d## cannot all be real.
I can do the question using brute force. First I multiply both the numerator and denominator by ##\sqrt{5} + \sqrt{3} - \sqrt{2}## then I simplify everything and rationalize again until no more square root in the denominator.
I want to ask if there is a trick to reduce the monstrous calculation...
Playing around with my calculator, I realized that if I do successive rooting operations on any positive non-zero number, I always get the number one.
Can I conclude that the infinite root of any positive number will always be zero?
If the statement is true, is there any synthesized formula to...
If we have a quadratic equation, ##px^2+qx+d## ,then the condition that the roots are rational is satisfied if our discriminant has the form ## q^2-4pd≥0## (also being a perfect square). Therefore we shall have,
##(c-a)^2-4(b-c)((a-b)≥0##
##(c-a)^2-4(ab-b^2-ac+bc)≥0##...
If we have a quadratic equation, ##px^2+qx+d## then it follows that for real roots; The discriminant
## D= q^2-4pd≥0## therefore on expanding ##(x-a)(x-b)=b^2## we get,
##x^2-bx-ax+ab-b^2=0##
##a^2+2ab+b^2-4ab+4b^2≥0##
##a^2-2ab+b^2+4b^2≥0##,
##(a-b)^2+4b^2≥0##
since, ##(a-b)^2 ≥0## and...
Hey guys,
Nice to be on here.
I have been banging my brain for the last two weeks trying to come up with an algebraic solution to the following question - to no avail.
Any input would be MUCH appreciated!
The problem is somewhat long but can be summarized as follows:
Begin with the following...
I am going to give up a bit more on the given problem. We start with polynomial ## x^27 -x ## over GF(3)[x] and we factorize it using a well known theorem it turns out it factorises into the product of monic polynomials of degree 1 and 3, 11 of them all together.
We then choose one of those...
AIUI, an algebraic is defined as a number that can be the solution (root) of some integer polynomial, and is any number that can be constructed via any binary arithmetic operation or unary root operation with arguments that are themselves algebraic numbers. I have been able to prove this for...
I need to find the values of ##\Omega## where ##(-\Omega^2 + i\gamma\Omega + \frac{2k}{3m})(-\Omega^2 + i\gamma\Omega + \frac{2k}{3m}) - (-i\gamma\Omega)(-i\gamma\Omega) = 0##
I get ##\Omega^4 -2i\gamma \Omega^3 - \frac{4k}{3m}\Omega^2 + i\frac{4k}{3m}\gamma\Omega + \frac{4k^2}{9m^2} = 0##
I...
How do we prove that every polynomial (with coefficients from C) of degree n has exactly n roots in C?
This is not a homework (I wish I was young enough to have homework) I guess this is covered in every typical undergraduate introductory algebra course but for the time being I can't find my...
Summary:: Hi guys, i can't seem to get the correct answer. I'm wondering where did I do wrong. Can someone help me to solve this? I think I need the correct formula to prove the answer :(
Given a root to 𝑥² + 𝑝𝑥 + 𝑞 = 0 is twice the multiple of another. Show that 2𝑝² = 9𝑞. The roots for 𝑥² +...
Hey! :giggle:
How can we calculate the number of natural numbers between $2$ and $n$ that have primitive roots?
Let $m$ be a positive integer.
Then $g$ is a primitive root modulo $m$, with $(g,m)=1$, if the modulo of $g\in (Z/m)^{\star}$ is a generator of the group.
We have that $g$ is a...
Hi,
I was trying to find roots of the following cubic polynomial and there are only two roots. I believe there should be three roots. Could you please guide me why there are only two roots?
If you say that the "1" repeats itself as a root then I'd say the same could be said of "0.9". Thank...
Let $p,\,q$ and $r$ be the distinct roots of the polynomial $x^3-22x^2+80x-67$. It is given that there exist real numbers $A,\,B$ and $C$ such that
$\dfrac{1}{s^3-22s^2+80s-67}=\dfrac{A}{s-p}+\dfrac{B}{s-q}+\dfrac{C}{s-r}$ for all $s\not \in \{p,\,q,\,r\}$. What is...
Hello everyone,
I'm currently doing some research about feedback systems in engineering and right now I'm playing around with special types of feedback matrices. In the process, I stumbled upon a potentially interesting polynomial, which is actually the characteristic polynomial of the system...
Given equation and conditions: ##\boldsymbol{x^2+2(k-3)x+9=0}##, with roots ##\boldsymbol{(x_1,x_2)}##. These roots satisfy the condition ##\boldsymbol{-6<x_1,x_2<1}##.
Question : ##\text{What are the allowable values for}\; \boldsymbol{k}?##
(0) Let me take care of the determinant first...
On simplifying the given equation we get, x^2-x-1=0 and using the quadratic formula we get x=(1+√5)/2 and x=(1-√5)/2
Now, as the formula suggests, there are two possible values for x which satisfies the given equation.
But now, if we follow a process in any general calculator by entering...
Let me start by pasting the question as it appears in the text :My Attempt :
Given equation : ##\boldsymbol{2x^2+mx+m^2-5 = 0}##.
For the roots of this equation to be real, the discriminant : ##m^2-8(m^2-5) \ge 0\Rightarrow 7m^2-40\le 0\Rightarrow -\sqrt{\frac{40}{7}} \le m \le...
Given : The equation ##2x^2-(a^3+8a-1)x+a^2-4a = 0## with roots of opposite signs.
Required : What is the value of ##a## ?
Attempt : The roots of the equation must be of the form ##\alpha, -\alpha##. The sum of the roots ##0 = a^3+8a-1##.
I do not know how to solve this equation.
However...
Given : Equation ##x^2+(2m+1)x+(2n+1) = 0## where ##m \in \mathbb{Z}, n \in \mathbb{Z}##, i.e. both ##m,n## are integers.
To prove : If ##\alpha,\beta## be its two roots, then they are not rational numbers.
Attempt : The discriminant of the equation ##\mathscr{D} = (2m+1)^2 - 4(2n+1) =...
Given : The quadratic equation ##x^2+px+q = 0## with coefficients ##p,q \in \mathbb{Z}##, that is positive or negative integers. Also the roots of the equation ##\alpha, \beta \in \mathbb{Q}##, that is they are rational numbers. To prove that ##\boxed{\alpha,\beta \in \mathbb{Z}}##, i.e. the...
I was looking at this discussion of swapping roots of a polynomial causing the discriminant to loop around the origin.
https://www.akalin.com/quintic-unsolvability
Although it appears to be the case, has this mathematical fact ever been proven?
It seems that the formula for the discriminant...
It is given that ##x_1, x_2\; \text{and}\; x_3## are roots of the equation ##ax^2+bx+c=0##, which are pairwise distinct.
If indeed they are roots, we should have ##ax_1^2+bx_1+c= 0 = ax_2^2+bx_2+c= 0 = ax_3^2+bx_3+c= 0##.
On subtracting the first two, we obtain ##a(x_1^2-x_2^2)+b(x_1-x_2) =...
If $\alpha,\,\beta,\,\gamma$ are the roots of the equation $x^3+ax+1=0$, where $a$ is a positive real number and $\dfrac{\alpha}{\beta},\,\dfrac{\beta}{\gamma},\,\dfrac{\gamma}{\alpha}$ be the roots of the equation $x^3+bx^2+cx-1=0$, find the minimum value of $\dfrac{|b|+|c|}{a}$.
I am using square roots, however, I am confused over how many significant figures (s.f.) to keep.
Suppose I have ##\sqrt{3.0}##, which has 2 s.f.
From three different sources, I'll put a summary in brackets:
https://www.kpu.ca/sites/default/files/downloads/signfig.pdf
(if 2 s.f. in the data...
Hi , I had to solve a quadratic equation , i got two roots as an answer ( ans= x1 / x2) , and now i need to use one of those answers to complete further tasks like finding y from x+y=c so i need to use x1 and x2 from roots , i was wondering if that's possible and how