How to solve $x^4+8x-1=0 $

Any idea how to solve the following equation? $$x^4+8x-1=0$$ I tried to find some obvious roots of this equation so it could help me to find the other roots (if it has more then $1$) but I had no success, and I would really appreciate some help. Answer There will be a real root a … Read more

if x≠0x \neq 0 and x=√4xy−4y2x = \sqrt{4xy – 4y^2}, then how does expressing xx in terms of yy mean x=2yx = 2y?

I have the equation x=√4xy−4y2, and I know that x=2y when expressed in terms of y, but I’m not sure of the process to get there. I know that √4xy−4y2=√4y(x−y)=2√y(x−y)=2(xy−y2)12 but pretty stumped. Answer x=√4xy−4y2 x2=4xy−4y2 x2−4xy+4y2=0 (x−2y)2=0 x−2y=0 x=2y AttributionSource : Link , Question Author : maudulus , Answer Author : Adi Dani

A simple proof for non-rationality of number n√1+qn\sqrt[n]{1+q^n}

I’m going to prove the following fact (preferably with an elementary simple method): Let n>2 be a natural number. Then for all q∈Q+∖{0} we have n√1+qn∉Q+∖{0}. Assuming the contrary, we then have (1+anbn)=cndn, where q=ab and n√1+qn=cd. Now multiplying both sides of the above equation by bndn we get a contradiction according to Fermat Last … Read more

Convergence of a series with radicals: \sum\limits_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n+1}}{n}\sum\limits_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n+1}}{n}

\sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n+1}}{n} Where a_n=\frac{\sqrt{n+2}-\sqrt{n+1}}{n} I want to check whether this sum converges… At quick glance this series seems to converge because n>\sqrt{n+2}-\sqrt{n+1} The ratio test, root test, integral test did not work in this case, so I tried the limit comparison test with b_n=\frac{1}{n^2} b_n=\frac{1}{n^3} Basically any p-series with p=2,3,4,5… i.e. an integer, gives \lim_{n\to\infty}\frac{a_n}{b_n}=\infty Then … Read more

Find minimal value √x2−5x+25+√x2−12√3x+144 \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144} without derivatives.

Find minimal value of √x2−5x+25+√x2−12√3x+144 without using the derivatives and without the formula for the distance between two points. By using the derivatives I have found that the minimal value is 13 at x=4023(12−5√3). Answer If we complete the squares we get √(x−52)2+754+√(x−6√3)2+36. This is the distance from the point P(x,0) to the point Q(5/2,−5√3/2) … Read more

Limit lim\lim_{x \to 2^{-}} \left ( \frac{1}{\sqrt[3]{x^{2} -3x+2}} + \frac{1}{\sqrt[3]{x^{2} -5x+6}} \right )

\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right ) I’ve tried using the A^3-B^3 identity, but that doesn’t help. Also, I tried multiplying every fraction with \sqrt[3]{A^2} to get rid of the roots in the denominator, but that doesn’t help either. Can someone suggest a solution? Thanks. Answer Let x-2=t. Hence, \lim_{x … Read more

How to multiply the radical expressions and simplify your answer?

I am trying to multiply and simplify the following radical expression. (\sqrt{x}+5 – 4)(\sqrt{x}+5+4) According to the book, the answer is x – 11 However, I am confused about how this even works. I tried using the following calculator that shows all the steps. However, the answer is completely different when using the calculator … Read more

How can I solve this absolute value equation?

This is the equation: $|\sqrt{x-1} – 2| + |\sqrt{x-1} – 3| = 1$ Any help would be appreciated. Thanks! Answer Let $a =\sqrt{x-1}$, $|a-2|+|a-3|=1$ Check for solutions in the different regions for $a$. Region 1: $a<2$. Then we have $(2-a)+(3-a)=1$, i.e. $5-2a=1$, so that $a=2$. Region 2: $2\leq a\leq 3$. We have $(a-2)+(3-a)=1$. This is … Read more

Prove ⌈(3+√5)n⌉\lceil(3 + \sqrt{5})^n\rceil is divisible by 2n2^n for all natural numbers nn [duplicate]

This question already has answers here: divisibility problem using induction (3 answers) Prove that the next integer greater than (3+√5)n is divisible by 2n where n is a natural number. (1 answer) Closed 2 years ago. This sub-question is part of a larger question: If Sn=(3+√5)n+(3−√5)n, show that Sn is an integer. Also prove that … Read more

Limits and infinity minus infinity

I understand that we need to rationalize when we have infinity minus infinity like here \lim_{x\to \infty}\left(\sqrt{x^2 + 1} – \sqrt{x^2 + 2}\right) My question is why can I not just split the limits like this \lim_{x\to \infty}\left(\sqrt{x^2 + 1}\right) – \lim_{x\to \infty}\left(\sqrt{x^2 + 2}\right) and then \lim_{x\to \infty}\sqrt{x^2} * \lim_{x\to \infty}\sqrt{1 + \frac 1x} … Read more