## 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

## Find real roots of the equation

Find all real solutions to √x+12+√2−x−√x2−x+22+√−x2+x+1=x3−x2−x+1 This question is very similar to one of my previous problem, except I cannot find a monotonic function as has been done in the solution to that problem. Any help will be appreciated. Thanks. Answer This really is similar in some way to your previous problem. √x+12+√2−x−√x2−x+22+√−x2+x+1=x3−x2−x+1 Firstly, for … 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

## U-substitution of 2x in trigonometric substitution

Find \int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x. The text says to use substitution of u = 2x. How did they get u = 2x and not u = x^3? Answer When u = 2x, du = 2dx, \int^{3\sqrt3/2}_0\frac{x^3}{(4x^2+9)^{3/2}}dx = \int^{3\sqrt{3}}_0\frac{u^3/8}{(u^2+9)^{3/2}}\frac{du}{2}, this allows us to do further substitution u = 3 \tan t to get rid of the root sign … Read more

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