Constructible Solutions

We know that if a cubic equation with a rational coefficients has a constructible root, then the equation has a rational root. Now let; x3−2x+2√2=0 Could √2t be a viable substitution? Answer You might find this page interesting. It is proving that if p(x) has no rational roots, then it does not have any constructible … Read more

Help on solving the equation √a+x√a+√a+x=√a−x√a−√a−x\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}

Could you give me some help on finding the roots (if any) of the following equation: √a+x√a+√a+x=√a−x√a−√a−x I tried to apply some classic approaches, but I had no luck… Could you lend me a hand? Thanks in advance! Answer √a+x√a+√a+x=√a−x√a−√a−x √1+xa√1+√1+xa=√1−xa√1−√1−xa xa↦y √1+y1+√1+y=√1−y1−√1−y √1+y−√1−y2=√1−y+√1−y2 √1+y−√1−y=2√1−y2 1+y+1−y−2√1−y2=4(1−y2) 4y2−2=2√1−y2 2y2−1=√1−y2 4y4−4y2+1=1−y2 4y4−3y2=0 y2(4y2−3)=0 y=0,y=±√32 x=0,x=±√32a AttributionSource : … Read more

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

Location of the roots of the equation f(x)+2f′(x)+f″f(x)+2f'(x)+f”(x)=0 if \alpha, \beta\alpha, \beta and \gamma\gamma are the roots of f(x)=0f(x)=0

Problem Statement:- If the cubic equation f(x)=0 has three real roots \alpha, \beta and \gamma such that \alpha\lt\beta\lt\gamma, show that the equation f(x)+2f'(x)+f”(x)=0 has a real root between \alpha and \gamma. Attempt at a solution:- If f(x) is a quadratic equation whose roots are \alpha, \beta and \gamma, then f(x)=(x-\alpha)(x-\beta)(x-\gamma) From this we get f'(x) … Read more

Let P(x)=4×2+6x+4P(x)=4x^2+6x+4 and Q(x)=4y2−12y+25Q(x) =4y^2-12y+25. If x,yx,y satisfy the equation P(x)Q(y)=28P(x)Q(y)=28, then find the value of 11y−26x11y-26x. [duplicate]

This question already has answers here: Find the unique pair of real numbers (x,y) that satisfy P(x)Q(y)=28 (2 answers) Closed 5 years ago. Let P(x)=4×2+6x+4 and Q(y)=4y2−12y+25. If x,y satisfy the equation P(x)Q(y)=28, then find the value of 11y−26x. I know this question has been asked earlier here but the answer given states “Note that … Read more

Some questions about trigonometric polynomials, Vieta’s relations and correct integer parameters for roots

I was trying to compute: P=sin(2π/7)sin(4π/7)sin(8π/7) So I managed to analyse the equation sin(7x)=0, thus x=kπ/7, when k is an integer. The problem arrives when the equation sin(7x)=7sin(x)−56sin3(x)+112sin5(x)−64sin7(x)=0 is derived, because I can’t exactly know what are its roots. Assuming sin(x)≠0, the equation becomes 7 −56sin2(x)+112sin4(x)−64sin6(x)=7−56t2+112t4−64t6=0 Where t=sin(x). But, to solve my problem, how can I … Read more

Calculate \lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}

Calculate:\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n} Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom. My own solution \lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}~\frac{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}} \lim_{n\to\infty} \frac{5n^2+4~-~(5n^2+n)}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}~=~\lim_{n\to\infty} \frac{4-n}{n\left(\sqrt{5+\frac4{n^2}}~+~\sqrt{5+\frac1{n}}\right)} =~\frac{-1}{2\sqrt{5}}~=~-\frac{\sqrt{5}}{10} Answer Two alternative ideas: (1) Write \sqrt{5n^2+4} – \sqrt{5n^2+n} = \sqrt 5n(\sqrt{1 … Read more

Relation between roots of function and roots of its derivative

I’m reading a section my Calculus book that is about the relation between the roots of a polynomial function and the roots of its derivative. So: Notice that if $x_1$ and $x_2$ are roots of $f$, so that $f(x_1) = f(x_2) = 0$, then by Rolle’s Theorem there is a number $x$ between $x_1$ and … Read more