My Math Forum Number of solutions

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 May 26th, 2019, 10:07 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 729 Thanks: 98 Number of solutions Given equation(with respect to x): $\displaystyle x^n +x=nx$ , how many real solutions the equation has ? $\displaystyle x\geq 0 \;$ and $\displaystyle n\in \mathbb{N}$.
 May 27th, 2019, 01:55 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,029 Thanks: 2259 2 Thanks from idontknow
 May 28th, 2019, 07:07 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 729 Thanks: 98 By factorization of the equation : $\displaystyle n\cdot x$ is totally independent to a constant . So the number of real solutions is 2.
 May 28th, 2019, 11:13 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,029 Thanks: 2259 Your reasoning is unclear. Does it always apply? Thanks from idontknow
May 29th, 2019, 12:42 AM   #5
Senior Member

Joined: Dec 2015
From: somewhere

Posts: 729
Thanks: 98

Quote:
 Originally Posted by skipjack Your reasoning is unclear. Does it always apply?
It is enough to just factorize the polynomial and then equal the factors with 0.

 May 29th, 2019, 05:42 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,029 Thanks: 2259 How exactly do you proceed in that way to your conclusion?

 Tags number, solutions

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post arybhatta01 Calculus 1 May 25th, 2019 08:26 PM idontknow Number Theory 5 December 31st, 2018 06:00 PM Nathalia Trigonometry 3 September 29th, 2016 10:22 PM ishaanmj007 Algebra 4 May 16th, 2015 04:38 PM earth Math Events 3 July 8th, 2009 09:14 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top