My Math Forum help needed.

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 18th, 2015, 01:14 PM #1 Newbie   Joined: May 2015 From: Israel Posts: 4 Thanks: 0 help needed. Hi all new around here. could u please help me solve the next eq. sqrt(x-(1)/(x)) - sqrt(1-(1)/(x)) = (x-1)/(x)
May 18th, 2015, 01:41 PM   #2
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,885
Thanks: 1504

Quote:
 Originally Posted by DreamFall Hi all new around here. could u please help me solve the next eq. sqrt(x-(1)/(x)) - sqrt(1-(1)/(x)) = (x-1)/(x)
You sure the left side of the equation is correct? If so ...

$\displaystyle \sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}} = 0$

so, $\displaystyle \frac{x-1}{x} = 0$

$\displaystyle x = 1$

 May 18th, 2015, 02:00 PM #3 Newbie   Joined: May 2015 From: Israel Posts: 4 Thanks: 0 I didnt understand the steps could u elebarete? and btw there are 2 solutions. 1 of them is 1.
 May 18th, 2015, 03:37 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Your equation is $\displaystyle \sqrt{x- \frac{1}{x}}- \sqrt{1- \frac{1}{x}}= 0$. Adding $\displaystyle \sqrt{1- \frac{1}{x}}$ to both sides: $\displaystyle \sqrt{x- \frac{1}{x}}= \sqrt{1- \frac{1}{x}}$ Squaring both sides, $\displaystyle x- \frac{1}{x}= 1- \frac{1}{x}$. Add $\displaystyle \frac{1}{x}$ to both sides: $\displaystyle x= 1$. I do NOT see a second solution. What reason do you have to say there are two solutions? Last edited by Country Boy; May 18th, 2015 at 03:40 PM.
May 18th, 2015, 04:17 PM   #5
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,138
Thanks: 872

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by DreamFall I didnt understand the steps could u elebarete? and btw there are 2 solutions. 1 of them is 1.
Are you perhaps looking at the graph? (See below.) It appears there is a solution at x = 0 as well as at x = 1, but we have the exclude the x = 0 because 1/x --> 1/0 in the equation doesn't exist.

-Dan
Attached Images
 Sqrt.jpg (24.9 KB, 2 views)

May 18th, 2015, 05:28 PM   #6
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,885
Thanks: 1504

Quote:
 Originally Posted by skeeter You sure the left side of the equation is correct? If so ... $\displaystyle \sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}} = 0$ so, $\displaystyle \frac{x-1}{x} = 0$ $\displaystyle x = 1$

$\displaystyle \sqrt{x-\frac{1}{x}} - \sqrt{x-\frac{1}{x}}$

instead of what you (and I) typed ...

$\displaystyle \sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}}$

May 19th, 2015, 03:55 AM   #7
Newbie

Joined: May 2015
From: Israel

Posts: 4
Thanks: 0

Quote:
 Originally Posted by skeeter my error ... I read $\displaystyle \sqrt{x-\frac{1}{x}} - \sqrt{x-\frac{1}{x}}$ instead of what you (and I) typed ... $\displaystyle \sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}}$
and it isnt = 0 as well. it = 1+x/x

sqrt(x-(1)/(x)) - sqrt(1-(1)/(x)) = (x-1)/(x)

 May 19th, 2015, 04:40 AM #8 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 the answers are: 1 and phi=golden ratio
May 19th, 2015, 04:41 AM   #9
Newbie

Joined: May 2015
From: Israel

Posts: 4
Thanks: 0

Quote:
 Originally Posted by islam the answers are: 1 and phi=golden ratio
Could u show the steps? cant figure it out

 Tags needed

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post cfg Algebra 8 August 21st, 2013 12:47 AM mathnoob99 Calculus 1 July 1st, 2010 10:27 AM nicky Algebra 7 June 20th, 2010 06:40 PM mathnoob99 Calculus 4 June 10th, 2010 06:40 PM mohit.choudhary Advanced Statistics 3 June 11th, 2009 01:52 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top