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#1 2005-08-14 18:39:47

sunny
Member
Registered: 2005-08-14
Posts: 5

Need help

I need to know how to simplify:

x^(3/4) + 5/x^(1/4)        (you may assume x>0) 

And how do I find the absolute inequality corresponding to: x<=-22/3 OR x>=11/3 (And what exactly is the absolute inequality?)

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#2 2005-08-14 18:57:10

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,534

Re: Need help

Multiply the first term by x^(1/4)/x^(1/4) :

x^(3/4) * x^(1/4)/x^(1/4) + 5/x^(1/4) 

= x^((3/4)+(1/4))/x^(1/4) + 5/x^(1/4) 

= x^1/x^(1/4) + 5/x^(1/4) 

= x / x^(1/4) + 5/x^(1/4)

= (x+5) / x^(1/4)

Absolute Inequality I think refers to something like |x|<3, therefore -3<x<3, so we need an absolute value (in other words it is the distance from 0, ignoring plus or minus)

x<=-22/3 OR x>=11/3 could possibly be solved by taking the difference of the two absolute values:

|-22/3| - |11/3| = 11/3

Then halve that and put it in a formula: |x+11/6|<11/3+11/6  ==> |x+11/6|<11/2

Why did I do this? Because I wanted to know how to find the "middle" of those two numbers. (Try putting dots on the number line to see what I mean).


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#3 2005-08-15 18:24:59

sunny
Member
Registered: 2005-08-14
Posts: 5

Re: Need help

At first I was confused, but when I tried your number line suggestion I got your meaning. Thank you so much for you help! But now I'm having trouble with this problem:

Solve for X:

x^3+2x^2-9x=18
x^3+2x^2-x=-2 Is this right? If it is then, I can't seem to go any further

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#4 2005-08-15 20:39:56

ganesh
Moderator
Registered: 2005-06-28
Posts: 13,145

Re: Need help

x³ + 2x² - 9x = 18

x³ + 2x² - 9x - 18 = 0
Putting x=3, we find that it is one of the solutions.
Therefore,
(x-3)(x² + 5x + 6) = 0
x² + 5x + 6 = (x+3)(x+2)
Therfore,
(x-3)(x+3)(x+2) = 0
which gives the following values of x :- 3, -3, -2

PS:_ A cubic equation, that is an equation of degree three, can be solved only by trial and error, as far as I know.


Character is who you are when no one is looking.

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#5 2005-08-16 11:02:23

sunny
Member
Registered: 2005-08-14
Posts: 5

Re: Need help

Oh, I came out with a different answer...so is this wrong?:

x^3 + 2x^2 - 9x = 18
x^3 + 2x^2 - 9x-18 = 0
x^2(x+2)-9(x+2)=0
(x+2)(x^2-9)=0
x=-2, 3

Check: -2
-2^3+2(-2)^2-9(-2)=18
-8+8+18=18
18=18

Check: 3
3^3 + 2(3)^2 - 9(3) = 18
27+18-27=18
18=18

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#6 2005-08-16 11:07:28

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Need help

What you've done is correct, just incomplete.
For the x²-9=0 bit, that means that x²=9, so x=√9.
Square roots always have two answers, positive and negative, so x=±3.
That means that your solutions are -2, -3 and 3, which agrees with ganesh's workings.


Why did the vector cross the road?
It wanted to be normal.

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#7 2005-08-23 22:16:30

ahgua
Guest

Re: Need help

Solve this:

2sinx.cosx + cos^2x = 1

#8 2005-08-25 13:39:18

ajp3
Member
Registered: 2005-08-25
Posts: 9

Re: Need help

2sinxcosx + cos^2x = 1

=>

2sinxcosx - sin^2x = 0 (bc of the pythagorean identity)

=>

(2cosx - sinx)sinx = 0

=>

2cosx=sinx or sinx = 0

=>

tan x = 2 or sin x = 0

so x = arctan 2 or x = n(pi), where n is any integer.

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