Author Topic: math doubts [NEW]  (Read 16028 times)

zara

  • Guest
math doubts [NEW]
« on: September 26, 2009, 03:00:03 pm »
Find the real roots of the equation 18/x4 + 1/x2 =4

can summon plz explain?
                                          
« Last Edit: October 31, 2009, 11:38:20 am by Z.J. »

Offline astarmathsandphysics

  • SF Overlord
  • *********
  • Posts: 11271
  • Reputation: 65534
  • Gender: Male
  • Free the exam papers!
Re: functions
« Reply #1 on: September 26, 2009, 03:18:32 pm »
Start by multiplying by x4. Home in 4 hours.

Offline slvri

  • SF Master
  • ******
  • Posts: 1773
  • Reputation: 1735
  • Gender: Male
Re: functions
« Reply #2 on: September 26, 2009, 03:28:38 pm »
18/x4 + 1/x2 =4
multiply by x4
18+x2=4x4
4x4-x2-18=0
4(x2)2-x2-18=0
u can see that this is a quadratic equation in x2
factorising we have
4x4-9x2+8x2-18=0
x2(4x2-9)+2(4x2-9)=0
(4x2-9)(x2+2)=0
either 4x2-9=0 or x2+2=0
4x2=9 or x2=-2
x2=9/4 (x2=-2 has no real roots, complex roots are +-sqrt(2)i)
so from the first one
x=sqrt(9/4)
x=+-(3/2)
hope this helped zara :)
« Last Edit: September 26, 2009, 08:10:57 pm by slvri »
i hate A level...........

nid404

  • Guest
Re: functions
« Reply #3 on: September 26, 2009, 03:40:07 pm »
 zara slvri has done it for ya and it's right....Thanks slvri :)

zara

  • Guest
Re: functions
« Reply #4 on: September 26, 2009, 05:36:28 pm »
18/x4 + 1/x2 =4
multiply by x4
18+x2=4x4

4x4-x2-18=0
4(x2)2-x2-18=0
u can see that this is a quadratic equation in x2
factorising we have
4x4-9x2+8x2-18=0
x2(4x2-9)+2(4x2-9)=0
4(x2-9)(x2+2)=0
either 4x2-9=0 or x2+2=0
4x2=9 or x2=-2
x2=9/4 (x2=-2 has no real roots, complex roots are +-sqrt(2)i)
so from the first one
x=sqrt(9/4)
x=+-(3/2)





the ones in red i didnt get it....plz elaborate... :-\
y is 4x2-9=0? wont it be x2-9=0?
« Last Edit: September 26, 2009, 05:38:33 pm by Z.J. »

zara

  • Guest
Re: functions
« Reply #5 on: September 26, 2009, 05:42:11 pm »
another question

The function h is defined by

h:x > 6x-x2 for x>=3

Express 6x-x2 in the form a-(x-b)2, where a and b are positive constants.

zara

  • Guest
Re: functions
« Reply #6 on: September 26, 2009, 05:46:22 pm »
3) f is defined by:  f:x > 3x-2 for x E R

   g is defined by: g:x >6x-x2 for x E R

Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9.


i did the expression part...buh im confused in the second part...how to show it? :-\
n yea one more thing....what does this mean "x E R"?
i don't remember our sir explaining this to us....n its der in our worksheet!

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: functions
« Reply #7 on: September 26, 2009, 05:48:56 pm »
ok as he is not online...i wud like to explain it to u..

the first one...u multiply it by x^4 so that u can simplify it and solve easily...as the divisor can be cancelled..

wen u multiply x^4 with 18/x^4 + 1/x^2 = u get => 18 (as the x^4 gets cancelled) + x^4/x^2 = 18 + x^2

next...

(4x^2 - 9)(x^2 + 2) = 0

this means that either (4x^2 - 9) = 0...or (x^2 +2) = 0....as something HAS to be multiplied by 0 to get a 0.

so 4x^2 - 9 = 0

4x^2 = 9..

and so on...

hope that helped :)
divine intervention!

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: functions
« Reply #8 on: September 26, 2009, 05:50:39 pm »
x E R means that the values of x belong to the set of real values...


PS: wud try to solve thos.e
divine intervention!

zara

  • Guest
Re: functions
« Reply #9 on: September 26, 2009, 05:55:43 pm »
thanks adi!!
tht helped!
n yea sure go ahead solve it!
its nt for specific people...whoever can is free to do! =)

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: functions
« Reply #10 on: September 26, 2009, 06:00:48 pm »
x^2 + 6x = 9 - (x-3)^2

therefore a = 9
             b = 3
divine intervention!

zara

  • Guest
Re: functions
« Reply #11 on: September 26, 2009, 06:12:55 pm »
x^2 + 6x = 9 - (x-3)^2

therefore a = 9
             b = 3
im lost adi..:\
i need the steps if possible?!

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: functions
« Reply #12 on: September 26, 2009, 06:17:04 pm »
the general rule

x2 + nx = (x + n/2)2 - (n/2)2

n = 6

therefore n/2 = 3

here its  -x2 + 6x = (x-3)2 - (3)2 = -x2 + 6x = 9 - (x-3)2

« Last Edit: September 26, 2009, 07:45:55 pm by eddie_adi619 »
divine intervention!

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: functions
« Reply #13 on: September 26, 2009, 06:30:31 pm »
did u get g(f(x)) = -3x2 +18x - 2????
divine intervention!

Offline astarmathsandphysics

  • SF Overlord
  • *********
  • Posts: 11271
  • Reputation: 65534
  • Gender: Male
  • Free the exam papers!
Re: functions
« Reply #14 on: September 26, 2009, 07:28:11 pm »
the general rule

x2 + nx = (x + n/2)2 - (n/2)2

n = 6

therefore n/2 = 3

here its  -x2 + 6x = (x-3)2 - (3)2 = -x2 + 6x = (x-3)2 - 9



Should be (x+3)2-9