Author Topic: math doubts [NEW]  (Read 16158 times)

Offline Ghost Of Highbury

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Re: functions
« Reply #15 on: September 26, 2009, 07:43:36 pm »
how sir?

9 - (x-3)2 is correct right?

(i edited it now)

(x+3)^2 - 9 doesnt give 6x - x^2
« Last Edit: September 26, 2009, 07:47:14 pm by eddie_adi619 »
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Offline slvri

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Re: functions
« Reply #16 on: September 26, 2009, 08:12:11 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?

sory i made a mistake there
and for the first one, u multiply each term in the equation by x^4 to turn it into a expression that u can factorise easily
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zara

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Re: functions
« Reply #17 on: September 26, 2009, 08:14:59 pm »
sory i made a mistake there
and for the first one, u multiply each term in the equation by x^4 to turn it into a expression that u can factorise easily
yea Thanks...adi expalined it...=)


Offline slvri

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Re: functions
« Reply #18 on: September 26, 2009, 08:15:52 pm »
yea Thanks...adi expalined it...=)


i hope u understood :)
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zara

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Re: functions
« Reply #19 on: September 26, 2009, 08:26:44 pm »
yes slvri i did!

okay one more ques...

function f and g are defined by:
f(x): k-x    k is a constant
g(x): 9/x+2   whr x nt = 2   both x E R

Find the values of k for whch the eqtn f(x)=g(x) has two equal roots and solve the equation f(x)=g(x) in these cases.



Offline slvri

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Re: functions
« Reply #20 on: September 26, 2009, 08:36:00 pm »
yes slvri i did!

okay one more ques...

function f and g are defined by:
f(x): k-x    k is a constant
g(x): 9/x+2   whr x nt = 2   both x E R

Find the values of k for whch the eqtn f(x)=g(x) has two equal roots and solve the equation f(x)=g(x) in these cases.



f(x)=g(x)
k-x=9/x+2
(k-x)(x+2)=9
kx-x2+2k-2x=9
x2+(2-k)x+9-2k=0
in this equation a=1, b=2-k, c=9-2k
the condition for the equation f(x)=g(x) to have two equal roots is b2-4ac=0
(2-k)2-4(1)(9-2k)=0
4-4k+k2-36+8k=0
k2+4k-32=0
k2+8k-4k-32=0
k(k+8)-4(k+8)=0
(k+8)(k-4)=0
k=-8 or k=4
when k=-8
x2+(2-k)x+9-2k=0
x2+(2--8)x+9-2(-8)=0
x2+10x+25=0
(x+5)2=0
x+5=0
x=-5
but when k=4
x2+(2-k)x+9-2k=0
x2+(2-4)x+9-2(4)=0
x2-2x+1=0
(x-1)2=0
x-1=0
x=1
there u go
« Last Edit: September 27, 2009, 08:46:44 am by slvri »
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zara

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Re: functions
« Reply #21 on: September 26, 2009, 08:49:01 pm »
function f and g are defined as

f(x): x2-2x
g(x):2x+3     both x E R


1) find the set of values of x for which f(x)>15
2)find the range of f...

im bad at this...had nly 2 or 3 classes of it...n didnt get much :-\

Offline Ghost Of Highbury

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Re: functions
« Reply #22 on: September 26, 2009, 08:58:12 pm »
i guess the 1st one is -3>x>5

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Offline astarmathsandphysics

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Re: functions
« Reply #23 on: September 26, 2009, 08:58:47 pm »
function f and g are defined as

f(x): x2-2x
g(x):2x+3     both x E R


1) find the set of values of x for which f(x)>15
2)find the range of f...

im bad at this...had nly 2 or 3 classes of it...n didnt get much :-\


x^2-2x>1.5
so
x^2-2x-1.5>0
Complete the square
(x-1)^2-1-1.5>0
(x-1)^2>2.5 so x-1>sqrt(2.5) sox>1+sqrt(2.5) or x-1<-sqrt(2.5) so x<1-sqrt(2.5)

Offline slvri

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Re: functions
« Reply #24 on: September 26, 2009, 09:02:11 pm »
function f and g are defined as

f(x): x2-2x
g(x):2x+3     both x E R


1) find the set of values of x for which f(x)>15
2)find the range of f...

im bad at this...had nly 2 or 3 classes of it...n didnt get much :-\

its ok with practice u will get better :)
1) f(x)>15
x2-2x>15
x2-2x-15>0
x2-5x+3x-15>0
x(x-5)+3(x-5)>0
(x-5)(x+3)>0
(x-5))(x-(-3))>0
now theres a rule u need to keep in mind (i cant illustrate it here using a diagram so u will have to memorise it for the time being)
whenever an expression of the form(x-a)(x-b)>0 occurs where a<b then the solution is x<a and x>b
over here a is 5 and b is -3(5 is greater than -3)
so the solution is x<-3 and x>5
2)x2-2x
=(x2-2x+1)-1
=(x-1)2-1
now u can see that when x=1
f(1)=(1-1)2-1=-1
see for yourself that whatever value u put for x, the answer will always be greater than -1(-1 is the minimum value of f(x))
so the range of f(x) is f(x)>=-1

« Last Edit: September 27, 2009, 08:48:26 am by slvri »
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Offline Ghost Of Highbury

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Re: functions
« Reply #25 on: September 26, 2009, 09:05:46 pm »
shudnt -3>x>5
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Offline slvri

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Re: functions
« Reply #26 on: September 27, 2009, 12:52:26 am »
Hw cnw b greater than 5 and less at the same time?
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Offline slvri

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Re: functions
« Reply #27 on: September 27, 2009, 12:55:54 am »
I meant to say hw cn x b greater than 5 and less than -3 at the same time?
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Offline Ghost Of Highbury

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Re: functions
« Reply #28 on: September 27, 2009, 07:13:46 am »
yeah..then..how can it be less than -5 and greater than 3 at the same time??
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nid404

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Re: functions
« Reply #29 on: September 27, 2009, 07:15:29 am »