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lakeoris (m)
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x2+(x2-1)1/2=1
best of luck
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Ehido (m)
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Hi Poster,
your question shoulde be simplify the expression
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femi4 (m)
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@ EHIDO, It is not an expression,the poster is right.It is an equation because of the equality sign.
To start with, square both sides to get polynomials in X^4+, =1 and solve the polynomials for different values of x by examination
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donlet (m)
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check this out
x^2+ ( x^2 - 1)^1/2 = 1
(x^2+(x^2-1)^1/2)^1/2 = 1^1/2
x + ( X^2 - 1 ) = 1
x + x^2 - 1 = 1
x^2 + x - 2 = 0
.
, (x+2) = 0
(x - 1) = 0
X = -2 or 1.
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Ehido (m)
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The equality sign was not there yesterday, trust me I know an equation when I see one
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olayhincar (m)
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The equality sign was not there yesterday, trust me I know an equation when I see one
but i just solved it today and i got 1 to be my final answer.
I put Logx to both sides
Remember LogxX = 1
X^2 + ( X^2 - 1 ) ^ 1/2 = 1
logx X^2 +( logx( X^2 - 1 ) ^ 1/2) = logx X^1
2logx X + 1/2( logx( X^2 - 1 )) = 1logx X
2logx X + (1/2(logx X^2) - (1/2 logx 1 )) = 1logx X
2logx X + ((1logx X) - (1/2(logx1) ) = 1logx X
2logx X - (1/2(logx 1) ) = 1logx X - 1logx X
2logx X - (1/2(logx 1) ) = 0
2logx X = (1/2(logx 1) )
logx X^2 = logx 1^(1/2)
Divide both sides by Logx
u are left with
X^2 = 1^(1/2)
X^2 = Sq rt(1)
X^2 = 1
X = Sq rt(1)
X = 1
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olayhincar (m)
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@napolie, use logarithm to simplify
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napolie (m)
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The equality sign was not there yesterday, trust me I know an equation when I see one
but i just solved it today and i got 1 to be my final answer.
I put Logx to both sides
Remember LogxX = 1
X^2 + ( X^2 - 1 ) ^ 1/2 = 1
logx X^2 +( logx( X^2 - 1 ) ^ 1/2) = logx X^1
2logx X + 1/2( logx( X^2 - 1 )) = 1logx X
2logx X + (1/2(logx X^2) - (1/2 logx 1 )) = 1logx X
2logx X + ((1logx X) - (1/2(logx1) ) = 1logx X
2logx X - (1/2(logx 1) ) = 1logx X - 1logx X
2logx X - (1/2(logx 1) ) = 0
2logx X = (1/2(logx 1) )
logx X^2 = logx 1^(1/2)
Divide both sides by Logx
u are left with
X^2 = 1^(1/2)
X^2 = Sq rt(1)
X^2 = 1
X = Sq rt(1)
X = 1
i never read the question carefully before attempting it  . i just went ahead and differentiate it. you're right olayhincar
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gen2dan (m)
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Now the question
x2 + (x2 - 1)1/2= 1
(x2 - 1)1/2 = 1 - x2
Squaring both sides and expanding we get
x2 - 1 = x4 - 2x2 + 1
Collecting like terms and simplifying we get
x4 - 3x2 + 2 = 0
Now this is a polynomial of the 4th degree in even consecutive indices to solve this we divide with a quadrant,
so using the trial method lets take x2 - 1
Dividing with this we get,
x2 - 2 as the answer
therefore,
(x2 - 1)(x2 - 2)
In which case
x = +/-1 or the square root of 2
Check:
x2 + (x2 - 1) = 1
Substituting x = 21/2
(21/2)2 + ({21/2}2 - 1)1/2
Gives,
2 +/- 11/2
which gives,
2 +/- 1
2-1=1
You can also check the other value yourself for confirmation
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M quadrant (m)
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different efiko nd ò je wè with different method nd answer.
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olalekan1 (m)
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Now the question
x2 + (x2 - 1)1/2= 1
(x2 - 1)1/2 = 1 - x2
Squaring both sides and expanding we get
x2 - 1 = x4 - 2x2 + 1
Collecting like terms and simplifying we get
x4 - 3x2 + 2 = 0
Now this is a polynomial of the 4th degree in even consecutive indices to solve this we divide with a quadrant,
so using the trial method lets take x2 - 1
Dividing with this we get,
x2 - 2 as the answer
therefore,
(x2 - 1)(x2 - 2)
In which case
x = +/-1 or the square root of 2
Check:
x2 + (x2 - 1) = 1
Substituting x = 21/2
(21/2)2 + ({21/2}2 - 1)1/2
Gives,
2 +/- 11/2
which gives,
2 +/- 1
2-1=1
You can also check the other value yourself for confirmation
you are the only one getting it but x4 - 3x2 + 2 = 0 , it has to be x4 - 3x2 - 2 = 0 |
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Ronnykay (f)
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@gen2dan
igi iwe
you are too good!!!!!!!!
omo la toro iwe la bi
bata re a dun ko ko ka
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M quadrant (m)
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@ronnykay, uhmmmnmm!!! Lol.[color=][/color]
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Uzzyan (f)
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@gen2dan
U are blessed!!!
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M quadrant (m)
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Na wa o! Pastor. God go bless us jare. Lol.
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bayedero (m)
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Solve thie:
3x + 4x = 5x
** Double dare you guys
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Free_rhyme (m)
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Solve thie:
3x + 4x = 5x
** Double dare you guys
Professor of Maths, Na u Know Maths Pass Everybody, A1 Perpendicular na e u get for secondary school. |
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kokoA (m)
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these are secondary school mathematics. Boys
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napolie (m)
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Professor of Maths, Na u Know Maths Pass Everybody, A1 Perpendicular na e u get for secondary school.
i will really appreciate it if the poster of this question could give the solution with workings since no efiko online has been able to solve for x in the eqn. looking forward to hear from you POSTER.  |
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Big Star (m)
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what's the shit? X= -2 or x=1.
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oseh4ril
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x^2 + (x^2 - 1)^1/2 = 1
squaring both sides of d eqn
we have : x^4 + x^2 - 1 = 1
x^4 + x^2 = 1+1
x^2(x^2+1) = 2
i.e x^2 = 2 or x^2+1 = 2
x=sqrt(2) or x^2=2-1
x=1.414 or x^2=1
x=1.414 or x=sqrt(1)
x=1.414 or x=1
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pongwa (m)
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pls u pple shld include your schools when solving problems so we kno d quality of your lecturers and school as well
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candylips (m)
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Solve thie:
3x + 4x = 5x
** Double dare you guys
this is high school math. just looking at the problem will tell you that X =2 can any body solve this simple Differential Equation y '' - 4y ' + 13y = 0. obtain the value of y y '' = d 2y/dx 2y ' = dy/dx |
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