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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Oct 2008 19:14:22 IST
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solve the equation
 = 9x-3
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5 Oct 2008 20:50:45 IST
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see tht thing inside roots..
get tht 2 inside root...
root (a) - root(b) = a -b
=> a =b or a & b are simultaneously..zero...
surely.. a=b ..as b cannot be zero..
x = 1/3...
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5 Oct 2008 20:58:33 IST
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i couldn't understand how could this a and b be equal and cancelled out to b zero????
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5 Oct 2008 21:48:04 IST
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@Decoder
x=1/3 yields -2.2=0.
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"The journey of a thousand miles begins with one step."
- Lao Tzu
Like and unlike the proverb above, the solution to a problem begins (and continues, and ends) with simple, logical steps. But as long as one steps in a firm, clear direction, with long strides and sharp vision, one would need far, far less than the million of steps needed to journey a thousand miles. And mathematics, being abstract, has no physical constraints; one can always restart from a scratch, try new avenues of attack, or backtrack at an instant's notice. One does not always have these luxuries in other forms of problem-solving (e.g. trying to go home if you are lost). |
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6 Oct 2008 12:33:10 IST
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@ lakshya..
mistakes..happen everybody does 2+2 =5....( myself saw Jee rank 5 doin it)
stupids laugh at it..while smart ones take sum lessons..
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6 Oct 2008 20:18:38 IST
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(4x^2+5x+1)^1/2 - 2(x^2-x+1)^1/2=9x-3
(4x^2+5x+1)^1/2 - (9x-3) = 2(x^2-x+1)^1/2
squaring both d sides we get,
4x^2+5x+1 - 2*(4x^2+5x+1)^1/2*(9x-3) + (9x-3)^2=4x^-4x+4
further, 9x-3 - 2*(4x^2+5x+1)^1/2*(9x-3) + (9x-3)^2=0
therefore, 9x-3=0 or 9x-2=2*(4x^2+5x+1)^(1/2)
therefore, x=1/3 and on solving other equation we get another two value of x. x=0 or x=56/65. both of these don't satisfy d eqn.
Hence there is only one solution i.e x=1/3.
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6 Oct 2008 20:25:59 IST
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the question is  = 9x-3
And you have taken (4x^2+5x+1)^1/2 - 2(x^2-x+1)=9x-3
this should be (4x^2+5x+1)^1/2 - 2[(x^2-x+1)]^1/2=9x-3
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7 Oct 2008 15:23:30 IST
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thanks for this
i've corrected .
now see is it correct??
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