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Algebra

Hot goIITian

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26 Dec 2007 17:23:32 IST

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Can anyone solve this?

None

Solve for x:

|x²+x-1|=2x-1

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Comments (9)

Decoder

Blazing goIITian

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26 Dec 2007 17:39:23 IST

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just square both sides..u will directly get forur roots from there...else solve equation by taking r.h.s as 2x-1 and -(2x-1)

Himanshu

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26 Dec 2007 17:40:02 IST

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ans is 1,-1,0,-2

use + 1ce and - the other time...dats it

vasanth_mech pilani

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26 Dec 2007 17:41:35 IST

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basically we must try to avoid squaring in these questions:

takin cases:

x² + x - 1 = +(2x-1) ,x> (-1+sqrt5)/2 or x<(-1- sqrt5)/2

x² -x =0 , so; x = 0,1, but we must exclude 0

case2

x² + x - 1 = -(2x-1) , (-1- sqrt5)/2 < x<(-1+ sqrt5)/2

x² + 3x - 2 =0

so x=[-3 + sqrt.(17)]/2 and [-3 - sqrt.(17)]/2

but [-3 - sqrt.(17)]/2 doesnt satisfy the first condition

sp final solution

x = 1, [-3 + sqrt.(17)]/2

abhishek sinha

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26 Dec 2007 17:52:19 IST

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See , we get only 2 soln s

for both of the cases x > 1/2

Now solve taking +ve & -ve sign

giving acceptable solns as x = 1 (taking + sign )

and x = ( -3 + sqrt (17 ) ) /2 ( - sign )

abhishek sinha

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26 Dec 2007 17:56:18 IST

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among the Solns himanshu mentioned -1 , 0 , -2 does not satisfy the given eqn

Tarin Bansal

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26 Dec 2007 18:32:25 IST

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Yup!,
feynmann is right. On keeping -1,0,-2. RHS comes out to be -ve and as it is equal to a quantity in modulus, it cannot be -ve as mod is always positive.

Merk anand

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26 Dec 2007 18:32:34 IST

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only x=1 satisfy..

abhishek sinha

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26 Dec 2007 19:34:42 IST

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No ,the other one mentioned by me also satisfies . Try it .

abhishek sinha

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26 Dec 2007 19:38:38 IST

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For the latter one all we need is x> 1/2 & x^2 + x -1 < 0 . Both the conditions are satisfied so, x = { -3 + sqrt(17 ) } /2 is also an acceptable soln.

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