Algebra

Level 2

Number of real solution of the equation 201 2 x + 201 4 x = 201 5 x + 201 1 x 2012^x+2014^x = 2015^x+2011^x


The answer is 2.

Jagdish Singh by jagdish singh , 33, India

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2 solutions

Shuhhank Saxena
Dec 17, 2013

Only two cases are possible 1.If X=0 2.IF X=1

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Dushyant Sharma
Dec 17, 2013

2012^x+2014^x=2015^x+2011^x => 2012^x-2011^x=2015^x-2014^x now by hit and trial method when we put x=1 we get the eqn 1=1 when we put x=2 we get the eqn 2012^2-2011^2=2015^2-2014^2 => (2012+2011)(2012-2011)=(2015-2014)(2015+2014) =>4026=4026

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x=0 is a solution as well ...

houssam kherraz - 7 years, 5 months ago

o ?

Bhosxz Polly - 7 years, 5 months ago

"Number of real solution of the equation " read it carefully dude the solution is 1,0 but there are 2 solution so u should write 2 in it

Đăng Bùi Minh - 7 years, 5 months ago

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