1057x=2011 is a obvious solution.....rest i dont think deres any other....
\hspace{-16}$Find real values of $\mathbf{x}$ in $\mathbf{x^2-(1+[x]).x+2011=0}$\\\\ Where $\mathbf{[\;.\;]=}$ Greatest Integer function.
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5 Answers
1057
71After Simplifying we obtain x ( { x } -1 ) = -2011.
Ketan, Could you please try solving after this, possibly by setting x = a + f and using some logic in the quadratic thus obtained.
1057@vivek if wat u posted is correct...
since 2011 is a integer either x or ({x}-1) has to be a integer.....coz two fractions when multiplied cannot result in a integer....
therefore x has to be a integer itself bcoz if its not den {x}-1 nd x both will be fractions nd de dere product will never result to -2011 or any integer for dat matter....
therefore x is a integer and {x}=0
therefore x=2011 is the only solution....
1what vivek wrote can be obtained like this,
x2-x-x[x]+2011 = x(x-1)-x[x]+2011 = x(x-[x]-1)+201 = x({x}-1}+2011
