clearly the roots of the eqn must lie in (-2 , 3)
and x2 must be an integer .
thus x can take the values - { -√2 , -1 , 0 , 1, √2, √3 ,2 ,√5 ,√6, √7 ,√8)
checking for the above values we get only -√2 and √6 as the solutions.
\hspace{-16}$Find all Real values of $\bf{x}$ that satisfy the equation\\\\ $\bf{x^2=4+[x]}$\\\\
Ans = - √2 and x = √6
I have solved it using very Lengthy Method
can anyone have a analytical Method without Using Graph
take :
{x}=-x2+x+4
or, 0≤-x2+x+4<1. then progress.
Moreover in the Q , RHS is an Integer so L.HS.must also be an Integer.
clearly the roots of the eqn must lie in (-2 , 3)
and x2 must be an integer .
thus x can take the values - { -√2 , -1 , 0 , 1, √2, √3 ,2 ,√5 ,√6, √7 ,√8)
checking for the above values we get only -√2 and √6 as the solutions.