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\[[x^2]+[2x]=3x\]where [ . ] -->greatest integer function Find the values of x which satisfy this equation \[x \in [0,6]\]
2.666 is wrong :) please tell ur method too (:
oh ya... just 0,1 then?
ehh.......its the probably the most bone headed way to approach it lol :)
RHS has to lie between 0 and 18. so i can substitute integer values for LHS which give me less than 18. so thats 0,1,2,3. if the RHS has to be an integer then, x=0,0.333,0.666.....3. sooo.. ya...i actually tried all of them
(: ok here is my approach- the greatest integer function will always give out an integer :D so [x^2] -->an integer [2x]---->an integer [x^2]+[2x]--->integer+integer =integer :D so 3x has to be an integer so x has to be integer so x can be either 0,1,2,3,4,5,6 so because x is an integer we can write \[[x^2]+[2x]=x^2+2x\] so now we are left with a quadratic \[x^2+2x=3x \]solving it we get x=1,0 :)
but x need not be an integer, i.e 0.3333*3=1 (ish)
we take exact values B)
do u agree that the LHS will give an integer?
and LHS=RHS so RHS will also be an integer?
not necessarily,...if that was the case then mentioning greatest integer function would have been pointless.
greatest integer function was just to confuse (:
@imqwerty aagh...i think 2.666 would work. greatest integer function rounds down so [2.66]=2 therefore LHS=8 and 3*2.666=8 (almost). i guess the answer to this is matter of opinion aswell XD
XD but it can never be an integer haha :D it is almost close tho ahaha
\(x=5/3\) is also a solution http://www.wolframalpha.com/input/?i=solve+floor%28x%5E2%29%2Bfloor%282x%29%3D3x%2C+0%3C%3Dx%3C%3D6
5/3 =1.666 [5/3]=1 so LHS=3 RHS=5
x=2.6666=8/3, but would not work. [x^2]+[2x] =[(8/3)^2]+[2*8/3] =[64/9]+[16/3] =7+5 =12 3x=3(8/3)=8 \(\ne12\) So x=8/3 does not work. But 5/3 works!
hmmm..so while doing such questions we must also search for rational solutions with the denominator which gets cancelled ... :)