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Junior | Next Rank: 30 Posts
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by spanlength » Tue Aug 05, 2008 4:49 am
Let 2^m = y. Hence the above equation becomes : y^(-2) + y^(-1) -6 = 0
Mulitplying y^2 to both L.H.S and R.H.S we have
1+ y - 6y^2 = 0
1+3y-2y -6y^2 = 0
Solving we have : (1 + 3y)(1-2y) = 0
Hece y = -1/3 , 1/2
Hence the value of (2^m) is -1/3, 1/2.
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Junior | Next Rank: 30 Posts
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by m51v50 » Tue Aug 05, 2008 7:42 am
Thaks@@@@
Have 790 or die
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