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If the roots of the equation 2^m x^2 + 8x + 32^m = 0 are

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If the roots of the equation 2^m x^2 + 8x + 32^m = 0 are [#permalink] New post 26 Jan 2004, 13:43
If the roots of the equation 2^m x^2 + 8x + 32^m = 0 are equal and real, then what is the value of m?
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 [#permalink] New post 26 Jan 2004, 16:18
2/3, we r given that b^2-4ac=0 so 8^2 - 4* 2^m *32^m=0 we get 2^6=2^6m+2 so 6=6m+2, m=2/3
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 [#permalink] New post 26 Jan 2004, 16:28
:good
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 [#permalink] New post 26 Jan 2004, 16:29
:good
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 [#permalink] New post 29 Jan 2004, 12:48
Hi sunnyboy,

Who's is the babe
Anand.

roots are
( -8 (+or-) sqrt( 64-4*32^m*2^m) ) /2

for the roots to be real we have to have
4*32^m*2^m < 64
so
2^(6m+2) < 2^6
so 6m < 4

But the second condition is that the roots are equal the only way this can happen in the value under sqrt is zero
so we have 6m = 4 or m = 2/3 :-D
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 [#permalink] New post 29 Jan 2004, 12:57
That hot babe my friend, is Sienna Miller. :sex
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  [#permalink] 29 Jan 2004, 12:57
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If the roots of the equation 2^m x^2 + 8x + 32^m = 0 are

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