Find the number which exceeds its reciprocal by8/3. plzz fast.....
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let the number be x
a/q,
x-1/x = 8/3
x^2 - 1 = 8/3x
x^2 - 8/3x -1= 0
3x square - 8x - 3 = 0
3x square -9x+x-3 =0
3x(x-3)+1(x-3)
(x-3)(3x+1)
x = 3, -1/3
a/q,
x-1/x = 8/3
x^2 - 1 = 8/3x
x^2 - 8/3x -1= 0
3x square - 8x - 3 = 0
3x square -9x+x-3 =0
3x(x-3)+1(x-3)
(x-3)(3x+1)
x = 3, -1/3
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ankit0000rocks:
I don't understand !!
can u solve it on a page?? and send me
yes sure!
i have posted it.
you may see and tell if u still have any confusion.
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