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Question
`2/x^2-5/x+2=0`
Solution
`2/x^2-5/x+2=0`
`(2-5x+2x^2)/x^2=0`
`2x^2-5x+2=0`
`4x^2-10x+4=0` (Multiplying both sides by 2)
`4x^2-10x=-4`
`(2x)^2-2xx2x xx5/2+(5/2)^2=-4+(5/2)^2` [Adding `(5/2)^2` on both sides ]
`(2x-5/2)^2=-4+25/4=9/4=(3/2)^2`
`2x-5/2=+-3/2` (Taking square root on both sides)
`2x-5/2=3/2 or 2x-5/2=-3/2`
`2x=3/2+5/2=8/2=4` or `2x=-3/2+5/2=2/2=1`
`x=2 or x=1/2`
Hence, 2 and `1/2` are the roots of the given equation.
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