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प्रश्न
`3x^2-x-2=0`
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उत्तर
`3x^2-x-2=0`
`9x^2-3x-6=0` (Multiplying both sides by 3)
`9x^2-3x=6`
`(3x)^2-2xx3x xx1/2+(1/2)^2=6(1/2)^2` [Adding `(1/2)^2` on both sides]
`(3x-1/2)^2=6+1/4=25/4=(5/2)^2`
`3x-1/2=+-5/2` (Taking square root on both sides)
`3x-1/2=5/2 or 3x-1/2=-5/2`
`3x=5/2+1/2=6/2=3 or =-5/2+1/2=-4/2=-2`
`x=1 or x=-2/3`
Hence, 1 and `-2/3` are the roots of the given equation.
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