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प्रश्न
Solve the following quadratic equations by factorization:
`sqrt2x^2-3x-2sqrt2=0`
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उत्तर
We have been given
`sqrt2x^2-3x-2sqrt2=0`
`sqrt2x^2-4x+x-2sqrt2=0`
`sqrt2x(x-2sqrt2)+1(x-2sqrt2)=0`
`(x-2sqrt2)(sqrt2x+1)=0`
Therefore,
`x-2sqrt2=0`
`x=2sqrt2`
or,
`sqrt2x+1=0`
`sqrt2x=-1`
`x=(-1)/sqrt2`
Hence, `x=2sqrt2` or `x=(-1)/sqrt2`
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