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प्रश्न
Solve the following quadratic equations by factorization:
`3x^2-2sqrt6x+2=0`
उत्तर
We have been given
`3x^2-2sqrt6x+2=0`
`3x^2-sqrt6x-sqrt6x+2=0`
`sqrt3x(sqrt3x-sqrt2)-sqrt2(sqrt3x-sqrt2)=0`
`(sqrt3x-sqrt2)(sqrt3x-sqrt2)=0`
Therefore,
`sqrt3x-sqrt2 = 0`
`sqrt3x=sqrt2`
`x=sqrt(2/3)`
or,
`sqrt3x-sqrt2 = 0`
`sqrt3x=sqrt2`
`x=sqrt(2/3)`
Hence, `x=sqrt(2/3)` or `x=sqrt(2/3)`
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