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Question
`sqrt3x^2-2sqrt2x-2sqrt3=0`
Solution
The given equation is `sqrt3x^2-2sqrt2x-2sqrt3=0`
Comparing it with `ax^2+bx+c=0`
`a=sqrt3,b=-2sqrt2 and c=-2sqrt3`
∴ Discriminant,` D=b^2-4ac=(-2sqrt2)^2-4xxsqrt3xx(-2sqrt3)=8+24=32>0`
So, the given equation has real roots.
Now, `sqrtD=sqrt32=4sqrt2`
∴ α =`(-b+sqrt(D))/(2a)=(-(-2sqrt2)+4sqrt(2))/(2xxsqrt(3))=(6sqrt2)/(2sqrt3)=sqrt6`
β =`(-b+sqrt(D))/(2a)=(-(-2sqrt2)+4sqrt(2))/(2xxsqrt(3))=(-2sqrt(2))/(2sqrt(3))=-sqrt6/3`
Hence, `sqrt6` and `-sqrt(6)/3` are the root of the given equation.
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