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Question
Solve `sqrt3x^2-2sqrt2x-2sqrt3=0`
Solution
`sqrt3x^2-2sqrt2x-2sqrt3=0`
⇒`sqrt3^2-3sqrt2x+sqrt2x-2sqrt3=0`
⇒`sqrt3x(x-sqrt6)+sqrt2 (x-sqrt6)=0`
⇒`(x-sqrt6) (sqrt3+sqrt2)=0`
⇒`x-sqrt6=0 or sqrt3x+sqrt2=0`
⇒`x=sqrt6 or x=-sqrt2/sqrt3=-sqrt6/3`
Hence,` sqrt6 and -sqrt6/3` are the roots of the given equation.
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Then we can take another root as any number like a positive or negative number or zero. Here I am taking another root of the quadratic equation as 2.
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