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प्रश्न
`1/x-1/(x-2)=3,x≠0,2`
उत्तर
The given equation is
`1/x-1/x-2=3,x≠0,2`
⇒ `(x-2-x)/(x(x-2))=3`
⇒`-2/(x^2-2x)=3`
⇒`-2=3x^2-6x`
⇒`3x^2-6x+2=0`
This equation is of the form `ax^2+bx+c=0` Where `a=3``, b=-6` and `c=2`
∴Discriminant, `D=b^2-4ac=(-6)^2-4xx3xx2=36-24=12>0`
So, the given equation has real roots.
Now, `sqrtD=sqrt12=2sqrt3`
∴ `α=(-b+sqrt(D))/(2a)=(-(-6)+2sqrt(3))/(2xx3)=(6+2sqrt(3))/6=(3+sqrt(3))/3`
`β= (-b-sqrt(D))/(2a)=(-(-6)-2sqrt(3))/(2xx3)=(6-2sqrt(3))/6=(3-sqrt(3))/3`
Hence, `(3+sqrt3)/3` and`(3-sqrt3)/3` are the roots of the given equation.
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