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प्रश्न
`x-1/x=3,x≠0`
उत्तर
The given equation is
`x-1/x=3,x≠0`
⇒`(x^2-1)/x=3`
⇒`x^2-1=3x`
⇒`x^2-3x-1=0`
This equation is of the form `ax^2+bx+c=0` where` a=1,b=-3 and c=-1`
∴ Discriminant, `D=b^2-4ac=(-3)^2-4xx1xx(-1)=9+4=13>0`
So, the given equation has real roots.
Now, `sqrtD=sqrt13`
∴ `α=(-b+sqrt(D))/(2a)=(-(-3)+sqrt(13))/(2xx1)=(3+sqrt(13))/2`
β=`(-b+sqrt(D))/(2a)=(-(-3)+sqrt(13))/(2xx1)=(3+sqrt(13))/2=(3-sqrt(13))/2`
Hence, `(3+sqrt(13))/3` and `(3-sqrt(13))/3`are the roots of the given equation.
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