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प्रश्न
`x^2-(sqrt3+1)x+sqrt3=0`
उत्तर
The given equation is `x^2-(sqrt3+1)x+sqrt3=0`
Comparing it with `ax^2+bx+c=0` we get
`a=1, b=-(sqrt3+1) and c=sqrt3`
∴ Discriminant,
`D=b^2-4ac=[-(sqrt3+1)]^2-4xx1xxsqrt3=3+1+2sqrt3-4sqrt3=3-2sqrt3+1=-2sqrt3+1=(sqrt3-1)^2>0`
So, the given equation has real roots.
Now, `sqrtD=sqrt(sqrt3-1)^2=sqrt3-1`
∴ `α =(-b+sqrt(D))/(2a)=(-[-sqrt(3)+1]+[sqrt(3)-1])/(2xx1)=(sqrt(3)+1+sqrt(3)-1)/2=(2sqrt(3))/2=sqrt3`
β=`(-b-sqrt(D))/(2a)=(-[-sqrt(3)+1]+[sqrt(3)-1])/(2xx1)=(sqrt(3)+1-sqrt(3)-1)/2=2/2=1`
Hence, `sqrt3` and `1` are the roots of the given equation.
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