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प्रश्न
`x^2-4x-1=0`
उत्तर
Given:
`x^2-4x-1-=0`
On comparing it with `ax^2+bx+c=0`
`a=1,b=-4 and c=-1`
Discriminant D is given by:
`D=(b^2-4ac)`
=`(-4)^2-4xx1xx(-1)`
=`16+4`
=`20`
=`20>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
α=`(-bsqrt(D))/(2a)=(-(-4)+sqrt(20))/(2xx1)=(4+2sqrt(5))/2=(2(2+sqrt(5)))/2=(2+sqrt(5))`
β=`(-b-sqrt(D))/(2a)=(-(-4)-20)/2=(4-2sqrt5)/2=(2(2-sqrt(5)))/2=(2-sqrt(5))`
Thus, the roots of the equation are` (2+sqrt(5)) and (2-sqrt(5))`
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