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प्रश्न
`16x^2+2ax+1`
उत्तर
Given:
`16x^2+24x+1`
⇒ `16x^2-24x-1=0`
On comparing it with `ax^2+bx+x=0`
a =16,b= -24 and c= -1
Discriminant D is given by:
`D=(b^2-4ac)`
=`(-24)^2-4xx16xx(-1)`
=`576+(64)`
=`640>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
`α=(-b+sqrt((D)))/(2a)=(-(-24)+sqrt(640))/(2xx16)=((24+8sqrt(10)))/(32)=(8(3+sqrt(10)))/32=((3+sqrt(10)))/4`
`β=(-b-sqrt((D)))/(2a)=(-(-24)-sqrt(640))/(2xx16)=((24-8sqrt(10)))/(32)=(8(3-sqrt(10)))/32=((3-sqrt(10)))/4`
Thus, the roots of the equation are `((3+sqrt(10)))/4` and `((3-sqrt(10)))/4`
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