Question

`16x^2+2ax+1`

Answer 1

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`