Question

If roots of `6x^2+3x+2=0` are `a` and `b`, what is the value of `(1-a)/(1+a)+(1-b)/(1+b)`?

Answer 1

`(1-a)/(1+a)+(1-b)/(1+b)=8/5=1 3/5`

Explanation:

In an equation `px^2+qx+r=0`, if its roots are `a` and `b`

then `a+b=-q/p` and `ab=r/p`

As roots of `6x^2+3x+2=0` are `a` and `b`, we have

`a+b=-3/6=-1/2` and `ab=2/6=1/3`

Hence `(1-a)/(1+a)+(1-b)/(1+b)`

= `((1-a)(1+b)+(1-b)(1+a))/((1+a)(1+b))`

= `(1-a+b-ab+1-b+a-ab)/(1+a+b+ab)`

= `(2-2ab)/(1+a+b+ab)`

= `(2-2/3)/(1-1/2+1/3)`

= `((6-2)/3)/((6-3+2)/6)`

= `(4/3)/(5/6)`

= `4/3xx6/5=8/5=1 3/5`