If #2016x^2+2017x+1=0# produces values #x_1 and x_2 and x_1>x_2#, then #63(x_1/x_2)#= ?

Question

722 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-14T06:42:12

#x_1 and x_2 # are two roots of #2016x^2+2017x+1=0# , So we can write

#x_1+x_2= -2017/2016......[1]#

and

#x_1*x_2=1/2016.....[2]#

Now #(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2#

#=>(x_1-x_2)^2=(- 2017/2016)^2-4/2016#

#=>(x_1-x_2)^2=( 2017^2-4xx2016)/2016^2=2015^2/2016^2#

#=>(x_1-x_2)=2015/2016......[3]#

Adding [1} and [3]

#2x_1=-2/2016#

#=>x_1=-1/2016#

So #x_2=-1#

Hence #63*(x_1/x_2)=63/2016=1/32#

If #2016x^2+2017x+1=0# produces values #​​x_1 and x_2 and x_1>x_2#, then #63(x_1/x_2)#= ?