If positive numbers #a^-1, b^-1, c^-1# are in AP, then the product of the roots of the equation #x^2-kx+2b^101-a^101-c^101=0#, (#k# is real) is greater than 0, less than 0 or equal to 0?

Question

5746 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-22T10:35:54

See below.

If #a^-1,b^-1,c^-1# are in AP then

#b^-1=a^-1+r# and #c^-1=a^-1+2r# or

#b = a/(ar+1)# and #c=a/(2ar+1)#

Now, regarding the the product of the roots of

#x^2-kx+2b^101-a^101-c^101=0# this product is given by

#2b^101-a^101-c^101# or

#2( a/(ar+1))^101-a^101-(a/(2ar+1))^101# or

#a^101(2(1/(ar+1))^101-(1/(2ar+1))^101-1)# but

#2(1/(ar+1))^101-(1/(2ar+1))^101 < 1# then

#2(1/(ar+1))^101-(1/(2ar+1))^101-1<0#

NOTE:

#2/(ar+1)-1/(2ar+1)=(3ar)/(3ar+1+2a^2r^2) < 1#