If the zeroes of the polynomial x^3 - 3x^2+x+1 are a-b, a, a+b, find a and b (or) what is the value of a and b?

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Answer 1 · 2017-12-03T07:47:51

#a=-1# and #b=+-sqrt2#

As coefficient of highest power #x^3# is#1#, if three roots are #alpha,beta# and #gamma#, we have

#x^3-3x^2+x+1=(x-alpha)(x-beta)(x-gamma)#

= #x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x+alphabetagamma#

Now let us compare coefficients of similar powers on each side. First comparing sum of roots from coefficient of #x^2#, we have

#a-b+a+a+b=-3# i.e. #3a=-3# i.e. #a=-1#.

Also coefficients of #x# give us

#a(a-b)+a(a+b)+(a+b)(a-b)=1#

i.e. #a(a-b+a+b)+a^2-b^2=1# or #3a^2-b^2=1#

and as #a=-1#, #b^2=3xx(-1)^2-1=2# and #b=+-sqrt2#

Further, products of roots is #(a-b)a(a+b)=1#.

and as #a(a^2-b^2)=(-1)(1-2)=1#, this is true as #alphabetagamma=1#

Hence, #a=-1# and #b=+-sqrt2#

Note that the two values of #b# give the same set of roots.