Question

if the zeros of polynomial x cube minus 3 X square + X + 1 are a minus b, a a + b find the value of a and b

Answer 1

given,
polynomial p(x)=x^3-3x^2+x+1
let α,βand¥ be the zeroes of the polynomial p(x)
given,
a-b, a, a+b are the zeroes of the polynomial p(x)
I.e., α=a-b, β=a, ¥=a+b
we know that
α+β+¥=-coefficient of x^2/coefficient of x^3
a-b+a+a+b=-(-3)/1
3a=3
a=1 ----------- eq 1
α*β*¥=-constant/coefficient of x^3
(a-b)(a)(a+b)=-(1)/1
(a)(a-b)(a+b)=-1
$( x + y)(x - y) = {x}^{2} - {y}^{2}$
a(a^2-b^2)=-1
now substitute a=1 in above equation
1(1^2-b^2)=-1
1-b^2=-1
b^2=1+1
b^2=2
take square root on both sides
b=√2
therefore a=1,b=√2