Maths Help (1 Viewer)

[Farhanthestudent005]

How do I complete these questions???

 

[B1andB2]

a) let the roots be alpha, beta, alpha+beta
then use sum of roots and stuff to solve simultaneously
b) let the roots be alpha, 1/alpha, beta
first use product of roots to find beta and then sum of roots etc and simultaneous

it should work out fine from there

 

[Eagle Mum]

I thought I’d set out the solution for Q1 as @B1 and B2 has described:

Given that the sum of two roots is the third root, let the roots be a, b, a+b
Then the factors of this cubic polynomial are (x-a)(x-b)(x-(a+b))

(x-a)(x-b)(x-(a+b)) = x^3 - 4x^2 + 2x + 4

Expanding the LHS:
[x^2 - (a+b)x + ab] [x-(a+b)] = x^3 -(a+b)x^2 +abx -(a+b)x^2 + (a+b)^2 x -ab(a+b)

= x^3 -2(a+b)x^2 + ((a+b)^2+ab)x -ab(a+b)

Matching the coefficients of the LHS and RHS:
#1: -2(a+b) = -4
#2: (a+b)^2+ab = 2

Dividing #1 by -2: a+b = 2
b = 2-a
Substituting in #2: 2^2 + a(2-a) = 2
4 + 2a -a^2 = 2
a^2 -2a -2 = 0
Solving this quadratic: a = (2 +/- 2sqrt3)/2 = 1 +/- sqrt(3)

Therefore, the three roots are 1+sqrt(3), 1-sqrt(3) and 2