Question

The roots of x^3+Px^2+Qx-19=0 are each one more than the roots of x^3-Ax^2+Bx-C=0. Find the value of A+B+C.

Answer 1

I know you are supposed to use Viete's equations. And I know how to set it up, but I just can't seem to solve - all I get are variables.

Answer 2

If no one can help with that one, then:
2) Determine two possible values for a so that one of the zeroes of \[P(x)=ax^2-30x+27\] will be the square of the other.

Answer 3

It is a bit tedious, but things cancel out nicely to give a single number.
if we call the roots of the first polynomial x^3+Px^2+Qx-19=0 a,b,c
then the roots of the 2nd polynomial x^3-Ax^2+Bx-C=0 are (a-1), (b-1), (c-1)

I assume you know Viete's equations (see http://en.wikipedia.org/wiki/Vieta's_formulas#Example )

for the first polynomial we get
a+b+c= -P
ab+bc+ac= Q
abc= 19

for the 2nd polynomial we get
(a-1) + (b-1) + (c-1) = A
(a-1)(b-1) + (a-1)(c-1) + (b-1)(c-1) = B
(a-1)(b-1)(c-1) = C

we can expand all of that to get
(a+b+c) - 3 = A
(ab+ac+bc) - 2(a+b+c) + 3 = B
abc - (ab+ac+bc) + (a+b+c)-1 = C

sub in the values from the first polynomial and add up to get A+B+C

Answer 4

If no one can help with that one, then: . . .
A Mathematica solution to the above is attached.

Answer 5

adding these up
(a+b+c) - 3 = A
(ab+ac+bc) - 2(a+b+c) + 3 = B
abc - (ab+ac+bc) + (a+b+c) -1 = C
--------------------------------
abc -1= A+B+C
and abc is 19
so A+B+C= 18