The roots of the eqn, x^4 + px^3 + qx^2 + rx + s = 0 where p, q, r, s are constants and s does not equal to 0, are a, b, c, d.
(i) a^2 + b^2 + c^2 + d^2 = p^2 -2q (in terms of p & q)
(ii) 1/a + 1/b + 1/c + 1/d = -r/s
(in terms of r & s)
(iii) using the above results, or otherwise, express a^3 + b^3 + c^3 + d^3 in terms of p, q & r.
This looks like modern algebra, not high school algebra. I'm not sure how much time I'll devote to this.
You should know that p is sigma-1(roots) and q is sigma-2(roots). If you know how to work with the symmetric polynomials then the result follows immediately for (i).
If you add the terms in (ii) you get -sigma-3(roots) / sigma-4(roots) and the result follows immediately
For the last part a^3 + b^3 + c^3 + d^3 =
(sigma-1)^3-3(sigma-1)(sigma-2)+3(sigma-3) all of them of the roots
p=-sigma-1, q=sigma-2, r=sigma-3 so (iii) is
(-p)^3-3(-p)q+3r=a^3+b^3+c^3+d^3
To solve this problem, we can use Vieta's formulas, which relate the coefficients of a polynomial equation to its roots.
Let's start by understanding the given equation: x^4 + px^3 + qx^2 + rx + s = 0. The roots of this equation are a, b, c, and d.
(i) To find a^2 + b^2 + c^2 + d^2, we need to express it in terms of p and q.
We know that the sum of the squares of the roots is related to the coefficients of the equation. This sum can be written as:
a^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd)
We can substitute the values of p and q into this expression, using Vieta's formulas:
a + b + c + d = -p
ab + ac + ad + bc + bd + cd = q
Therefore, substituting these values, we get:
a^2 + b^2 + c^2 + d^2 = (-p)^2 - 2q
= p^2 - 2q
So, the expression for a^2 + b^2 + c^2 + d^2 in terms of p and q is p^2 - 2q.
(ii) To find 1/a + 1/b + 1/c + 1/d, we need to express it in terms of r and s.
We know that the sum of the reciprocals of the roots is also related to the coefficients of the equation. This sum can be written as:
1/a + 1/b + 1/c + 1/d = (a + b + c + d)/(abcd)
Again, we can substitute the values of r and s into this expression:
a + b + c + d = -p
abcd = -s
Therefore, substituting these values, we get:
1/a + 1/b + 1/c + 1/d = (-p)/(-s)
= p/s
So, the expression for 1/a + 1/b + 1/c + 1/d in terms of r and s is p/s.
(iii) To find a^3 + b^3 + c^3 + d^3, we can use the identity: a^3 + b^3 + c^3 + d^3 = (a + b + c + d)(a^2 + b^2 + c^2 + d^2 - ab - ac - ad - bc - bd - cd) + 3(abc + abd + acd + bcd).
From part (i), we know that a^2 + b^2 + c^2 + d^2 = p^2 - 2q.
From part (ii), we know that 1/a + 1/b + 1/c + 1/d = p/s.
Substituting these values into the above identity, we get:
a^3 + b^3 + c^3 + d^3 = (-p)(p^2 - 2q - 3(ab + ac + ad + bc + bd + cd)) + 3(abc + abd + acd + bcd)
Expanding and simplifying this expression, we have:
a^3 + b^3 + c^3 + d^3 = -p^3 + 3pq + 3r
So, the expression for a^3 + b^3 + c^3 + d^3 in terms of p, q, and r is -p^3 + 3pq + 3r.