## Find sum and product of zeros of equations

In the previous post, Our IB Maths Tutors have discussed how to solve a quadratic polynomial using the Quadratic formula. Here I will tell you about different relationships based on the sum and product of quadratic polynomial, cubic polynomial, and bi-quadratic polynomials.

ax2 + bx + c = 0

Sum of the roots = −b/a

The product of the roots = c/a

If we know the sum and product of the roots/zeros of a quadratic polynomial, then we can find that polynomial using this formula

x2 − (sum of the roots)x + (product of the roots) = 0

## Cubic:

Now let us look at a Cubic (one degree higher than Quadratic):

ax3 + bx2 + cx + d=0

if α, β and γ are the zeros of this cubic polynomial then

α+β+γ=-b/a

αβ+βγ+γα=c/a

αβγ=-d/a
If we know these relationships of polynomials then we cal also calculate the polynomial using this formula:

x³-(α+β+γ)x²+(αβ+βγ+γα)x-(α+β+γ)=0

If we are given a bi-quadratic polynomial with degree 4 like:

ax 4+bx³+cx²+dx+e=0

and its roots/zeros are α, β, γ, and δ then

α+β+γ+δ=-b/a

αβ+αγ+αδ+βγ+βδ+γδ=c/a

αβγ+βγδ+γδα+αβδ=-d/a

αβγδ=e/a
using these formulas of sum and product of zeros of polynomials, we can find a lot of relationships in zeros of polynomials. Usually, we are asked to find these types of relationships in zeros.
Question: If α and β are the zeros of polynomial x²-px+q=0 then find the following relationships.
i) 1/α+1/β   ii) α²β+αβ²   iii) α²+β²    iv) α/β+β/α   v) α³+β³
Ans: To find these relationships we convert every value either in sum (α+β)  or in the product (αβ) of zeros. For this conversion, we use the following Mathematical Tricks
1)Try to take common
2) Try to take L.C.M
3) Try to make a Perfect Square
4) Use algebraic identities wherever required
If we use the above steps properly, in most of the cases we are able to convert everything either in sum or in the product of zeros
If we compare the given equation with std. form ax²+bx+c=0 then
a=1, b=-p and c=q
sum of zeros α+β=-b/a=-(-p/a)=p

product of zeros    αβ=c/a=q/1=q                                                     (i) 1/α+1/β=β+α/αβ              [By L.C.M]

=p/q
(ii) α²β+αβ²  = αβ(α+β)               [By common]
=q.p
(iii) α²+β² = (α²+β²+2αβ)-2αβ          [add and subtract 2αβ, make it a perfect square ]

=(α+β)²-2αβ

=p²-2q
(iv) α/β+β/α = β²+α²/βα          [By taking L.C.M]

we have already found β²+α² that is p²-2q so
α/β+β/α=(p²-2q)/q

(v) (α+β)³=(α+β)³-3αβ(α+β)                [Direct algebraic identity]

=p³-3qp