## Find Sum And Product of Zeros of Equations

In the previous post, our IB Maths Tutors 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 polynomials, cubic polynomials, 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 see 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 correctly, we can usually convert everything either to a sum or to a product of zeros.
If we compare the given equation with the 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