# Vieta’s Formula in Quadratic Equations

Given a quadratic equation $f(x)=ax^2+bx+c$, let and q be the roots of this equation then, we can relate the sum of the roots as well as the product of the roots by the following formula. This is called Viete’s formula.

$p+q=\displaystyle\frac{-b}{a}$
$pq=\displaystyle\frac{c}{a}$

The following formula was discovered by François Viète. The formula is not just for quadratic but for other polynomial equations with higher degree but we will just deal with quadratic equations for now.

Worked Problem 1:
Let p and q be the roots of equation $2x^2-x+4=0$, what is the value of $\displaystyle\frac{2}{q} +\displaystyle\frac{2}{p}$

Solution:
Using Viete’s formula, we know that

$p+q=\displaystyle\frac{1}{2}$
$pq= 2$

We are asked to find the value of the following

$\displaystyle\frac{2}{q} +\displaystyle\frac{2}{p}$
$=2(\displaystyle\frac{1}{q} +\displaystyle\frac{1}{p})$ by common factor
$=2(\displaystyle\frac{p+q}{pq})$
$=2(\displaystyle\frac{\frac{1}{2}}{2})$
$=\displaystyle\frac{1}{2}$

Worked Problem 2:

Let a and b be the roots of equation $x^2-3x+1$. Find the value of $a^3+b^3$.

Solution:
By Viete’s formula,

$a+b=3$ and $ab=1$

We are asked to find the value of $a^3+b^3$

$a^3+b^3=(a+b)(a^2-ab+b^2)$ by factoring special product.

We already have (a+b) we just to find $a^2+b^2$

$(a+b)^2=a^2+2ab+b^2$
$a^2+b^2=(a+b)^2-2ab$

Going back here

$a^3+b^3=(a+b)((a+b)^2-2ab-ab)$ since $a^2+b^2=(a+b)^2-2ab$
$a^3+b^3=(a+b)((a+b)^2-3ab)$
$a^3+b^3=(3)(3^2-3(1))$
$a^3+b^3=18$

Worked Problem 3:

Find the value of K such that the sum of the roots of quadratic $kx^2+(k+1)x+1=0$ is -2.

Solution:
Let p and q are the roots of equation.
By Vieta’s formula,

$p+q=\displaystyle\frac{-b}{a}$
$-2=\displaystyle\frac{(k+1}{-k}$
$2k=k+1$
$k=1$

### 2 Responses

1. Anonymous says:

sir dan wala ba sa cubic , quartic or even a general Vieta’s formula for all types of degree 🙂

2. Dan Lang says:

We will discuss that shortly.