Example
The roots of $2x^2+3x-1=0$   are α and β.  Find the values of
(a) $(\alpha +1)(\beta +1)$
(b) $\frac{1}{\alpha }+\frac{1}{\beta }$
(c) ${\alpha }^2\beta +\alpha {\beta }^2$
(d) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

[Clue: ${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$   ]

2.6 Finding the Sum of Roots (SoR) and Product of Roots (PoR)

Example
Find the sums and products of the roots of the following equations.
a. $x^2+7x-<!--\; -\; -->3=0$
b. $x(x-<!--\; -\; -->1)=5(1-<!--\; -\; -->x)$

Answer:
(a)
$\begin{array}{l}{x}^2+7x-3=0\\ a=1,b=7,c=-3\\ \\ \text{Sum of Roots}\\ \alpha +\beta =-\frac{b}{a}=-\frac{7}{1}=-7\\ \text{Product of Roots}\\ \alpha \beta =\frac{c}{a}=\frac{-3}{1}=-3\end{array}$

(b)
$\begin{array}{l}x(x-1)=5(1-x)\\ x^2-x=5-5x\\ x^2-x+5x-5=0\\ x^2+4x-5=0\\ a=1,b=4,c=-5\\ \\ \text{Sum of Roots}\\ \alpha +\beta =-\frac{b}{a}=-\frac{4}{1}=-4\\ \\ \text{Product of Roots}\\ \alpha \beta =\frac{c}{a}=\frac{-5}{1}-5\end{array}$

The quadratic equation $a{x}^2+bx+c=0$ can be solved by using the quadratic formula

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Example 
Use the quadratic formula to find the solutions of the following equations.
a. $x^2+5x-<!--\; -\; -->24=0$
b. $x\left(x+4\right)=10$

Answer
(a)
For the equation $x^2+5x-<!--\; -\; -->24=0$
$\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(5)\pm \sqrt{{(5)}^2-4(1)(-24)}}{2(1)}\\ x=\frac{-5\pm \sqrt{121}}{2}\\ x=-8\text{ or }x=3\end{array}$

(b)
$\begin{array}{l}x\left(x+\text{ 4}\right)=\text{ 10}\\ {\text{x}}^2+4x-10=0\\ \\ a=1,b=4,c=-10\\ \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(4)\pm \sqrt{{(4)}^2-4(1)(-10)}}{2(1)}\\ x=\frac{-4\pm \sqrt{56}}{2}\\ x=1.742\text{ or }x=-5.742\end{array}$

Solving Quadratic Equations

To solve a quadratic equation means to find all the roots of the quadratic equations.

Example:
Find the roots of the quadratic equations
a. $x^2=9$
b. $2x^2-<!--\; -\; -->98=0$

Answer:
 (a)
$\begin{array}{l}{x}^2=9\\ x=\pm <!--\; \pm \; -->\sqrt{9}\\ x=\pm <!--\; \pm \; -->3\end{array}$

(b)
$\begin{array}{l}2x^2-<!--\; -\; -->98=0\\ 2x^2=98\\ x^2=\frac{98}{2}=49\\ x=\pm <!--\; \pm \; -->\sqrt{49}=\pm <!--\; \pm \; -->7\end{array}$

A quadratic equation may be solve by using one of the following method

1. Factorisation
2. Completing the square
3. Using quadratic formula

Remember : Roots of a quadratic equation are the values of variables/unknowns that satisfy the equation.

Example
(a) Given x = 3 is the root of the quadratic equation $x^2+2x+p=0$ , find the value of p.
(b) The roots of the quadratic equation $3x^2+hx+k=0$   are -2 and 4. Find the value of h and k.

General Form of Quadratic Equation

General form of a quadratic equation is
$a{x}^2+bx+c=0$
where a, b, and c are constants and a≠0.

*Note that the highest power of an unknown of a quadratic equation is 2.

Example (Find the values of a, b and c)
Rewrite the following into the general form of a quadratic equation. Find the values of a, b, and c.
(a) ${(3x-5)}^2=0$
(b) (x - 8)(x + 8) = 0

Solution

What is a Quadratic Equation?

1. A quadratic equation is a polynomial equation of the second degree. 
2. A quadratic equation has only one variable 
3. The highest power of the variable is 2.

Example of Quadratic Equation
The followings are some examples of quadratic equations

• $2x^2+3x+4=0$
• $t^2=\frac{2}{5}$
• $y(6y-<!--\; -\; -->3)=5$
• $a-<!--\; -\; -->3y+2y^2=2$ where a is a constant.