Quadratic Functions - A function that can be expressed in the form \(f(x)=a{x^2} + bx + c \)
If f(x) = 0, solutions (roots) of the function can be solved by:
The exponent (n) of the polynomial function \(f(x)=0=a{x^n} + b{x^{n-1}} + ...\) denotes the number of solutions. Thus, for a quadratic equation \(0=a{x^2} + bx + c \), it has 2 solutions or roots.
We can check if there is a solution or roots on a given quadratic equation by its discriminant.
The discriminant of the quadratic
\({b^2} - 4ac \) , is called the discriminant
If \({b^2} - 4ac = 0\) , there is one real solution
If \({b^2} - 4ac < 0\) , there are no real solutions
If \({b^2} - 4ac > 0\) , there are two real solutions
Example:
1. What is the solution of the equation 2x2 + 2x - 5 = 0?
Solution 01: By Factoring
We will factor this using the principle of completing the square.
\(\frac{2x^2+2x-5 =0}{2}\)
\(x^2 +x -\frac{5}{2} = 0\)
\(x^2 + x + (\frac{1}{2})^2 = \frac{5}{2} +(\frac{1}{2})^2\)
\((x + \frac{1}{2})^2 = \frac{11}{4}\)
\(\sqrt{(x + \frac{1}{2})^2} = \sqrt{ \frac{5}{2} +(\frac{1}{2})^2}\)
\(x = \frac{\sqrt{11}}{2} - \frac{1}{2}\)
\(x_1 = \frac{-1 +\sqrt{11}}{2}\)
\(x_2 = \frac{-1 -\sqrt{11}}{2}\)
Solution 02: By Quadratic Formula
\(x_1 = \frac{-b +\sqrt{b^2 -4ac}}{2a}\)
\(x_2 = \frac{-b +\sqrt{b^2 -4ac}}{2a}\)
a = 2 : b =2 : c = -5
\(x_1 = \frac{-2 +\sqrt{2^2 -4(2)(-5)}}{2(2)}\)
\(x_1 = \frac{-1+\sqrt{11}}{2}\)
\(x_2 = \frac{-2 -\sqrt{2^2 -4(2)(5)}}{2(2)}\)
\(x_2 = \frac{-1-\sqrt{11}}{2}\)