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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.

$${b^2} - 4ac$$ , 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.

$$\frac{2x^2+2x-5 =0}{2}$$

$$x^2 +x -\frac{5}{2} = 0$$

$$x^2 + x + (\frac{1}{2})^2 = \frac{5}{2} +(\frac{1}{2})^2$$

$$(x + \frac{1}{2})^2 = \frac{11}{4}$$

$$\sqrt{(x + \frac{1}{2})^2} = \sqrt{ \frac{5}{2} +(\frac{1}{2})^2}$$

$$x = \frac{\sqrt{11}}{2} - \frac{1}{2}$$

$$x_1 = \frac{-1 +\sqrt{11}}{2}$$

$$x_2 = \frac{-1 -\sqrt{11}}{2}$$

$$x_1 = \frac{-b +\sqrt{b^2 -4ac}}{2a}$$

$$x_2 = \frac{-b +\sqrt{b^2 -4ac}}{2a}$$

a = 2 : b =2 : c = -5

$$x_1 = \frac{-2 +\sqrt{2^2 -4(2)(-5)}}{2(2)}$$

$$x_1 = \frac{-1+\sqrt{11}}{2}$$

$$x_2 = \frac{-2 -\sqrt{2^2 -4(2)(5)}}{2(2)}$$

$$x_2 = \frac{-1-\sqrt{11}}{2}$$