*ax² + bx + c = 0*, is a non-linear (

*2*degree polynomial,

^{nd}*a*≠ 0) equation that

**always**has

*two roots*as the solution. Sometimes the roots are different, sometimes they're twins. Sometimes they all have

*real*numbers or

*complex*numbers, or just

*imaginary*number.

Methods

To find the roots (the solution), we can choose between these four different methods :

- Factoring
- Completing the square
- Using quadratic formulation
- Drawing the graph : this can be achieved by writing your own loop-n-plot computer script, or using pre-programmed Mathematical software, or just using a pencil and a paper then create a table of values, continued with plotting the values on the Cartesian plane (2D x-y axis).

A glimpse about the function

*f(x)*graph

Discriminant (∇)

This is the expression that will tell us if a quadratic function crosses an axis. Because this is

*f(x)*(the function of

*x*values axis), then the discriminant here will tell us whether a function crosses the

*x-axis*, touching the

*x-axis*on one point, or floating above/below the

*x-axis*.

For the quadratic equation

**, the discriminant is defined as :**

*ax² + bx + c = 0*∇

*= b² - 4ac*

The constant numbers

**and**

*a***are the numbers (value) that represent the sum of the variables**

*b***and**

*x²***, respectively. And the**

*x***is the constant number of a quadratic equation.**

*c*Also, with this discriminant expression, we can find out if a quadratic function graph (or the equation) has two real numbers roots, two complex numbers (or just imaginary) roots, or twin real numbers roots.

For example :

**has**

*2x² + 7x + 5 = 0***,**

*a = 2**and*

**b = 7****.**

*c = 5*So the discriminant of that thingy is

∇ = b² - 4ac

∇ = (7)² - 4 * 2 * 5

∇ = 49 - 40

∇ = 9(∇ > 0: this quadratic equationcrosses the x-axis, it hastwo real numbers roots)

The discriminant values :

**∇ > 0**: the equation has**two**, the function graph*real numbers*roots*crosses*the x-axis**∇ < 0**: the equation has**two**, the function graph*complex numbers*roots*doesn't touch*the x-axis.**∇ = 0**: the equation has twins real numbers roots, it*touches*the x-axis*on exactly one point*.

The

*a*constant

**a > 0***function. The*

**y = x² - x - 2****constant of that function, the sum of**

*a**, is*

**x²***. Therefore,*

**1***or*

**a = 1****. It means the graph shape will have a valley, also means, it has a**

*a*> 0*minimum*value.

On this quadratic

*non-linear*equation (function), we can't have the

*as*

**a***. Why? Because, that's why.*

**0**OK, let's take a look :

**, if we substitute the**

*a*x² +*b*x +*c*= 0*with*

**a***, then it will become*

**0****.**

*b*x +*c*= 0The last equation loses its manhood, that is its quadratic thingy, it becomes

*a linear equation.*

So there, because.

The values of

**a****a****> 0**: the graph has a*valley*(pointing downward - opening up) with a*minimum*value**a****< 0**: the graph has a*peak*(pointing upward - opening down) with a*maximum*value

Methods to find the roots

The general formulation to get the roots (the solution) of a quadratic thingy is using :

For example, find the roots of this equation :

Let's solve that using quadratic formulation :

The general formulation to get the roots (the solution) of a quadratic thingy is using :

For example, find the roots of this equation :

**► with***2x² + 7x + 5 = 0***,***a = 2**and***b = 7****.***c = 5*Let's solve that using quadratic formulation :

*x = [ -7 ± √( [7]² - [4 * 2 * 5] ) ] / 2 * 2*

x = [ -7 ± √(49 - 40) ] / 4x = [ -7 ± √(49 - 40) ] / 4

*x = [ -7 ± √9 ] / 4*

*x = [ -7 ± 3 ] / 4*

*x*

_{1}= [ -7 + 3 ] / 4 ∨ x_{2}= [ -7 - 3 ] / 4*x*

_{1}= [ -4 ] /4 ∨ x_{2}= [ -10 ] / 4

**x**_{1}= -1 ∨ x_{2}= -(5/2)
∴ The roots of

**are***2x² + 7x + 5 = 0***or***x*_{1}= -1**, simplified as***x*_{2}= -(5/2)*x = {-(5/2) , -1}*I'll show you another one, the

*factoring*method.

Factoring

Let's take an example from the previous one :

**with**

*2x² + 7x + 5 = 0***,**

*a = 2**and*

**b = 7****.**

*c = 5*Continue

So we have

*and*

**2***there, we put this general formulation to solve that :*

**5**Keep in mind that the equation is

**has**

*2x² + 7x + 5 = 0**a = 2*.

And also, we have already got

*, and*

**value1 = 2****value2 = 5**.

*Substitute*those variables with the numbers we have :

Then we can simplify that as :

*(x + 1)(2x + 5)*

We put it back to the complete equation, so :

2x² + 7x + 5 = 0

(x + 1)(2x + 5) = 0

x_{1}= 0 - 1 or x_{2}= (0 - 5)/2

xor_{1}= -1x_{2}= -(5/2)

*The solution, x = {-(5/2) , -1}**

**the same result compared with using quadratic formulation earlier*

Anotha example :

Let's find the roots of this quadratic equation

*-5x² + 24x + 5 = 0*

*1st step*

Find two numbers which if multiplied will have the same value ofa*c, that is(-5)*5=-25, and if added will give the same value ofb, that is24.From the image above we find that theandvalue1 = (-1)value2 = 25.

*2nd step*

Put those values into the basic factorization formula :

So :

Simplify that into :

(-5x - 1)(x - 5)

*Last step*

Put that back in to the original complete equation :

-5x² + 24x + 5 = 0

(-5x - 1)(x - 5) = 0

x_{1}= (0 + 1)/(-5) or x_{2}= 0 + 5

xor_{1}= -(1/5)x_{2}= 5The solution, x = {-(1/5) , 5}

Other forms

*"shortcuts"*:

*ax² + bx = 0 ► x (ax + b) = 0 ►***x**_{1}= 0 ∨ x_{2}= -(b/a)*ax² - bx = 0 ► x (ax - b) = 0 ►***x**_{1}= 0 ∨ x_{2}= b/a*-ax² - bx = 0 ► (-x)(ax + b) = 0 ►***x**_{1}= 0 ∨ x_{2}= -(b/a)*-ax² + bx = 0 ► x (-ax + b) = 0 ►***x**_{1}= 0 ∨ x_{2}= b/a*ax² - c = 0 ► ( √a·x + √c )( √a·x - √c ) = 0 ►***x = ±( √c / √a )***-ax² + c = 0 ► ( √a·x + √c )( -√a·x + √c ) = 0 ►***x = ±( √c / √a )**- etc

Try out these :

###
First Problem : *x² - 81 = 0*

☺
###
Second Problem : *x² + 8x + 7 = 0*

☺
###
Third Problem : *3x² + 5x + 2 = 0*

#####
*hint: type fraction like -1/2 or 2/7 etc*

☺
###
Fourth Problem : *-2x² + 8x = 0 *

☺
###
Last Problem : *x² + 169 = 0*

#####
*hint: use i for imaginary number, for instance 2 - 3i or 7i*

☻
Have fun... I hope.

## No comments

## Post a Comment