Math: Finding the Roots of a Quadratic Equation

The quadratic equation, ax² + bx + c = 0, is a non-linear (2^nd degree polynomial, 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 ax² + bx + c = 0, the discriminant is defined as :
∇= b² - 4ac
The constant numbers a and b are the numbers (value) that represent the sum of the variables x² and x, respectively. And the c is the constant number of a quadratic equation.

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 :
2x² + 7x + 5 = 0 has a = 2, b = 7 and c = 5.
So the discriminant of that thingy is

∇ = b² - 4ac
∇ = (7)² - 4 * 2 * 5
∇ = 49 - 40
∇ = 9 ( ∇ > 0 : this quadratic equation crosses the x-axis, it has two real numbers roots)

The discriminant values :

∇ > 0 : the equation has two real numbers roots, the function graph crosses the x-axis
∇ < 0 : the equation has two complex numbers roots, the function graph 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

As you can see on the graph above, it plots the y = x² - x - 2 function. The a constant of that function, the sum of x², is 1. Therefore, a = 1 or a > 0. It means the graph shape will have a valley, also means, it has a minimum value.

On this quadratic non-linear equation (function), we can't have the a as 0. Why? Because, that's why.
OK, let's take a look : ax² + bx + c = 0, if we substitute the a with 0, then it will become bx + c = 0.
The 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 :
2x² + 7x + 5 = 0 ► with a = 2, b = 7 and c = 5.

Let's solve that using quadratic formulation :
x = [ -7 ± √( [7]² - [4 * 2 * 5] ) ] / 2 * 2
x = [ -7 ± √(49 - 40) ] / 4

x = [ -7 ± √9 ] / 4

x = [ -7 ± 3 ] / 4

x₁ = [ -7 + 3 ] / 4 ∨ x₂ = [ -7 - 3 ] / 4

x₁ = [ -4 ] /4 ∨ x₂ = [ -10 ] / 4

x₁ = -1 ∨ x₂ = -(5/2)

∴ The roots of 2x² + 7x + 5 = 0 are x₁ = -1 or x₂ = -(5/2), simplified as x = {-(5/2) , -1}

I'll show you another one, the factoring method.

Factoring

solving quadratic equation with factorization

Let's take an example from the previous one : 2x² + 7x + 5 = 0 with a = 2, b = 7 and c = 5.

example 1 solving quadratic equation with factorization

Continue
So we have 2 and 5 there, we put this general formulation to solve that :

quadratic equation factorization formula

Keep in mind that the equation is 2x² + 7x + 5 = 0 has a = 2.
And also, we have already got value1 = 2, and 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
x₁ = 0 - 1 or x₂ = (0 - 5)/2
x₁ = -1 or x₂ = -(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 of a*c, that is (-5)*5 = -25, and if added will give the same value of b, that is 24.
From the image above we find that the value1 = (-1) and 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
x₁ = (0 + 1)/(-5) or x₂ = 0 + 5
x₁ = -(1/5) or x₂ = 5
The solution, x = {-(1/5) , 5}

Other forms "shortcuts" :

ax² + bx = 0 ► x (ax + b) = 0 ► x₁ = 0 ∨ x₂ = -(b/a)
ax² - bx = 0 ► x (ax - b) = 0 ► x₁ = 0 ∨ x₂ = b/a
-ax² - bx = 0 ► (-x)(ax + b) = 0 ► x₁ = 0 ∨ x₂ = -(b/a)
-ax² + bx = 0 ► x (-ax + b) = 0 ► x₁ = 0 ∨ x₂ = 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

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Second Problem : x² + 8x + 7 = 0

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Third Problem : 3x² + 5x + 2 = 0

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

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Fourth Problem : -2x² + 8x = 0

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Last Problem : x² + 169 = 0

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

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Have fun... I hope.

Monkey Raptor

Monkey Raptor

Thursday, November 21, 2013