ISSUE

Bibliography

01

MONTHLY JOURNAL OF

MONTH NOV 4 YEAR 2013

MATH

ry

Math Times Issue 01 Month 11 Year 2013

technology c ons ulting

Babylonian Mathematicians, as early as 2000 BC (displayed on Old Babylonian Clay Tablets) could solve problems relating the areas and sides of rectangles. By 1545 Gerolamo Cardano compiled the works related to the quadratic equations. The quadratic formula covering all cases was first obtained by Simon Stevin in 1594. In 1637 Rene Descartes published La Geometrie containing the quadratic formula in the form we know today. The first appearance of the general solution in the modern mathematical literature appeared in a 1896 paper by Henry Heaton.

A quadratic equation is defined as an equation in which one or more of the terms is squared but raised to no higher power. The general form is ax2 + bx + c = 0, where a, b and c are constants. We learned how to solve linear - quadratic systems algebraically and graphically. With our new found knowledge of quadratics, we are now ready to attack problems that cannot be solved by factoring, and problems with no real solutions.

A Linear Equation is an equation for a line.

A Quadratic Equation is the equation of a parabola and has at least one variable squared

(such as x2)

And together they form a System of a Linear and Quadratic Equation

(such as x2)

this issue Quadratic Linear Equations Process Solutions Bibliography

How to Solve using Algebra

A System of those two equations can be solved (find where they intersect), either:

Graphically

    

Make both equations into "y =" format Simplify into "= 0" format (like a standard Quadratic Equation)

There are three possible cases:

Solve the Quadratic Equation! Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers

No real solution (happens when they never intersect)

One real solution (when the straight line just touches the quadratic)

Two real solutions (like the example above)

An example will help: Example: Solve these two equations:

Solutions

Set them equal to each other

 

or using Algebra

2

y = x - 5x + 7 y = 2x + 1

Make both equations into "y=" format: They are both in "y=" format, so go straight to next step Set them equal to each other 2

x - 5x + 7 = 2x + 1

Simplify into "= 0" format (like a standard Quadratic Equation) 2

Subtract 2x from both sides: x - 7x + 7 = 1 2

Subtract 1 from both sides: x - 7x + 6 = 0

2

Rewrite -7x as -x-6x: x - x - 6x + 6 = 0 Then: x(x-1) - 6(x-1) = 0 Then: (x-1)(x-6) = 0 Which gives us the solutions x=1 and x=6 Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers The matching y values are (also see Graph):

  

for x=1: y = 2x+1 = 3 for x=6: y = 2x+1 = 13 Our solution: the two points are (1,3) and (6,13)

I think of it as three stages:

Combine into Quadratic Equation ⇒ Solve the Quadratic ⇒ Calculate the points