Is there a way around the Quadratic Formula

1
Is there a way around the Quadratic Formula?

• How to solve a quadratic equation without the use
  of the
• Quadratic Formula

2
The possibilities available to us for solving a
quadratic equation

• Graphing
• Factoring
• Substitution
• Completing the Square
• and of course the Quadratic Formula (which we are
  trying not to use!)

3
Graphing

• The first way to solve a quadratic equation is by
  graphing the equation
• (for middle/high school students this is very
  helpful as it gives them a good visual
  representation of what they are working with)
• Be careful! This is not always the most accurate
  method and can lead to answers that are not as
  close as you may need them to be
• Also, keep in mind that when we are graphing we
  are using two variables, typically a y and an x
  and when finding the solutions of the quadratic
  equation we are letting y 0 (this gives us the
  solution points where the graph intercepts the
  x-axis)
• In general, the graph will let the someone know
  if the quadratic equation has one, two or no
  solutions (providing that the graph is relatively
  accurate) with just a quick sketch from a small
  T-chart of values and give approximate values for
  the solution points
• Most graphing calculators can also achieve this
  process
• with the TI calculators a very good
  approximation can be found using the 2nd trace
  button (calc) and then 2 on the list (finds the
  zeros of the equation)

4
Factoring

• This method is not always going to work and is
  somewhat complicated when the equation involves
  large numbers in the coefficient and constant
  terms
  (y ax2 bx c where a, b and c are
  numbers or constants)
• There are many formulas for solving the special
  cases of quadratic equations
• (i.e. y ax2 b ? y (vax vb)(vax vb)
  where a and b are perfect squares)
• Solving a quadratic by factoring involves several
  steps, an understanding of how to combine the
  numbers (a, b and c) and lots of practice

5
Substitution

• This is the method that Dr. Jim so nicely
  explained how to do last week, it goes something
  like this

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• Lets begin with the equation
• 4x2 4x - 24 0
• Next we need to find the value for
• where b -4 and a 4
• Simplifying gives us
• and so we have the following x (y ) which
  we
• will substitute into our original equation as
  follows
• 4(y )2 4(y ) 24 0
• We will now need to use some algebra to simplify
• this equation as follows

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• 4(y )2 - 4(y ) - 24 0
• 4(y2 y ) - 4(y ) - 24 0 squaring the
  first set of ( )
• 4y2 4y 1 - 4y - 2 - 24 0
• using the distributive property and simplifying
• 4y2 4y 1 - 4y - 2 24 0 cancel like terms
• 4y2 1 - 2 - 24 0 combine like terms
• 4y2 - 25 0
• Next solve for y
• 4y2 25 add 25 to both sides
• y2 divide both sides by 4
• y take the square root of both sides
• and finally you get y which we can
  substitute
• back into our equation x ( ) and now we
• will have our two solutions for the value of x
• which are x 3 or x
  -2

8
Completing the Square

• This is another method that can be used to solve
  a quadratic equation without having to use the
  Quadratic Formula
• This method will work for all quadratic equations
  and can also be used when finding the equations
  of the other conics such as the circle, ellipse
  and hyperbola
• This is a very powerful method that helps you to
  find the vertex as well as the distance to the
  directrix and focus

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How it works!

• Ex. y x2 6x - 1
• y (x2 6x __) 5 - __
• we need to find a number that will make the part
  of the equation in the ( ) a perfect square
• this is easier to do than you might think!
• to do this we need only take the b term (-6)
  divide it by 2 and then square that result
• this value will then be placed in the __ of the
  equation (by both adding inside the ( ) and
  subtracting outside of the ( ) we are
  essentially adding zero and not changing the
  value of the initial equation)
• y (x2 - 6x 9) - 1 - 9
• y (x - 3)2 - 10 (after simplifying, it just
  so happens that the number that you
• are going to put in the ( ) with the x
  is the result when you divided the previous b
  term by 2)
• This tells us that the vertex of the parabola is
  (3,-10), the focus is at (3, -3.75) and the
  directrix is y -10.25
• the focus can be found by the following (h, k
  1/(4a)) where a is the value in front of the __
  (x 3) piece of the equation and the directrix
  can be found using the equation y k - 1/(4a)

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Completing the Square a Visual

• Starting with our previous equation
• y x2 - 6x - 1
• and letting y 0 then subtracting 1 from
  both sides gives us x2 - 6x 1
• We can then divide the -6x rectangle by 2
  and get two rectangles with areas -3x
• we then need to find the area of
  the small square that is formed
• This new missing square will
• have an area
• Now we can rewrite the equation
  with the new information
• after adding 9 (as found from the
  missing square) to both sides
• x2 - 6x 9 1 9 x2
  - 6x 9 10
• (x 3)2 10

x
-6
x
x2
-6x
x
-3
-3
9
-3
-3x
-3
-3x
x
x2
11

• We can now see how the original equation
  becomes the new equation after the
  transformations
• (x 3)2 10
• This is the positive version of the visual
  solution for the equation only!

x
-3
-3
-3x
-3x
x
x2
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The Quadratic Formula

• Of course there is still the Quadratic Formula
• y ax2 bx c
• so x
• I think you know how this goes from here

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In conclusion

• While there is still a little bit more to explain
  about the completing the square method, you have
  seen the basics
• There are several methods for solving quadratic
  equations that dont involve using the quadratic
  formula

14
Questions

• While answering this question I thought Are
  there still other methods of solving quadratic
  equations that I have not seen?

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Connections

• I think that I will definitely use this
  presentation in my high school classes as an
  introduction to quadratic equations.
• While some of these methods can be used in the
  middle school curriculum, I think that some of
  these methods may be a little bit more difficult
  to implement unless your students are ready for
  them.

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I picked this question because

• I thought I might give others a better view of
  the methods that can be used to solve a quadratic
  equation.