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The “Generic” Quadratic Equation

Module by: Kenny M. Felder. E-mail the author

Summary: Finding a general quadratic equation.

OK, let’s say I wanted to solve a quadratic equation by completing the square. Here are the steps I would take, illustrated on an example problem. (These steps are exactly the same for any problem that you want to solve by completing the square.)

Note that as I go along, I simplify things—for instance, rewriting 312+9312+9 size 12{3 { {1} over {2} } +9} {} as 12121212 size 12{"12" { {1} over {2} } } {}, or 12121212 size 12{ sqrt {"12" { {1} over {2} } } } {} as 5252 size 12{ { {5} over { sqrt {2} } } } {} . It is always a good idea to simplify as you go along!

Table 1
Step Example
The problem itself 2x 2 3x 7 = 9x 2x 2 3x 7 = 9x size 12{2x rSup { size 8{2} } - 3x - 7=9x} {}
Put all the xx size 12{x} {} terms on one side, and the number on the other 2x 2 12 x = 7 2x 2 12 x = 7 size 12{2x rSup { size 8{2} } - "12"x=7} {}
Divide both sides by the coefficient of x2x2 size 12{x rSup { size 8{2} } } {} x 2 6x = 3 1 / 2 x 2 6x = 3 1 / 2 size 12{x rSup { size 8{2} } - 6x=3 {1} slash {2} } {}
Add the same number to both side. What number? Half the coefficient of xx size 12{x} {}, squared. (The coefficient of xx size 12{x} {} is -6. Half of that is -3. So we add 9 to both sides.) x 2 6x + 9 ̲ = 3 1 2 + 9 ̲ x 2 6x + 9 ̲ = 3 1 2 + 9 ̲ size 12{x rSup { size 8{2} } - 6x {underline {+9}} =3 { {1} over {2} } {underline {+9}} } {}
Rewrite the left side as a perfect square ( x 3 ) 2 = 12 1 2 ( x 3 ) 2 = 12 1 2 size 12{ \( x - 3 \) rSup { size 8{2} } ="12" { {1} over {2} } } {}
Square root—but with a “plus or minus”! (*Remember, if x2x2 size 12{x rSup { size 8{2} } } {} is 25, xx size 12{x} {} may be 5 or -5 x 3 = ± 12 1 2 = ± 25 2 = ± 5 2 x 3 = ± 12 1 2 = ± 25 2 = ± 5 2 size 12{x - 3= +- sqrt {"12" { {1} over {2} } } = +- sqrt { { {"25"} over {2} } } = +- { {5} over { sqrt {2} } } } {}
Finally, add or subtract the number next to the xx size 12{x} {} x = 3 ± 5 2 x = 3 ± 5 2 size 12{x=3 +- { {5} over { sqrt {2} } } } {} ( . 5, 6 . 5 ) ( . 5, 6 . 5 ) size 12{ \( approx - "." 5, 6 "." 5 \) } {}

Now, you’re going to go through that same process, only you’re going to start with the “generic” quadratic equation:

ax 2 + bx + c = 0 ax 2 + bx + c = 0 size 12{ ital "ax" rSup { size 8{2} } + ital "bx"+c=0} {}
(1)

As you know, once we solve this equation, we will have a formula that can be used to solve any quadratic equation—since every quadratic equation is just a specific case of that one!

Walk through each step. Remember to simplify things as you go along!

Exercise 1

Put all the xx size 12{x} {} terms on one side, and the number on the other.

Exercise 2

Divide both sides by the coefficient of x2x2 size 12{x rSup { size 8{2} } } {}.

Exercise 3

Add the same number to both sides. What number? Half the coefficient of xx size 12{x} {}, squared.

  • What is the coefficient of xx size 12{x} {}?
  • What is ½ of that?
  • What is that squared?

OK, now add that to both sides of the equation.

Exercise 4

>This brings us to a “rational expressions moment”—on the right side of the equation you will be adding two fractions. Go ahead and add them!

Exercise 5

Rewrite the left side as a perfect square.

Exercise 6

Square root—but with a “plus or minus”! (*Remember, if x2=25x2=25 size 12{x rSup { size 8{2} } ="25"} {}, xx size 12{x} {} may be 5 or –5!)

Exercise 7

Finally, add or subtract the number next to the xx size 12{x} {}.

Did you get the good old quadratic formula? If not, go back and see what’s wrong. If you did, give it a try on these problems! (Don’t solve these by factoring or completing the square, solve them using the quadratic formula that you just derived!)

Exercise 8

4x 2 + 5x + 1 = 0 4x 2 + 5x + 1 = 0 size 12{4x rSup { size 8{2} } +5x+1=0} {}

Exercise 9

9x 2 + 12 x + 4 = 0 9x 2 + 12 x + 4 = 0 size 12{9x rSup { size 8{2} } +"12"x+4=0} {}

Exercise 10

2x 2 + 2x + 1 = 0 2x 2 + 2x + 1 = 0 size 12{2x rSup { size 8{2} } +2x+1=0} {}

Exercise 11

In general, a quadratic equation may have two real roots, or it may have one real root, or it may have no real roots. Based on the quadratic formula, and your experience with the previous three problems, how can you look at a quadratic equation ax2+bx+c=0ax2+bx+c=0 size 12{ ital "ax" rSup { size 8{2} } + ital "bx"+c=0} {} and tell what kind of roots it will have?

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