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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 3 1 2 + 9 size 12{3 { {1} over {2} } +9} {} as 12 1 2 size 12{"12" { {1} over {2} } } {} , or 12 1 2 size 12{ sqrt {"12" { {1} over {2} } } } {} as 5 2 size 12{ { {5} over { sqrt {2} } } } {} . It is always a good idea to simplify as you go along!

Step Example
The problem itself 2x 2 3x 7 = 9x size 12{2x rSup { size 8{2} } - 3x - 7=9x} {}
Put all the x size 12{x} {} terms on one side, and the number on the other 2x 2 12 x = 7 size 12{2x rSup { size 8{2} } - "12"x=7} {}
Divide both sides by the coefficient of x 2 size 12{x rSup { size 8{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 x size 12{x} {} , squared. (The coefficient of x 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 ̲ 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 size 12{ \( x - 3 \) rSup { size 8{2} } ="12" { {1} over {2} } } {}
Square root—but with a “plus or minus”! (*Remember, if x 2 size 12{x rSup { size 8{2} } } {} is 25, x size 12{x} {} may be 5 or -5 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 x size 12{x} {} x = 3 ± 5 2 size 12{x=3 +- { {5} over { sqrt {2} } } } {} ( . 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 size 12{ ital "ax" rSup { size 8{2} } + ital "bx"+c=0} {}

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!

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

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

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

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

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

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>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!

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Rewrite the left side as a perfect square.

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

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

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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!)

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

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

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

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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 ax 2 + 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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Source:  OpenStax, Advanced algebra ii: activities and homework. OpenStax CNX. Sep 15, 2009 Download for free at http://cnx.org/content/col10686/1.5
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