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This module teaches the method of solving quadratic equations by factoring.

Consider the equation 4x 2 + 14 x 60 = 0 size 12{4x rSup { size 8{2} } +"14"x - "60"=0} {} . This is not an algebraic generalization, but an equation to be solved for x : that is, it asks the question “What x value, or values, will make this equation true?” We will be solving such equations in three different ways. The fastest and easiest is by factoring.

Using the techniques discussed above, we can rewrite this problem as follows. (Try it for yourself!)

4x 2 + 14 x 60 = 0 size 12{4x rSup { size 8{2} } +"14"x - "60"=0} {} Original form

2 2x 5 x + 6 = 0 size 12{2 left (2x - 5 right ) left (x+6=0 right )} {} Factored form

The second form may look more complicated than what we started with. But consider what this equation says. There are three numbers: 2, 2x 5 size 12{2x - 5} {} , and x + 6 size 12{x+6} {} . The equation says that when you multiply these three numbers, you get 0. Ask yourself this crucial question: How can you multiply numbers and get the answer 0 ?

The only way it can happen is if one of the numbers is 0. Take a moment to convince yourself of this: if several numbers multiply to give 0, one of those numbers must be 0.

So we have three possibilities.

2 = 0 size 12{2=0} {} 2x 5 = 0 size 12{2x - 5=0} {} x + 6 = 0 size 12{x+6=0} {}
( it just isn't ) size 12{ \( "it just isn't" \) } {} x = 2 1 2 size 12{x=2 { { size 8{1} } over { size 8{2} } } } {} x = 6 size 12{x= - 6} {}

The moral of the story is: when a quadratic equation is factored, it can be solved easily. In this case, the equation 4x 2 + 14 x 60 = 0 size 12{4x rSup { size 8{2} } +"14"x - "60"=0} {} has two valid solutions, x = 2 1 2 size 12{x=2 { { size 8{1} } over { size 8{2} } } } {} and x = 6 size 12{x= - 6} {} .

Consider this example:

x 2 9x + 20 = 6 size 12{x rSup { size 8{2} } - 9x+"20"=6} {}

A common mistake is to solve it like this.

x 2 9x + 20 = 6 size 12{x rSup { size 8{2} } - 9x+"20"=6} {} , solved incorrectly

  • ( x 4 ) ( x 5 ) = 6 size 12{ \( x - 4 \) \( x - 5 \) =6} {}
  • x 4 = 6 size 12{ left (x - 4 right )=6} {}
    • x = 10  ✗ size 12{x="10"} {}
  • x 5 = 6 size 12{ left (x - 5 right )=6} {}
    • x = 11  ✗ size 12{x="11"} {}

All looks good, doesn’t it? The factoring was correct. But if you try x = 10 size 12{x="10"} {} or x = 11 size 12{x="11"} {} in the original equation, you will find that neither one works. What went wrong?

The factoring was correct, but the next step was wrong. Just because ( x 4 ) ( x 5 ) = 6 size 12{ \( x - 4 \) \( x - 5 \) =6} {} does not mean that either x 4 size 12{ left (x - 4 right )} {} or x 5 size 12{ left (x - 5 right )} {} has to be 6. There are lots of ways for two numbers to multiply to give 6. This trick only works for 0!

x 2 9x + 20 = 6 size 12{x rSup { size 8{2} } - 9x+"20"=6} {} , solved correctly

  • x 2 9x + 14 = 0 size 12{x rSup { size 8{2} } - 9x+"14"=0} {}
  • x 7 x 2 = 0 size 12{ left (x - 7 right ) left (x - 2 right )=0} {}
  • x 7 = 0 size 12{ left (x - 7 right )=0} {}
    • x = 7  ✓ size 12{x=7} {}
  • x 2 = 0 size 12{ left (x - 2 right )=0} {}
    • x = 2  ✓ size 12{x=2} {}

You may want to confirm for yourself that these are the correct solutions.

Moral: When solving quadratic equations, always begin by moving everything to one side of the equation , leaving only a 0 on the other side. This is true regardless of which of the three methods you use.

x 2 + 14 x + 49 = 0 size 12{x rSup { size 8{2} } +"14"x+"49"=0} {}

  • x + 7 2 = 0 size 12{ left (x+7 right ) rSup { size 8{2} } =0} {}
  • x = 7 size 12{x= - 7} {}

Moral : If the left side factors as a perfect square, the quadratic equation has only one solution.

Not all quadratic functions can be factored. This does not mean they have no solutions! If the function cannot be factored, we must use other means to find the solutions.

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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