Math 2

Math 2
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  • Foil That!
  • Now, remember that (a-6)^2 is the same as (a-6)(a-26). This means you have to use FOIL to create a standard form equation.
  • After using FOIL, I got a=a^2-12a+36
  • Set it to 0
  • Subtract a from both sides?
  • Now, you'll need to set it to 0.
  • Correct!
  • Find the X's
  • I'm going to factor it and then set it to 0. 
  • Good thinking!
  • Now, solve for your x's. You may use whatever method you want.
  • Work  √a=a-6  ( √a)^2=(a-6)^2  a=(a-6)(a-6) -> a=a^2-12a+36
  • Time to Substitute
  • Let's substitute the new values in for a.
  • Work  √a=a-6  ( √a)^2=(a-6)^2  a=(a-6)(a-6) -> a=a^2-12a+36 a-a=a^2-12a+36-a -> 0=a^2-13a+36
  • Wow That's Interesting
  • What is an extraneous solution?
  • Work  √a=a-6  ( √a)^2=(a-6)^2  a=(a-6)(a-6) -> a=a^2-12a+36 a-a=a^2-12a+36-a -> 0=a^2-13a+36 -> 0=(a-9)(a-4)   a-9+9=0+9-> a=9  a-4+4=0+4 -> a=4                    
  • The End 
  • a=9 is the actual solution which means it worksand a=4 is the extraneous solution.
  •  √9=9-6 -> 3=3 ✓  √4=4-6 -> 2=-2 ❌
  • √9=9-6 -> 3=3 ✓  √4=4-6 -> 2=-2 ❌
  •  a-9+9=0+9-> a=9  a-4+4=0+4 -> a=4         a=9 a=4  √a=a-6   √9=9-6 -> 3=3 ✓  √4=4-6 -> 2=-2 ❌ Not an actual solution since it does't sastify the original equation.
  • I plugged in both values but a=4 does not work. Why is that?
  • This the part where I got confused.
  • Don't worry Mike, that's what we mathmeticians like to call an extraneous solution.
  • You got this.
  • Professor Smith explains what an extraneous solution is and they depart.
  • Thanks Professor,  I was very confused.
  • No need to thank me. Anytime Mike.
  • It is just a solution that does not work when substituted into your original equation. It happens but you would want to distinguish it from your actual solution which in this case is a=9 since it works.
  • Mike explains the problem that was once haunting him and he explained it to his math class. He received extra credit from his math teacher and applause from his classmates.
  • An extraneous solution is a solution that does not satisfy the original equation. It needs to be distinguished from the actual solution.
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