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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.