Solving Quadratic Equations: Is My Analysis Correct?

  • Thread starter Imparcticle
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    Quadratic
In summary, during a discussion on quadratic equations, a student realized there was an error in their teacher's method of simplifying an equation. The student suggested factoring out a 2 from the numerator before canceling it out with the 2 in the denominator, while the teacher simply canceled out the 6 and 2 at once. The student questioned the validity of their teacher's method and asked for a proof of the quadratic equation. Another participant chimed in, stating that many math teachers often cover up their lack of understanding by saying "that's just the way it is" or leaving the proof as an exercise for the student.
  • #1
Imparcticle
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4
We were recently reviewing quadratic equations in my algebra 1 class. As my teacher simplified equation after equation on the board, I began to get this nagging feeling there was something incorrect.
I have pin pointed where I believe an error was made.

At this point in solving a quadratic equation, (6 +/- 2 root24)/2, my teacher simply cancels out the 6 and the 2 at once. I disagree here. It is a rule that you cannot cancel each component of an equation where a term is separated by a + or - sign (of course, one can cancel the two since it is being multiplied with the "root24"). Instead, it I believe one must factor out a 2 from the numerator, then cancel out the 2 in the denominator.
my way:

1.)
6 +/- 2 root24 2(3 +/- root24)
-------------- = ---------------- = 3 +/- root24
2 2

2.)
The way my teacher does it:

6 +/- 2 root24 6/2 +/- 2/2 root24 = 3 +/- root24
--------------=
2


I realize that essentially, when you factor (as I did) , you are dividing each term, seperately by 2. However on the second example, one is dividing each term by the exact same integer.


is my analysis correct or incorrect?
 
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  • #2
It is very difficult to figure out your equations. This site has very nice equation capabilities. I suggest that you read this thread
 
  • #3
Factoring the 2 first or distributing the 1/2 first doesn't change things. Multiplication is distributive.

cookiemonster
 
Last edited:
  • #4
I wonder if using [ code ][ /code ] will help:

Code:
1.)
6 +/- 2 root24        2(3 +/- root24)
-------------- =    ---------------- = 3 +/- root24
      2                           2
   
2.)                                               
The way my teacher does it:

6 +/- 2 root24     6/2 +/- 2/2 root24 = 3 +/- root24
--------------=
       2
 
  • #5
How about

[tex]\frac{6 \pm 4\sqrt{6}}{2} = \frac{2(3 \pm 2\sqrt{6})}{2} = 3 \pm 2\sqrt{6}[/tex]

and

[tex]\frac{6 \pm 4\sqrt{6}}{2} = \frac{6}{2} \pm \frac{4\sqrt{6}}{2} = 3 \pm 2\sqrt{6}[/tex]

cookiemonster
 
  • #6
thank you for your clarification, cookie. I totally understand now.
 
  • #7
You know I’ve never seen how the quadratic equation is derived, or a proof for it. Would some one post (or link) one please?
 
  • #8
JonF said:
You know I’ve never seen how the quadratic equation is derived, or a proof for it. Would some one post (or link) one please?
It is just a generalisation of completing the square method: http://mathworld.wolfram.com/QuadraticEquation.html
 
  • #9
I feel silly for asking now…
 
  • #10
JonF said:
I feel silly for asking now…
Really, why?
It's not your fault that too many math teachers say "that's just the way it is" (usually to cover up their own ignorance/lack of understanding.)
 
  • #11
arildno said:
Really, why?
It's not your fault that too many math teachers say "that's just the way it is" (usually to cover up their own ignorance/lack of understanding.)

BRAVO!, right on the money there. lol
 
  • #12
My favorite: "The proof is left as an exercise to the reader." :)
 

1. How do I know if my solution to a quadratic equation is correct?

To ensure the accuracy of your solution, you can plug the values into the original equation and see if it satisfies the equation. You can also graph the equation and see if the solution(s) intersect with the parabola.

2. Can I use the quadratic formula for any type of quadratic equation?

Yes, the quadratic formula can be used to solve any type of quadratic equation, whether it is in standard form, vertex form, or intercept form.

3. How do I check for extraneous solutions when solving quadratic equations?

To check for extraneous solutions, you can plug the solution(s) into the original equation and see if it satisfies the equation. If it does not, then it is considered an extraneous solution and should be discarded.

4. Is there an easier way to solve quadratic equations other than using the quadratic formula?

Aside from factoring, the quadratic formula is the most efficient and reliable method for solving quadratic equations. However, you can also use the completing the square method, although it can be more time-consuming.

5. Can I use the quadratic formula to solve equations with a non-1 leading coefficient?

Yes, the quadratic formula can be used for equations with any leading coefficient. Just remember to substitute the correct values for a, b, and c in the formula.

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