How can you simplify the quadratic formula using completing the square?

[Robert1986]

IMO, about 80% of the confusion in this thread has been caused by your inability to use standard terminology to explain what you mean. If you don't mind, I would like to re-state what you have said, but in a way that math people might better understand:

Given the quadratic ax^2 + bx +c, we all know how to find the roots with the quadratic formula. The following method also works: let b' = -b and c' = -c and use this formula:

x = [b' +- sqrt(b'^2 + 4ac')]/[2a]

Proof:

x = [b' +- sqrt(b'^2 + 4ac')]/[2a] = [(-b)^2 +- sqrt((-b)^2 + 4a(-c))]/[2a] = x = [-b +- sqrt(b^2 -4ac)]/[2a] which is the quadratic formula associated with ax^2 + bx + c. QED.


If you had said that, I would have understood exactly what you mean much sooner. Now, you might not have used the same method to derive your formula, but that doesn't matter. Usually, when giving a proof, you give the most efficient and easily understood proof, which is what I did above. For example, Gauss would never really explain why something was true. He would just state something and then prove it. He felt that all written theorems should appear as though they just came from the brow of the author of the proof.


Now, let's move on to this thing about not having to set and equation equal to 0 (in your terms) to find the roots. When someone says to find the roots of a polynomial p(x) it means to find the value of x such that p(x) = 0. This is by definition. It confuses us when you write down a completley new polynomial, say p'(x), and then describe a method that does not find the roots of p'(x) but of the original p(x).


The polynomials p(x) and p'(x) don't have the same solutions so when you start with this:

ax^2 + bx + c = 0

then

ax^2 = bx + c

doesn't make sense. Saying that these two equations are equivilant (which, BTW, is no difference than saying they are equal) is simply not true.


Now, please explain how your method makes life easier.

 

[Robert1986]

I should also point out that in ax^2 + bx + c, b and c are not variables, they are constants so they are not "cycling" through anything.

 

[agentredlum]

Can I just get two questions answered, so that I understand a little more:

a) Do you think that the following equations have the same solution set:

ax^2 + bx + c = 0

and

ax^2 = bx + c

b)please post an example which uses this trick to make my life easier. I still do not see how finding the roots of ax^2 + bx + c is made easier by writting ax^2 = bx +c and then using your quadratic formula which does not give solutions to ax^2 = bx + c but, in fact, gives solutions to ax^2 + bx + c = 0. How do the added steps make it easier than just using the original quadratic formula from the get-go?

If a, b, c are particular values, example a = 1 b = 2 c = 3 then the equations above DO NOT have the same solution set. I never claimed the 2 forms above have the same solution set for PARTICULAR values.

If you consider a, b, c, abstractly, as running through all real numbers then the 2 forms above have the same solution set. TAKEN AS A WHOLE SET. I emphasize this last part because it is very important for the definition and derivation that I did.

To post 1 example would not be enough. I have to post 7, to show the way signs effect both formulas, a total of 14 calculations. I have been dreading this but i will do it. Give me some time.

My argument was the textbook definition and formula has more symbols than my version so that makes my version simpler.

 

[Robert1986]

If a, b, c are particular values, example a = 1 b = 2 c = 3 then the equations above DO NOT have the same solution set. I never claimed the 2 forms above have the same solution set for PARTICULAR values.

If you consider a, b, c, abstractly, as running through all real numbers then the 2 forms above have the same solution set. TAKEN AS A WHOLE SET. I emphasize this last part because it is very important for the definition and derivation that I did.

I don't even know what you mean by this. If by "solution set" you mean that the text-book quadratic formula associated with ax^2 + bx + c = 0 and your quadratic formula associated with ax^2 = bx +c give the same answers, then yes. But this is not what people mean by solution set.

And I have no idea what gave you the impression that I wasn't look at the whole solution set.

 

[agentredlum]

I will post an example that supports my case without question.

Problem: Solve for x, x^2 = 2x + 5

Teextbook version solution...

Step1) Move all terms to one side of the equation.

x^2 - 2x - 5 = 0

Step2) Identify a, b, c, a = 1 b = -2 c = -5

Step3) Plug into x = (-b +-sqrt(b^2 - 4ac))/(2a)

x = (-(-2) +-sqrt((-2)^2 - 4(1)(-5))/(2(1))

Step4) Simplify

x = (2 +-sqrt(4 + 20))/2

Let's stop here because the rest is the same for both formulas.

My version solution...

Step1) Isolate ax^2 term

No manipulations necessary

Step2) Identify a, b, c, a =1 b = 2 c = 5

Step3) Plug into x = (b +-sqrt(b^2 + 4ac))/(2a)

x = ((2) +-sqrt((2)^2 + 4(1)(5))/(2(1))

Step4) Simplify

x = (2 +-sqrt(4 + 20))/2

Let's stop here and compare the number of calculations at each step.

In step one textbook version you performed 2 subtractions and added an extra symbol, zero.

I did nothing in my version.

In step 2 you identified a = 1 b = -2 c = -5

In step 2 i identified a = 1 b = 2 c = 5

you now have 2 extra minus signs that you have to carry over to step 3. I don't have that problem.

In step 3 We both substituted I had positive numbers, you had a few negatives, your formula also contains an extra negative.

Step 4 you had to compute -(-2) i had to compute (2) you have 2 extra minus signs

you had to compute (-2)^2 so you had to square the minus sign i had to compute (2)^2

you had to compute 4(1)(-5) and then subtract from b^2 I had to compute 4(1)(5) and then add to b^2 again you had that pesky minus sign and subsequent subtraction forces you to do addition. I didn't have that problem.

So you did 3 extra things in step1, 2 extra in step 2, and at least 4 extra in step 4. I won't count step 3 cause i want to be fair to you.

Satisfied now?

 

[Robert1986]

OK, now I see where you are comming from.

IF you are given an equation in this form:

ax^2 = bx + c

then in some sense your way is easier (though, marginally so, anyone who does a lot of math wouldn't actually move the stuff around as you suggested, they would just know what a, b and c are by looking at the equation) but it requires one to memorize another quadratic formula when the text-book one works just fine given one has enough experience to make simple algebraic manipulations in one's head.


But, I assume you would still use the text-book quadratic formula if given this:

ax^2 + bx + c = 0, right?


Also, did you look what I posted about explaining this more clearly? Did what I write make sense to you, I mean, did my way of getting this result make sense to you?

For me, I don't really see a benefit. If you do, that's fine; I don't think you're insane if it helps you. For me, the extra negatives really don't pose much of a problem.

 

[agentredlum]

OK, now I see where you are comming from.

IF you are given an equation in this form:

ax^2 = bx + c

then in some sense your way is easier (though, marginally so, anyone who does a lot of math wouldn't actually move the stuff around as you suggested, they would just know what a, b and c are by looking at the equation) but it requires one to memorize another quadratic formula when the text-book one works just fine given one has enough experience to make simple algebraic manipulations in one's head. But, I assume you would still use the text-book quadratic formula if given this:

ax^2 + bx + c = 0, right?Also, did you look what I posted about explaining this more clearly? Did what I write make sense to you, I mean, did my way of getting this result make sense to you?

For me, I don't really see a benefit. If you do, that's fine; I don't think you're insane if it helps you. For me, the extra negatives really don't pose much of a problem.

Finally, but it took much more than half an hour LOL. That's because we're not in the same room talking.

I looked at what you wrote and I don't see a problem with it initially, except you put a b^2 where it didn't belong, but that doesn't matter to me because it did not affect your final result.

x = [b' +- sqrt(b'^2 + 4ac')]/[2a] = [(-b)^2 +- sqrt((-b)^2 + 4a(-c))]/[2a]

The rest of it seems fine to me but I must caution you, I am not as rigorous as others. To me, it seems a tiny bit like circular reasoning but I am no expert in logic. Let others more qualified make their assessment of your proof.

I wanted to avoid any connection to the textbook version, that's why i considered the 4 forms, made my arguments, definitions, and derivations.

If i had an equation to solve in the form ax^2 + bx + c = 0 Then I would use the textbook version.

If you gave me 10 equations to solve in the form ax^2 = bx + c then I would feel silly using the textbook version because i know a better version for this form. My version. Imagine... sitting there with pencil and paper, collecting terms on one side, carrying around un-necessary minus signs, transforming subtractions to additions, squaring negatives, on and on, for 10 equations. Or better yet, ask a professional mathematician, who doesn't know, or won't accept my version because it's 'trivial', to do 10 equations by hand, would you chuckle as you watch him struggle?

If you were throwing random forms at me, I wouldn't know what i was going to get next out of possible

form1 ax^2 + bx + c = 0

form2 ax^2 + bx = c

form3 ax^2 + c = bx

form4 ax^2 = bx + c

And you asked me to solve 100 random forms or 1000 or 10^6 random 2nd degree equations, Then I would use my version because I would expect 75% of the forms thrown at me would not be form1

Don't forget that the textbook version has an extra minus sign and subtraction instead of addition. Those 2 differences are going to cause more problems. Does that make sense?

Thank you for your reply, thank you for your comments.

 

[agentredlum]

I have 2 questions about -b and - 4ac in the formula...

x = (-b +-sqrt(b^2 - 4ac))/(2a)

If you are given 2nd degree equations at random (that can be solved using above formula)

Question1: What percent of the equations given would you expect to compute a double negative for -b?

Question2: What percent of the equations given would you expect to compute a triple negative for - 4ac?



For instance if b is -5 then -(-5) must be evaluated, how often would you expect to see 2 negatives?

If a is -2 and c is -3 then - 4(-2)(-3) must be evaluated, how often would you expect to see 3 negatives?

 

[Char. Limit]

Wait... so all of this is to avoid minus signs? Are people really that scared of a -4ac?

 

[micromass]

Wait... so all of this is to avoid minus signs? Are people really that scared of a -4ac?

Minusphobia...

 

[Robert1986]

Wait... so all of this is to avoid minus signs?

Yeah, and here's the funny thing: they don't actually go away in practice.

Are people really that scared of a -4ac?

People in general? No.

Agenredlum? Apparently very much so.

 

[Robert1986]

Finally, but it took much more than half an hour LOL. That's because we're not in the same room talking.

I looked at what you wrote and I don't see a problem with it initially, except you put a b^2 where it didn't belong, but that doesn't matter to me because it did not affect your final result.

x = [b' +- sqrt(b'^2 + 4ac')]/[2a] = [(-b)^2 +- sqrt((-b)^2 + 4a(-c))]/[2a]

My mistake.

The rest of it seems fine to me but I must caution you, I am not as rigorous as others.

Yes, I have observed.

To me, it seems a tiny bit like circular reasoning but I am no expert in logic. Let othe

It only seems that way because the result is trivial.

Don't forget that the textbook version has an extra minus sign and subtraction instead of addition. Those 2 differences are going to cause more problems. Does that make sense?

No, that doesn't make sense. I would say that writting down a completely new equation, and then using that to solve the original one will cause problems. I'm a math guy, a minus sign doesn't scare me this much.

 

[Hurkyl]

Okay, you've spent long enough trying to repeal the use of negative numbers to recognize that all of the various forms of a quadratic equation differ only superficially. Time to wrap things up.