what goes on here?

[allegansveritatem]

I tried to solve the following problem in two ways:

Problem:

I tried both the square root property solution and the quadratic formula:

solutions:

I then checked the answers with a CAS calculator. The calculator came back with the same solutions as the quadratic formula did. Now my question is: Was the square root property solution partially right? And if so, why didn't I get two solutions as I did with the quadratic formula?

 

[tkhunny]

I tried to solve the following problem in two ways:

Problem:View attachment 10941

I tried both the square root property solution and the quadratic formula:

solutions:View attachment 10942

I then checked the answers with a CAS calculator. The calculator came back with the same solutions as the quadratic formula did. Now my question is: Was the square root property solution partially right? And if so, why didn't I get two solutions as I did with the quadratic formula?

The "Square Root Property of Equals", which, in my opinion, is a little silly, states:

If A^2 = B^2, then A = +/-B.

You seem to have overlooked the +/- part.

Still, I see no value in this property. When would it EVER be necessary?

 

[pka]

I tried to solve the following problem in two ways:

Problem:View attachment 10941

I tried both the square root property solution and the quadratic formula:

solutions:View attachment 10942

I then checked the answers with a CAS calculator. The calculator came back with the same solutions as the quadratic formula did. Now my question is: Was the square root property solution partially right? And if so, why didn't I get two solutions as I did with the quadratic formula?

On first viewing your post, I thought of \(\displaystyle \sqrt{x^2}=|x|\)
I do not know what computer algebra system(CAS) you use, but look at this HERE & again HERE

 

[Deleted member 4993]

I tried to solve the following problem in two ways:

Problem:View attachment 10941

I tried both the square root property solution and the quadratic formula:

solutions:View attachment 10942

I then checked the answers with a CAS calculator. The calculator came back with the same solutions as the quadratic formula did. Now my question is: Was the square root property solution partially right? And if so, why didn't I get two solutions as I did with the quadratic formula?

Posts above have explained to you the reason you did not get two answers by equating square roots.

I would do the problem slightly differently.

(3x-8)^2 - (2x-5)^2 = 0

[(3x-8) + (2x-5)] * [(3x-8) - (2x-5)] = 0

[5x-13] * [x-3] = 0

thus we have:

5x - 13 = 0

x = 13/5

and

x - 3 =0

x = 3

 

[Steven G]

I tried to solve the following problem in two ways:

Problem:View attachment 10941

I tried both the square root property solution and the quadratic formula:

solutions:View attachment 10942

I then checked the answers with a CAS calculator. The calculator came back with the same solutions as the quadratic formula did. Now my question is: Was the square root property solution partially right? And if so, why didn't I get two solutions as I did with the quadratic formula?

So (3x-8)= (2x-5) and (3x-8)= -(2x-5)
So x=3 and 5x = 13. so x= 3 and x = 13/5

 

[topsquark]

View attachment 10942

FYI: If you had written line 3 on a paper for me I'd take marks off. This is not a valid equation. You should make sure to write such things on different lines.

-Dan

 

[allegansveritatem]

The "Square Root Property of Equals", which, in my opinion, is a little silly, states:

If A^2 = B^2, then A = +/-B.

You seem to have overlooked the +/- part.

Still, I see no value in this property. When would it EVER be necessary?

Well, No. 1, you are right, I for got the +/- dynamic of this procedure. No. 2 , I can't understand why you have so little regard for this technique in this case. I think, when done correctly, it is easier.

 

[allegansveritatem]

On first viewing your post, I thought of \(\displaystyle \sqrt{x^2}=|x|\)
I do not know what computer algebra system(CAS) you use, but look at this HERE & again HERE

I am using a Texas Instruments CAS Nspire. Thanks for the links. Interesting site. Shows the whole parade, so to speak, from problem to solution. I confess the second link flummoxed me however. I mean, how did the get that reversal, which they call "alternate form" in the second step?

 

[allegansveritatem]

Posts above have explained to you the reason you did not get two answers by equating square roots.

I would do the problem slightly differently.

(3x-8)^2 - (2x-5)^2 = 0

[(3x-8) + (2x-5)] * [(3x-8) - (2x-5)] = 0

[5x-13] * [x-3] = 0

thus we have:

5x - 13 =
x = 13/5

and

x - 3 =0

x = 3

That is a very nice procedure. I will try it out. Thanks

 

[allegansveritatem]

So (3x-8)= (2x-5) and (3x-8)= -(2x-5)
So x=3 and 5x = 13. so x= 3 and x = 13/5

Right. I forgot to, as it were, accentuate the negative.

 

[allegansveritatem]

FYI: If you had written line 3 on a paper for me I'd take marks off. This is not a valid equation. You should make sure to write such things on different lines.

-Dan

Yes. I see that. Actually, when I did that I had a sense that something was off about it but I was in rhino mode (i.e., why think when you came ram things home without it?)and carried on anyway.

 

[lookagain]

So (3x-8)= (2x-5) and (3x-8)= -(2x-5)
So x=3 and 5x = 13. so x= 3 and x = 13/5

So either (3x-8) = (2x-5) or (3x-8) = -(2x-5).

So either x = 3 or 5x = 13.

So either x = 3 or x = 13/5.


-- OR - -


So (3x-8) = (2x-5) or (3x-8) = -(2x-5).

So x = 3 or 5x = 13.

So x = 3 or x = 13/5.

 

[tkhunny]

Well, No. 1, you are right, I for got the +/- dynamic of this procedure. No. 2 , I can't understand why you have so little regard for this technique in this case. I think, when done correctly, it is easier.

It is a simple concept:

1. Solving the quadratic equation without the magic square root naturally leads to the TWO solutions.
2. Square Roots - One must REMEMBER to use the +/-. This is the part you forgot.

It my not be harder to be done, but it is harder to be done correctly.

And then, there's the thing about my signature.

 

[Harry_the_cat]

Well, No. 1, you are right, I for got the +/- dynamic of this procedure. No. 2 , I can't understand why you have so little regard for this technique in this case. I think, when done correctly, it is easier.

I agree this method is easier IF DONE CORRECTLY. Unfortunately, from my teaching experience, it is more often than not done incorrectly (as you did at first).