#### elseif

##### New member
I attached three files. Excuse the fact that I swapped M for X in the first two images.

P is the problem, S and S2 are the solutions.
S is the solution in the textbook from which the problem came.
S2 is the solution in photomath (a phone app).

Which is correct? As far as I can discern, S2 is; I believe this primarily because photomath shows me how to get that answer, but I want to double check.

#### Attachments

• 3.4 KB Views: 1
• 3.6 KB Views: 1
• 1.8 KB Views: 1

#### Harry_the_cat

##### Senior Member
S is the correct answer. Where did the 18 in s2 come from?

#### JeffM

##### Elite Member
If you need the steps, here they are:

$$\displaystyle x^{5/2} * \sqrt{289} = 17x^{5/2} = 17\sqrt{x^5} = 17\sqrt{x^2x^2x^1} = 17\sqrt{x^2} * \sqrt{x^2} * \sqrt{x^1} = 17x * x\sqrt{x} = 17x^2\sqrt{x}.$$

#### elseif

##### New member
That's what I thought; I'm glad I can trust Photomath. Thanks.

#### Dr.Peterson

##### Elite Member
Why do you say you can trust it when it gave you a wrong answer? Are you implying that you typed the wrong thing into it, so the 18 is your own fault?

By the way, your "p" is not a problem; it is an expression. The problem must have had some words, such as "Write in simplified radical form." Did it?

#### JeffM

##### Elite Member
That's what I thought; I'm glad I can trust Photomath. Thanks.

#### elseif

##### New member
Woops, I swapped the labels! s (not s2) is from photomath. Sorry for the confusion.

#### Dr.Peterson

##### Elite Member
So the textbook gave the wrong answer? That's problematic, though not unheard of.

Can you show us the entire problem as given, and verify that you looked at the right answer in the book?

#### lookagain

##### Elite Member
If you need the steps, here they are:

$$\displaystyle 17\sqrt{x^5} = 17\sqrt{x^2x^2x^1} = 17\sqrt{x^2} * \sqrt{x^2} * \sqrt{x^1} = 17x * x\sqrt{x} = ...$$
$$\displaystyle \sqrt{x^2} = |x|, \ but \ \sqrt{x^4 \ } = x^2$$.

With that in mind, the natural progression would be this:

$$\displaystyle 17\sqrt{x^5} = 17\sqrt{x^4x^1} = 17\sqrt{x^4}*\sqrt{x^1} = 17x^2\sqrt{x}$$

#### elseif

##### New member
So the textbook gave the wrong answer? That's problematic, though not unheard of.

Can you show us the entire problem as given, and verify that you looked at the right answer in the book?
Sure... Attached.
It's exercise #39.

#### Attachments

• 264.3 KB Views: 8
• 418.4 KB Views: 8

#### Jomo

##### Elite Member
Sure... Attached.
It's exercise #39.
I am a bit confused. Your only concern was whether the answer should contain 17 or 18?? Well what is the sqrt(289)? Is it 17 or 18?

#### JeffM

##### Elite Member
$$\displaystyle \sqrt{x^2} = |x|, \ but \ \sqrt{x^4 \ } = x^2$$.

With that in mind, the natural progression would be this:

$$\displaystyle 17\sqrt{x^5} = 17\sqrt{x^4x^1} = 17\sqrt{x^4}*\sqrt{x^1} = 17x^2\sqrt{x}$$
Actually, it is not the natural progression for a student in the first year of the algebra of real numbers.

$$\displaystyle x^{5/2)} \in \mathbb R \implies x \in \mathbb R \text { and } x \ge 0.$$

Actually, I thought briefly about how ro proceed if x is negative and the roots are complex, but quickly concluded that the OP was not at a stage where those complexities would be relevant. Once it is assumed that

$$\displaystyle x in \mathbb \text { and } x \ge 0, \text { then } x = \sqrt{x^2}.$$

Obviously you are completely correct that in general

$$\displaystyle x \in \mathbb R \implies \sqrt{x^2} = |\ x \ |.$$

But that would have been of no consequence to this student on this problem. But then not everyone is interested in helping students.

#### JeffM

##### Elite Member
I am a bit confused. Your only concern was whether the answer should contain 17 or 18?? Well what is the sqrt(289)? Is it 17 or 18?
Jomo

I too am bothered by why the OP concluded that showing steps validates an answer, but it is now obvious that confusion was inevitable when he mis-labeled the responses. We cannot know how closely he followed those steps.

#### elseif

##### New member
Jomo

I too am bothered by why the OP concluded that showing steps validates an answer, but it is now obvious that confusion was inevitable when he mis-labeled the responses. We cannot know how closely he followed those steps.
Conclude is a little strong; I posted here to be certain . But the steps seemed very reasonable, so that was reasonable evidence.

#### elseif

##### New member
Thanks to everyone for your help.

#### lookagain

##### Elite Member
Actually, it is not the natural progression for a student in the first year of the algebra of real numbers.
You are certainly wrong. For the square root, the radicand is naturally broken up into the largest perfect square multiplied by some leftover factor(s).

JeffM said:
But then not everyone is interested in helping students.
This quote of yours applies to helpers who do not acknowledge the natural
progression as I have explained.

Last edited:

#### lookagain

##### Elite Member
One of the problems with this assignment is that it is missing the equivalent of this
instruction:
"Assume all variables represent positive numbers."