Why is the answer a negative number?

einstein

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(-5a^3b^5)^2 / a^4b^3

My working out and answer:

(-5a^3b^5)(-5a^3b^5) / a^4b^3

25a^6b^10 / a^4 b^3

= 25a^2b^7

The book says the answer= -25a^2b^7
 
einstein said:
(-5a^3b^5)^2 / a^4b^3

My working out and answer:

(-5a^3b^5)(-5a^3b^5) / a^4b^3

25a^6b^10 / a^4 b^3

= 25a^2b^7

The book says the answer= -25a^2b^7

Following PEMDAS - both you and book are incorrect (according to the problem as posted).
 
einstein said:
(-5a^3b^5)^2 / a^4b^3
Book is correct IF this is the expression: -(5a^3b^5)^2 / (a^4b^3)
In other words: the - sign outside the brackets, PLUS denominator MUST be bracketed :shock:
 
Subhotosh Khan said:
einstein said:
(-5a^3b^5)^2 / a^4b^3

My working out and answer:

(-5a^3b^5)(-5a^3b^5) / a^4b^3

25a^6b^10 / a^4 b^3

= 25a^2b^7

The book says the answer= -25a^2b^7

Following PEMDAS - both you and book are incorrect (according to the problem as posted).

Can you please elaborate?
 
einstein said:
[Can you please elaborate?

Did you read Deniss's reply above??

If you did - read it again slowly and carefully ......
 
Subhotosh Khan said:
einstein said:
[Can you please elaborate?

Did you read Deniss's reply above??

If you did - read it again slowly and carefully ......

yes i read his reply diligently

he said the book is correct if there is a negative sign before the bracket and brackets around the last two terms... but there is not. i copied the the question from the book to the forum correctly and always do

so denis is saying IF the book did it this way then the books answer is right. But the book did not do it the way he wrote, so that means the books answer is wrong. i got that!

but you said BOTH the book's anwer was wrong (which, like i said, i get it) but you also said my answer was wrong. So i said please elaborate because i don't see how my answer is wrong. I believe i followed pemdas correctly. If you would kindly point out why MY answer was wrong i'd much appreciate it.
 
Your work:
25a^6b^10 / a^4 b^3
= 25a^2b^7

That is incorrect (sorry, man!)
As is (no brackets) 25a^6b^10 / a^4 b^3 means (25a^6b^10 / a^4) times b^3
You PEMDASed badly :wink:
 
Denis said:
Your work:
25a^6b^10 / a^4 b^3
= 25a^2b^7

That is incorrect (sorry, man!)
As is (no brackets) 25a^6b^10 / a^4 b^3 means (25a^6b^10 / a^4) times b^3
You PEMDASed badly :wink:

thanks.

so the answer to 25a^6b^10 / a^4 b^3 = 25a^2b^13
 
einstein said:
i copied the the question from the book to the forum correctly and always do

If your book displays the given expression as shown below, then you did not correctly copy the exercise.

\(\displaystyle \frac{(-5 a^3 b^5)^2}{a^4 b^3}\)

If your book displays the given expression the following way, instead, then you could clarify your typing with square brackets as shown in blue.

\(\displaystyle \frac{(-5 a^3 b^5)^2}{a^4} \; b^3\)

[(-5a^3b^5)^2/a^4] b^3

Either way, the factor of -1 is squared, so there is no negation sign in the answer.

 
Re:

mmm4444bot said:
einstein said:
i copied the the question from the book to the forum correctly and always do

If your book displays the given expression as shown below, then you did not correctly copy the exercise.

\(\displaystyle \frac{(-5 a^3 b^5)^2}{a^4 b^3}\)

If your book displays the given expression the following way, instead, then you could clarify your typing with square brackets as shown in blue.

\(\displaystyle \frac{(-5 a^3 b^5)^2}{a^4} \; b^3\)

[(-5a^3b^5)^2/a^4] b^3

Either way, the factor of -1 is squared, so there is no negation sign in the answer.


The book displayed the expression neither of the ways you mentioned above. It displayed it EXACTLY like this:

\(\displaystyle (-5a^3b^5)^2\div a^4b^3\)

which is what i wrote, so i copied it correctly. if the book displayed it either of the ways you wrote it i would have written it like that in the forum.
 
a / bc means a divided by b, the result multiplied by c
a / (bc) means a divided by the result of b times c

With similar problems (and similar books!), please try and "decide" yourself which is which;
we're here to help, not to guess :wink:
 
einstein said:
It displayed it EXACTLY like this:

\(\displaystyle (-5a^3b^5)^2\div a^4b^3\)

which is what i wrote

No, that is not what you posted.

I read the symbol \(\displaystyle \div\) as a grouping symbol.

I do not read the symbol / as a grouping symbol.

In mathematical formatting, the vinculum (i.e., fraction bar) is a grouping symbol. We text it as /, and it must be used with grouping symbols added around the denominator.

You typed no grouping symbols, in your original post.

If you still don't understand this error, you might want to ask for additional explanation because this particular issue (failure to text grouping symbols in conjunction with using a forward slash) is the bane of the majority of people asking for help here with exercises that contain rational expressions. 8-)

 
Re:

mmm4444bot said:
einstein said:
It displayed it EXACTLY like this:

\(\displaystyle (-5a^3b^5)^2\div a^4b^3\)

which is what i wrote

No, that is not what you posted.

I read the symbol \(\displaystyle \div\) as a grouping symbol.

I do not read the symbol / as a grouping symbol.

In mathematical formatting, the vinculum (i.e., fraction bar) is a grouping symbol. We text it as /, and it must be used with grouping symbols added around the denominator.

You typed no grouping symbols, in your original post.

If you still don't understand this error, you might want to ask for additional explanation because this particular issue (failure to text grouping symbols in conjunction with using a forward slash) is the bane of the majority of people asking for help here with exercises that contain rational expressions. 8-)


Okay then, i still dont understand and further explanation would be helpful because your explanation above may have confused me as it seems to contradict Subhotosh Khan's answer. But i appreciate your attempt to help.

So what you're saying is:

\(\displaystyle \frac{2+2}{2+2}= 2+2 / 2+2 = 1\)

Whereas:

\(\displaystyle 2 + 2 \div 2 + 2 = 5\)
 
So then what is the answer to:


\(\displaystyle (-5a^3b^5)^2\div a^4b^3\)

Is the answer 25a^2b^7 or is the answer 25a^2b^13
 
Re: Re:

einstein said:
So what you're saying is:

\(\displaystyle \frac{2+2}{2+2}= 2+2 / 2+2 = 1\)

Whereas:

\(\displaystyle 2 + 2 \div 2 + 2 = 5\)


mmm4444bot said:
In mathematical formatting, the vinculum (i.e., fraction bar) is a grouping symbol.
We text it as /, and it must be used with grouping symbols added around the denominator.

No, user einstein, look at the highlighted quote box for user mmm4444bot. For your example,
you have to have grouping symbols around the denominator, *and* also around the numerator,
*in this particular case*, because the numerator has more terms than a monomial.

So, your example must be as \(\displaystyle ** :\)


\(\displaystyle \frac{2 + 2}{2 + 2} \ = \ {(2 + 2)}/{(2 + 2)} \ = \ 4/4 \ = \ 1\)


\(\displaystyle 2 + 2 \div 2 + 2 \ = \ 2 + 1 + 2 \ = \ 5, \ \ and \ this \ is \ correct.\)



\(\displaystyle ** \ \ or \ square \ brackets \ or \ braces\)
 
Re: Re:

lookagain said:
einstein said:
So what you're saying is:

\(\displaystyle \frac{2+2}{2+2}= 2+2 / 2+2 = 1\)

Whereas:

\(\displaystyle 2 + 2 \div 2 + 2 = 5\)


mmm4444bot said:
In mathematical formatting, the vinculum (i.e., fraction bar) is a grouping symbol.
We text it as /, and it must be used with grouping symbols added around the denominator.

No, user einstein, look at the highlighted quote box for user mmm4444bot. For your example,
you have to have grouping symbols around the denominator, *and* also around the numerator,
*in this particular case*, because the numerator has more terms than a monomial.

So, your example must be as \(\displaystyle ** :\)


\(\displaystyle \frac{2 + 2}{2 + 2} \ = \ {(2 + 2)}/{(2 + 2)} \ = \ 4/4 \ = \ 1\)


\(\displaystyle 2 + 2 \div 2 + 2 \ = \ 2 + 1 + 2 \ = \ 5, \ \ and \ this \ is \ correct.\)



\(\displaystyle ** \ \ or \ square \ brackets \ or \ braces\)

Okay so i have to put quotes around both numerator and denominator in this forum if i use a forward slash character in an expression. Is that what you guys are saying?

So basically what youre saying is typing an expression like this: 4*2/2+1 is wrong, not valid, doesn't mean anything. I have to type it like this in the forum: (4*2)/(2+1).

But what if i actually want to say and mean: 4 TIMES 2 DIVIDED BY 2 PLUS 1. Then what am i supposed to do when typing that in the forum since i cannot use the forward slash without brackets like you have stated.
 
Re: Re:

einstein said:
Okay so i have to put quotes around both numerator and denominator in this forum if i use a forward slash character in an expression. Is that what you guys are saying? <-------YES!!! By George, I think he's got it.

So basically what youre saying is typing an expression like this: 4*2/2+1 is wrong, not valid, doesn't mean anything.<-----No, we're saying its VERY SPECIFIC meaning is "four times two, the result then divided by two, and THEN one added to that result." I have to type it like this in the forum: (4*2)/(2+1).<----Yes, if what you mean is to multiply four times 2, and then divided that result by the sum of 2 plus 1.

But what if i actually want to say and mean: 4 TIMES 2 DIVIDED BY 2 PLUS 1. Well, as you've been told MULTIPLE TIMES in this forum, either use LaTex to write it in fraction form as it probably appears in your text, OR use grouping symbols and write it as (4*2)/(2 + 1). Grouping symbols are not that difficult to type, and will clear up any ambiguity about WHAT belongs WHERE. Then what am i supposed to do when typing that in the forum since i cannot use the forward slash without brackets like you have stated.<----I guess that if typing grouping symbols is too much effort for you to make, then perhaps you could look for a different forum from which to seek help.
 
Re: Re:

Mrspi said:
einstein said:
So basically what youre saying is typing an expression like this: 4*2/2+1 is wrong, not valid, doesn't mean anything.<-----No, we're saying its VERY SPECIFIC meaning is "four times two, the result then divided by two, and THEN one added to that result."

I already knew this!! Go back to my original question. I did not use brackets because neither did the book. So i posted the question correctly. So what is all this fuss about? Why are people confusing me and telling me that i forget to use brackets, etc.

The book did not show brackets so why should i?? Like you said above, as i quoted, if there are no brackets you you simply solve the equation following pemdas from left to right. The expression has a different meaning without brackets.

But then people started chiming in saying i should have used brackets, blah bla blah and assuming that the book used brackets and i left them out when i posted the question on the forum. Probably because the books answer was close to being right *if* brackets were used. But the book did not use brackets so neither did i. I wrote the expression in the forum how the book wrote it in the book. If the book didnt use brackets, why should i? (That's a rhetorical question so dont answer it).

Mrspi said:
Well, as you've been told MULTIPLE TIMES in this forum, either use LaTex to write it in fraction form as it probably appears in your text, OR use grouping symbols and write it as (4*2)/(2 + 1). Grouping symbols are not that difficult to type, and will clear up any ambiguity about WHAT belongs WHERE.

Lol. I was never told multiple times to use latex. I asked how to use it and learned how to use it by myself. But that was after i posted this question. And as *i've* stated MULTIPLE TIMES the book did not use grouping symbols and i posted the question as it appeared in the book. Yes, i could have added grouping symbols myself but maybe then i would have changed the meaning of the equation. And i want to see what your response and asnwer is to the question exactly as the book printed it.

So from what i can see, based on your contemptuous explanation above all those people ("experts") who kept telling me that i did not copy the question/equation from the book to the forum correctly and use grouping symbols were wrong.

The book printed the question like this: \(\displaystyle (-5a^3b^5)^2\div a^4b^3\) and i wrote that in the forum like this: (-5a^3b^5)^2 / a^4b^3

So i was right all along.

If i had put grouping symbols around (a^4b^3) so as to make the equation (-5a^3b^5)^2 / (a^4b^3) when i posted it, i would have changed the equation from the way the book displayed it. So why would i do that?
 
Re: Re:

einstein said:
The book printed the question like this: \(\displaystyle (-5a^3b^5)^2\div a^4b^3\) and i wrote that in the forum like this: (-5a^3b^5)^2 / a^4b^3

So i was right all along. <<<< No you were NOT

You have been told (at least once) that:

\(\displaystyle a\div bc\) is not same as a\bc

\(\displaystyle a\div bc\) is same as a\(bc)

You are here to ask for help - not to argue "conventional meaning".

Argument is next door ..... Contradiction would be available door next to that....
 
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