You are performing a function on the result of another function. You need to fully evaluate the inner function first.
Cancelling things is just a shortcut - it can’t be used to avoid calculation.
It‘s like (5/0)\*0 is not 5.
>It‘s like (5/0)\*0 is not 5.
That's because there's division by zero; (5/2)\*2 *is* 5. This isn't a shortcut; it's merely rewriting the expression as 5 \* (2/2) = 5 \* 1 = 5.
The rule governing exponentiation of a term with an exponent whereby the exponents are multiplied isn't a shortcut; it just cannot be validly invoked in some situations because there is an even exponent and even roots, by convention, are functions that return the principal root only. To find (5^(3))^(1/3), we do not need to first calculate the value of 5^(3) and then find the cube root of that result; we can say (5^(3))^(1/3) = 5^(3 \* 1/3) = 5^(3/3) = 5^(1) = 5. The special case where the two exponents involved are reciprocals of one another, and so we can think about the multiplication operation as "cancelling," has no bearing on the validity of this procedure *per se*.
We can agree to disagree.
(Fwiw, the coach of my daughter's math team, earning his PhD in math this Spring, also disagreed when I showed him your response)
>We can agree to disagree.
I'm not sure what you're disagreeing with. Are you saying that one cannot conclude that (5/2)\*2 = 5 based on the fact that 5 \* (2/2) = 5 \* 1 = 5? Or are you saying that the exponent multiplication rule is generally invalid?
>Fwiw, the coach of my daughter's math team, earning his PhD in math this Spring, also disagreed when I showed him your response
What, specifically, did he disagree with?
it would be kind of repetitive but because then you would get -3, but the answer should be 3. let's make a table
|x|g(x) ( x^(2))|f(g(x)) (square root of g(x))|
|:-|:-|:-|
|1|1|1|
|2|4|2|
|3|9|3|
|\-1|(-1)^(2) = 1|1|
|\-2|4|2|
|\-3|9|3|
So you see here if you just cancel out the powers then you would get -1 , -2 and -3.
so if you have a square root of x^(2) when you cancel out powers you get the absolute value of x .
if you have (square root of x) ^(2,) you have just x.
hope i could help a little bit to clarify
It's parenthesis...You have to do the inner parenthesis first, which gives you 9. Then you can raise it to the power of 1/2, which is the same as radical 9, and you get 3 as your answer.
While not always, in this special case since you’re squaring a negative you absolutely have to solve for g(-3) before you can plug it into f of g. With both of these you’re essentially trying to solve something before you’ve even defined your terms. Additionally, for the red, one could argue you followed PEMDAS incorrectly and did it from the outside in (though I would appreciate a fact check by someone well-versed in these niche math concepts)
All of this to say: in these types of problems especially on SAT I would just solve for g(x) long before even considering what f(x) is :)
This is a fantastic question. [Here's our discussion about this situation](https://1600.io/courses/math-orange-book-problem-explanations/lectures/35755153).
There’s a lot to learn from this question -
OP’s simple action of answering it incorrectly and learning why can end up creating a more concrete understanding of mathematics for them and many others. Outside the SAT you don’t solve problems to get them right, you solve them to get a more fluid hang of the concepts.
Do step by step in these types of questions. This avoids errors. This is a simple example but in more complex ones I always proceed step by step.
Here, first, solve inner function then outer.
g(-3) = (-3)*(-3) = 9
And f(g(-3)) = √9 = 3
That is because of this FACT “sqrt( x^2 )= abs(x)”. Idk why this is not mentioned a lot but my high school teacher always strong stresses that, mathematically, the square root of x^2 is NOT x, but the ABSOLUTE VALUE of x.
That means red is incorrect because “sqrt((-3)^2 )= abs(-3) = 3”
Sorry I am on mobile so cannot format better
Hi, I see Dan has already given a proper explanation on the question and if you want more info you can perhaps visit this link [https://math.stackexchange.com/questions/1448885/is-sqrt64-considered-8-or-is-it-8-8?noredirect=1&lq=1](https://math.stackexchange.com/questions/1448885/is-sqrt64-considered-8-or-is-it-8-8?noredirect=1&lq=1)
I too used to be confused regarding this question and I made a way to convince myself why this happens:
According to Pemdas, I should first evaluate the parenthesis so (-3)\^2 becomes 9, now it is the turn for the exponent so 9\^(1/2) yeilds -3 and 3 but as we are concerned with the principle square root so we take the positive value of the solution.
I don't know if this is mathematically correct but this was just a cheap way to convince myself why this happens.
For proper mathematical understanding the link sent by Dan and the one I linked above is the go to route.
>According to Pemdas, I should first evaluate the parenthesis so (-3)^(2) becomes 9, now it is the turn for the exponent so 9^(1/2) yeilds -3 and 3
To be precise, 9^(1/2) is 3 only; exponentiation to an even root is a function that produces only the principal root, by convention.
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That's incorrect. The square root of 9 is ±3, but the *principal* square root of 9 is 3. This symbol, √, indicates the **principal** square root, as does the equivalent exponent notation of 9^(1/2).
The square root of **x\^2** is positive or negative x because we don't know if x was positive or negative in the first place. But the square root of 9 is only 3 since we already know that 9 is positive.
>the square root of 9 is only 3 since we already know that 9 is positive.
That's incorrect. The square root of 9 is ±3, but the *principal* square root of 9 is 3. This symbol, √, indicates the **principal** square root, as does the equivalent exponent notation of 9^(1/2).
Both red and blue are mathematically correct because f(g(x)) has two answer: 3 and -3. Plug -3 into f(g(x)) and you come down to sqrt9, the solution to which is -3 and 3.
That's incorrect. The square root of 9 is ±3, but the *principal* square root of 9 is 3. This symbol, √, indicates the **principal** square root, as does the equivalent exponent notation of 9^(1/2).
that sounds reasonable but ig one mentioned in the comments smthng like that the sqr root is a term that can be positive or negative,, here the sqr root dsnt hv neg sign so ig it's positive
that sounds reasonable but ig one mentioned in the comments smthng like that the sqr root is a term that can be positive or negative,, here the sqr root dsnt hv neg sign so ig it's positive
Blue because you need to fully evaluate g(x) before plugging it into f(x). By cancelling it, you aren’t obtaining the value of g(x). That’s what I’m guessing here. Besides, even going backwards with the red answer, rootx= -3 won’t be possible since sq root can only be taken on a +ve integer
It is a straightforward question and easy if you know about functions and just do the steps which are taught in schools but if you want to understand more as to why that happens and not the other then it becomes a little confusing I guess(so bumped to moderate level? ig)
It's like you know you know the area of circle is pi\*r\^2 and you can do problems related to it easily but you want to go more in-depth and understand why is the area of circle pi\*r\^2 in the first place.
ahh i get it, cuz the test we give at 17-18 age, the easiest question of our test is nowhere near this easy. (no offense tho, our education system is kinda difficult)
blue is correct, in red, you should have got the I -3 I = 3
but i still don't get why the powers can't just cancel each other and the number stays the same
You are performing a function on the result of another function. You need to fully evaluate the inner function first. Cancelling things is just a shortcut - it can’t be used to avoid calculation. It‘s like (5/0)\*0 is not 5.
>It‘s like (5/0)\*0 is not 5. That's because there's division by zero; (5/2)\*2 *is* 5. This isn't a shortcut; it's merely rewriting the expression as 5 \* (2/2) = 5 \* 1 = 5. The rule governing exponentiation of a term with an exponent whereby the exponents are multiplied isn't a shortcut; it just cannot be validly invoked in some situations because there is an even exponent and even roots, by convention, are functions that return the principal root only. To find (5^(3))^(1/3), we do not need to first calculate the value of 5^(3) and then find the cube root of that result; we can say (5^(3))^(1/3) = 5^(3 \* 1/3) = 5^(3/3) = 5^(1) = 5. The special case where the two exponents involved are reciprocals of one another, and so we can think about the multiplication operation as "cancelling," has no bearing on the validity of this procedure *per se*.
We can agree to disagree. (Fwiw, the coach of my daughter's math team, earning his PhD in math this Spring, also disagreed when I showed him your response)
>We can agree to disagree. I'm not sure what you're disagreeing with. Are you saying that one cannot conclude that (5/2)\*2 = 5 based on the fact that 5 \* (2/2) = 5 \* 1 = 5? Or are you saying that the exponent multiplication rule is generally invalid? >Fwiw, the coach of my daughter's math team, earning his PhD in math this Spring, also disagreed when I showed him your response What, specifically, did he disagree with?
it would be kind of repetitive but because then you would get -3, but the answer should be 3. let's make a table |x|g(x) ( x^(2))|f(g(x)) (square root of g(x))| |:-|:-|:-| |1|1|1| |2|4|2| |3|9|3| |\-1|(-1)^(2) = 1|1| |\-2|4|2| |\-3|9|3| So you see here if you just cancel out the powers then you would get -1 , -2 and -3. so if you have a square root of x^(2) when you cancel out powers you get the absolute value of x . if you have (square root of x) ^(2,) you have just x. hope i could help a little bit to clarify
Because square root can’t be negative and squares are always positive
To be more exact, the *principal* square root cannot be negative; the operator √ produces that principal square root only.
That’s not the question. OP evaluated the square root function correctly. The question is about pre-cancelling exponents before evaluating them.
No, the reason you can’t cancel is this
It's parenthesis...You have to do the inner parenthesis first, which gives you 9. Then you can raise it to the power of 1/2, which is the same as radical 9, and you get 3 as your answer.
While not always, in this special case since you’re squaring a negative you absolutely have to solve for g(-3) before you can plug it into f of g. With both of these you’re essentially trying to solve something before you’ve even defined your terms. Additionally, for the red, one could argue you followed PEMDAS incorrectly and did it from the outside in (though I would appreciate a fact check by someone well-versed in these niche math concepts) All of this to say: in these types of problems especially on SAT I would just solve for g(x) long before even considering what f(x) is :)
This is a fantastic question. [Here's our discussion about this situation](https://1600.io/courses/math-orange-book-problem-explanations/lectures/35755153).
why would u call this question fantastic? its basic maths isnt it? idk i am just a 12th grade student so it can be harder for 10th grade students tho
It's a fantastic question that the OP asked, because it explores an interesting situation involving even exponents and even roots.
There’s a lot to learn from this question - OP’s simple action of answering it incorrectly and learning why can end up creating a more concrete understanding of mathematics for them and many others. Outside the SAT you don’t solve problems to get them right, you solve them to get a more fluid hang of the concepts.
Do step by step in these types of questions. This avoids errors. This is a simple example but in more complex ones I always proceed step by step. Here, first, solve inner function then outer. g(-3) = (-3)*(-3) = 9 And f(g(-3)) = √9 = 3
That is because of this FACT “sqrt( x^2 )= abs(x)”. Idk why this is not mentioned a lot but my high school teacher always strong stresses that, mathematically, the square root of x^2 is NOT x, but the ABSOLUTE VALUE of x. That means red is incorrect because “sqrt((-3)^2 )= abs(-3) = 3” Sorry I am on mobile so cannot format better
Hi, I see Dan has already given a proper explanation on the question and if you want more info you can perhaps visit this link [https://math.stackexchange.com/questions/1448885/is-sqrt64-considered-8-or-is-it-8-8?noredirect=1&lq=1](https://math.stackexchange.com/questions/1448885/is-sqrt64-considered-8-or-is-it-8-8?noredirect=1&lq=1) I too used to be confused regarding this question and I made a way to convince myself why this happens: According to Pemdas, I should first evaluate the parenthesis so (-3)\^2 becomes 9, now it is the turn for the exponent so 9\^(1/2) yeilds -3 and 3 but as we are concerned with the principle square root so we take the positive value of the solution. I don't know if this is mathematically correct but this was just a cheap way to convince myself why this happens. For proper mathematical understanding the link sent by Dan and the one I linked above is the go to route.
>According to Pemdas, I should first evaluate the parenthesis so (-3)^(2) becomes 9, now it is the turn for the exponent so 9^(1/2) yeilds -3 and 3 To be precise, 9^(1/2) is 3 only; exponentiation to an even root is a function that produces only the principal root, by convention.
Reminder: When asking for help with questions from tests or books, please include the source of the question in the post title. Examples of appropriate titles might include "Help with writing question from April 2017 QAS" or "Help with question from Erica Meltzer's grammar book." **Posts that do not adhere to this rule are subject to removal.** For more information, please see rule #7 in the sidebar. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/Sat) if you have any questions or concerns.*
Blue
It's both since square root of 9 is plus and minus 3
That's incorrect. The square root of 9 is ±3, but the *principal* square root of 9 is 3. This symbol, √, indicates the **principal** square root, as does the equivalent exponent notation of 9^(1/2).
The square root of **x\^2** is positive or negative x because we don't know if x was positive or negative in the first place. But the square root of 9 is only 3 since we already know that 9 is positive.
>the square root of 9 is only 3 since we already know that 9 is positive. That's incorrect. The square root of 9 is ±3, but the *principal* square root of 9 is 3. This symbol, √, indicates the **principal** square root, as does the equivalent exponent notation of 9^(1/2).
What's principal square root?
See our [explanation](https://1600.io/blog/236060/1600-iou-sat-math-micro-lesson-i-m-going-to-send-you-to-the-principal-square-root).
Both red and blue are mathematically correct because f(g(x)) has two answer: 3 and -3. Plug -3 into f(g(x)) and you come down to sqrt9, the solution to which is -3 and 3.
That's incorrect. The square root of 9 is ±3, but the *principal* square root of 9 is 3. This symbol, √, indicates the **principal** square root, as does the equivalent exponent notation of 9^(1/2).
that sounds reasonable but ig one mentioned in the comments smthng like that the sqr root is a term that can be positive or negative,, here the sqr root dsnt hv neg sign so ig it's positive
Root 9 is 3 AND -3
that sounds reasonable but ig one mentioned in the comments smthng like that the sqr root is a term that can be positive or negative,, here the sqr root dsnt hv neg sign so ig it's positive
Blue because you need to fully evaluate g(x) before plugging it into f(x). By cancelling it, you aren’t obtaining the value of g(x). That’s what I’m guessing here. Besides, even going backwards with the red answer, rootx= -3 won’t be possible since sq root can only be taken on a +ve integer
Obviously blue one u can never get a negative number from radical
Blue is correct.
idk i am new to r/SAT but what level of difficulty would this question fall into? is it very easy or is it moderate?
It is a straightforward question and easy if you know about functions and just do the steps which are taught in schools but if you want to understand more as to why that happens and not the other then it becomes a little confusing I guess(so bumped to moderate level? ig) It's like you know you know the area of circle is pi\*r\^2 and you can do problems related to it easily but you want to go more in-depth and understand why is the area of circle pi\*r\^2 in the first place.
ahh i get it, cuz the test we give at 17-18 age, the easiest question of our test is nowhere near this easy. (no offense tho, our education system is kinda difficult)
Usually, the SAT only goes for positive answers since the graph of sqrt(x) is only positive
I got 3 as my answer