why does 4r multiplied by square root of 6r becone 4r^2 times square root of 6? [closed]
$begingroup$
Original question was to write the fraction in its simplest form:
Question: 4r/[(√6r) + 9]
I attempted to solve it by multiplying the denominator and numerator by the conjugates of the denominator, ((√6r) - 9).
My answer:
numerator: [4r*(√6r)] - 36r
denominator: 6r2 - 81
Actual answer:
numerator: [4r2*(√6)] - 36r
denominator: 6r2 -81
Why was the r under the √6r removed in the numerator?
Why and how did 4r become 4r2?
algebra-precalculus radicals fractions
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
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closed as unclear what you're asking by Andrei, lulu, Lord Shark the Unknown, KReiser, Shailesh Jan 3 at 8:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 1 more comment
$begingroup$
Original question was to write the fraction in its simplest form:
Question: 4r/[(√6r) + 9]
I attempted to solve it by multiplying the denominator and numerator by the conjugates of the denominator, ((√6r) - 9).
My answer:
numerator: [4r*(√6r)] - 36r
denominator: 6r2 - 81
Actual answer:
numerator: [4r2*(√6)] - 36r
denominator: 6r2 -81
Why was the r under the √6r removed in the numerator?
Why and how did 4r become 4r2?
algebra-precalculus radicals fractions
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
$endgroup$
closed as unclear what you're asking by Andrei, lulu, Lord Shark the Unknown, KReiser, Shailesh Jan 3 at 8:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Well, it's not clear (to me at least) whether (√6r) means $sqrt {6r}$ or $sqrt 6times r$. This should be clarified. to the question in the header: of course it doesn't. $4rtimes sqrt {6r}=4r^{3/2}times sqrt 6$.
$endgroup$
– lulu
Jan 2 at 14:48
$begingroup$
Do you mean $$frac{4r}{sqrt{6r}+9}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 14:50
$begingroup$
Why do you get $6r^2$ in the denominator? If you consider it as $sqrt{6r}$ surely the square is $6r$ (at least assuming $r$ positive). If it's $sqrt{6} r$ though then it's not clear why $4 r $ times $sqrt{6} r$ would not be $4 sqrt{6} r^2$ when $sqrt{6} r$ times $sqrt{6} r$ is $6 r^2$
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– quid♦
Jan 2 at 14:54
1
$begingroup$
Please format your question, so we can read it. Look at math.stackexchange.com/help/notation, to get an idea how to do that
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– Andrei
Jan 2 at 15:10
1
$begingroup$
Voting to close the question as it not clear what you are asking. As you can see from the comments, nobody is clear on what it is you are trying to write. Please edit for clarity.
$endgroup$
– lulu
Jan 2 at 15:12
|
show 1 more comment
$begingroup$
Original question was to write the fraction in its simplest form:
Question: 4r/[(√6r) + 9]
I attempted to solve it by multiplying the denominator and numerator by the conjugates of the denominator, ((√6r) - 9).
My answer:
numerator: [4r*(√6r)] - 36r
denominator: 6r2 - 81
Actual answer:
numerator: [4r2*(√6)] - 36r
denominator: 6r2 -81
Why was the r under the √6r removed in the numerator?
Why and how did 4r become 4r2?
algebra-precalculus radicals fractions
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
$endgroup$
Original question was to write the fraction in its simplest form:
Question: 4r/[(√6r) + 9]
I attempted to solve it by multiplying the denominator and numerator by the conjugates of the denominator, ((√6r) - 9).
My answer:
numerator: [4r*(√6r)] - 36r
denominator: 6r2 - 81
Actual answer:
numerator: [4r2*(√6)] - 36r
denominator: 6r2 -81
Why was the r under the √6r removed in the numerator?
Why and how did 4r become 4r2?
algebra-precalculus radicals fractions
algebra-precalculus radicals fractions
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
asked Jan 2 at 14:46
schoolwithschoolschoolwithschool
1914
1914
closed as unclear what you're asking by Andrei, lulu, Lord Shark the Unknown, KReiser, Shailesh Jan 3 at 8:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Andrei, lulu, Lord Shark the Unknown, KReiser, Shailesh Jan 3 at 8:31
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Well, it's not clear (to me at least) whether (√6r) means $sqrt {6r}$ or $sqrt 6times r$. This should be clarified. to the question in the header: of course it doesn't. $4rtimes sqrt {6r}=4r^{3/2}times sqrt 6$.
$endgroup$
– lulu
Jan 2 at 14:48
$begingroup$
Do you mean $$frac{4r}{sqrt{6r}+9}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 14:50
$begingroup$
Why do you get $6r^2$ in the denominator? If you consider it as $sqrt{6r}$ surely the square is $6r$ (at least assuming $r$ positive). If it's $sqrt{6} r$ though then it's not clear why $4 r $ times $sqrt{6} r$ would not be $4 sqrt{6} r^2$ when $sqrt{6} r$ times $sqrt{6} r$ is $6 r^2$
$endgroup$
– quid♦
Jan 2 at 14:54
1
$begingroup$
Please format your question, so we can read it. Look at math.stackexchange.com/help/notation, to get an idea how to do that
$endgroup$
– Andrei
Jan 2 at 15:10
1
$begingroup$
Voting to close the question as it not clear what you are asking. As you can see from the comments, nobody is clear on what it is you are trying to write. Please edit for clarity.
$endgroup$
– lulu
Jan 2 at 15:12
|
show 1 more comment
2
$begingroup$
Well, it's not clear (to me at least) whether (√6r) means $sqrt {6r}$ or $sqrt 6times r$. This should be clarified. to the question in the header: of course it doesn't. $4rtimes sqrt {6r}=4r^{3/2}times sqrt 6$.
$endgroup$
– lulu
Jan 2 at 14:48
$begingroup$
Do you mean $$frac{4r}{sqrt{6r}+9}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 14:50
$begingroup$
Why do you get $6r^2$ in the denominator? If you consider it as $sqrt{6r}$ surely the square is $6r$ (at least assuming $r$ positive). If it's $sqrt{6} r$ though then it's not clear why $4 r $ times $sqrt{6} r$ would not be $4 sqrt{6} r^2$ when $sqrt{6} r$ times $sqrt{6} r$ is $6 r^2$
$endgroup$
– quid♦
Jan 2 at 14:54
1
$begingroup$
Please format your question, so we can read it. Look at math.stackexchange.com/help/notation, to get an idea how to do that
$endgroup$
– Andrei
Jan 2 at 15:10
1
$begingroup$
Voting to close the question as it not clear what you are asking. As you can see from the comments, nobody is clear on what it is you are trying to write. Please edit for clarity.
$endgroup$
– lulu
Jan 2 at 15:12
2
2
$begingroup$
Well, it's not clear (to me at least) whether (√6r) means $sqrt {6r}$ or $sqrt 6times r$. This should be clarified. to the question in the header: of course it doesn't. $4rtimes sqrt {6r}=4r^{3/2}times sqrt 6$.
$endgroup$
– lulu
Jan 2 at 14:48
$begingroup$
Well, it's not clear (to me at least) whether (√6r) means $sqrt {6r}$ or $sqrt 6times r$. This should be clarified. to the question in the header: of course it doesn't. $4rtimes sqrt {6r}=4r^{3/2}times sqrt 6$.
$endgroup$
– lulu
Jan 2 at 14:48
$begingroup$
Do you mean $$frac{4r}{sqrt{6r}+9}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 14:50
$begingroup$
Do you mean $$frac{4r}{sqrt{6r}+9}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 14:50
$begingroup$
Why do you get $6r^2$ in the denominator? If you consider it as $sqrt{6r}$ surely the square is $6r$ (at least assuming $r$ positive). If it's $sqrt{6} r$ though then it's not clear why $4 r $ times $sqrt{6} r$ would not be $4 sqrt{6} r^2$ when $sqrt{6} r$ times $sqrt{6} r$ is $6 r^2$
$endgroup$
– quid♦
Jan 2 at 14:54
$begingroup$
Why do you get $6r^2$ in the denominator? If you consider it as $sqrt{6r}$ surely the square is $6r$ (at least assuming $r$ positive). If it's $sqrt{6} r$ though then it's not clear why $4 r $ times $sqrt{6} r$ would not be $4 sqrt{6} r^2$ when $sqrt{6} r$ times $sqrt{6} r$ is $6 r^2$
$endgroup$
– quid♦
Jan 2 at 14:54
1
1
$begingroup$
Please format your question, so we can read it. Look at math.stackexchange.com/help/notation, to get an idea how to do that
$endgroup$
– Andrei
Jan 2 at 15:10
$begingroup$
Please format your question, so we can read it. Look at math.stackexchange.com/help/notation, to get an idea how to do that
$endgroup$
– Andrei
Jan 2 at 15:10
1
1
$begingroup$
Voting to close the question as it not clear what you are asking. As you can see from the comments, nobody is clear on what it is you are trying to write. Please edit for clarity.
$endgroup$
– lulu
Jan 2 at 15:12
$begingroup$
Voting to close the question as it not clear what you are asking. As you can see from the comments, nobody is clear on what it is you are trying to write. Please edit for clarity.
$endgroup$
– lulu
Jan 2 at 15:12
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If r is not inside the under-root [this is suggested by your answer in denominator]
then
$frac{4r}{(sqrt{6}r+9)}= frac{(4r)(sqrt{6}r-9)}{(6r^2 -81)}=frac{4sqrt{6}r^2-36r}{6r^2-81}$
Which is the actual answer.
You are doing mistake while calculating.
edited Jan 2 at 16:40
Andrei
13.2k21230
answered Jan 2 at 15:53
BJKShahBJKShah
456
$endgroup$
1
$begingroup$
Please format your answer. See math.stackexchange.com/help/notation or look upMathJax
$endgroup$
– Andrei
Jan 2 at 15:57
$begingroup$
Thanks for the help as this is my first day of using this site
$endgroup$
– BJKShah
Jan 2 at 16:25
$begingroup$
To add the square root, use$sqrt{expression}$
$endgroup$
– Andrei
Jan 2 at 16:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If r is not inside the under-root [this is suggested by your answer in denominator]
then
$frac{4r}{(sqrt{6}r+9)}= frac{(4r)(sqrt{6}r-9)}{(6r^2 -81)}=frac{4sqrt{6}r^2-36r}{6r^2-81}$
Which is the actual answer.
You are doing mistake while calculating.
edited Jan 2 at 16:40
Andrei
13.2k21230
answered Jan 2 at 15:53
BJKShahBJKShah
456
$endgroup$
1
$begingroup$
Please format your answer. See math.stackexchange.com/help/notation or look upMathJax
$endgroup$
– Andrei
Jan 2 at 15:57
$begingroup$
Thanks for the help as this is my first day of using this site
$endgroup$
– BJKShah
Jan 2 at 16:25
$begingroup$
To add the square root, use$sqrt{expression}$
$endgroup$
– Andrei
Jan 2 at 16:39
add a comment |
$begingroup$
If r is not inside the under-root [this is suggested by your answer in denominator]
then
$frac{4r}{(sqrt{6}r+9)}= frac{(4r)(sqrt{6}r-9)}{(6r^2 -81)}=frac{4sqrt{6}r^2-36r}{6r^2-81}$
Which is the actual answer.
You are doing mistake while calculating.
edited Jan 2 at 16:40
Andrei
13.2k21230
answered Jan 2 at 15:53
BJKShahBJKShah
456
$endgroup$
1
$begingroup$
Please format your answer. See math.stackexchange.com/help/notation or look upMathJax
$endgroup$
– Andrei
Jan 2 at 15:57
$begingroup$
Thanks for the help as this is my first day of using this site
$endgroup$
– BJKShah
Jan 2 at 16:25
$begingroup$
To add the square root, use$sqrt{expression}$
$endgroup$
– Andrei
Jan 2 at 16:39
add a comment |
$begingroup$
If r is not inside the under-root [this is suggested by your answer in denominator]
then
$frac{4r}{(sqrt{6}r+9)}= frac{(4r)(sqrt{6}r-9)}{(6r^2 -81)}=frac{4sqrt{6}r^2-36r}{6r^2-81}$
Which is the actual answer.
You are doing mistake while calculating.
edited Jan 2 at 16:40
Andrei
13.2k21230
answered Jan 2 at 15:53
BJKShahBJKShah
456
$endgroup$
If r is not inside the under-root [this is suggested by your answer in denominator]
then
$frac{4r}{(sqrt{6}r+9)}= frac{(4r)(sqrt{6}r-9)}{(6r^2 -81)}=frac{4sqrt{6}r^2-36r}{6r^2-81}$
Which is the actual answer.
You are doing mistake while calculating.
edited Jan 2 at 16:40
Andrei
13.2k21230
answered Jan 2 at 15:53
BJKShahBJKShah
456
edited Jan 2 at 16:40
Andrei
13.2k21230
edited Jan 2 at 16:40
Andrei
13.2k21230
edited Jan 2 at 16:40
Andrei
13.2k21230
13.2k21230
answered Jan 2 at 15:53
BJKShahBJKShah
456
answered Jan 2 at 15:53
BJKShahBJKShah
456
answered Jan 2 at 15:53
BJKShahBJKShah
456
456
1
$begingroup$
Please format your answer. See math.stackexchange.com/help/notation or look upMathJax
$endgroup$
– Andrei
Jan 2 at 15:57
$begingroup$
Thanks for the help as this is my first day of using this site
$endgroup$
– BJKShah
Jan 2 at 16:25
$begingroup$
To add the square root, use$sqrt{expression}$
$endgroup$
– Andrei
Jan 2 at 16:39
add a comment |
1
$begingroup$
Please format your answer. See math.stackexchange.com/help/notation or look upMathJax
$endgroup$
– Andrei
Jan 2 at 15:57
$begingroup$
Thanks for the help as this is my first day of using this site
$endgroup$
– BJKShah
Jan 2 at 16:25
$begingroup$
To add the square root, use$sqrt{expression}$
$endgroup$
– Andrei
Jan 2 at 16:39
1
1
$begingroup$
Please format your answer. See math.stackexchange.com/help/notation or look up
MathJax$endgroup$
– Andrei
Jan 2 at 15:57
$begingroup$
Please format your answer. See math.stackexchange.com/help/notation or look up
MathJax$endgroup$
– Andrei
Jan 2 at 15:57
$begingroup$
Thanks for the help as this is my first day of using this site
$endgroup$
– BJKShah
Jan 2 at 16:25
$begingroup$
Thanks for the help as this is my first day of using this site
$endgroup$
– BJKShah
Jan 2 at 16:25
$begingroup$
To add the square root, use
$sqrt{expression}$$endgroup$
– Andrei
Jan 2 at 16:39
$begingroup$
To add the square root, use
$sqrt{expression}$$endgroup$
– Andrei
Jan 2 at 16:39
add a comment |
2
$begingroup$
Well, it's not clear (to me at least) whether (√6r) means $sqrt {6r}$ or $sqrt 6times r$. This should be clarified. to the question in the header: of course it doesn't. $4rtimes sqrt {6r}=4r^{3/2}times sqrt 6$.
$endgroup$
– lulu
Jan 2 at 14:48
$begingroup$
Do you mean $$frac{4r}{sqrt{6r}+9}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 14:50
$begingroup$
Why do you get $6r^2$ in the denominator? If you consider it as $sqrt{6r}$ surely the square is $6r$ (at least assuming $r$ positive). If it's $sqrt{6} r$ though then it's not clear why $4 r $ times $sqrt{6} r$ would not be $4 sqrt{6} r^2$ when $sqrt{6} r$ times $sqrt{6} r$ is $6 r^2$
$endgroup$
– quid♦
Jan 2 at 14:54
1
$begingroup$
Please format your question, so we can read it. Look at math.stackexchange.com/help/notation, to get an idea how to do that
$endgroup$
– Andrei
Jan 2 at 15:10
1
$begingroup$
Voting to close the question as it not clear what you are asking. As you can see from the comments, nobody is clear on what it is you are trying to write. Please edit for clarity.
$endgroup$
– lulu
Jan 2 at 15:12