Factoring out square roots from the numerator after rationalizing the denominator.
The question I was working on was to rationalize the denominator of $$frac{1}{sqrt{3}(sqrt{21}+sqrt{7})}$$
My answer was $frac{sqrt7}{42}(3-sqrt{3})$.
But both my book and Symbolab gave the answer as $frac{1}{42}(3sqrt{7}-sqrt{21})$.
Why shouldn't I factor out the $sqrt{7}$?
Thanks for your help.
algebra-precalculus
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
add a comment |
The question I was working on was to rationalize the denominator of $$frac{1}{sqrt{3}(sqrt{21}+sqrt{7})}$$
My answer was $frac{sqrt7}{42}(3-sqrt{3})$.
But both my book and Symbolab gave the answer as $frac{1}{42}(3sqrt{7}-sqrt{21})$.
Why shouldn't I factor out the $sqrt{7}$?
Thanks for your help.
algebra-precalculus
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
1
math.stackexchange.com/questions/2047349/… math.stackexchange.com/questions/1274936/…
– lab bhattacharjee
Dec 5 '18 at 12:55
I think that, since you need to rationalize the denominator, both your solution and Symbolab answer are perfectly fine.
– rafa11111
Dec 5 '18 at 13:00
add a comment |
The question I was working on was to rationalize the denominator of $$frac{1}{sqrt{3}(sqrt{21}+sqrt{7})}$$
My answer was $frac{sqrt7}{42}(3-sqrt{3})$.
But both my book and Symbolab gave the answer as $frac{1}{42}(3sqrt{7}-sqrt{21})$.
Why shouldn't I factor out the $sqrt{7}$?
Thanks for your help.
algebra-precalculus
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
The question I was working on was to rationalize the denominator of $$frac{1}{sqrt{3}(sqrt{21}+sqrt{7})}$$
My answer was $frac{sqrt7}{42}(3-sqrt{3})$.
But both my book and Symbolab gave the answer as $frac{1}{42}(3sqrt{7}-sqrt{21})$.
Why shouldn't I factor out the $sqrt{7}$?
Thanks for your help.
algebra-precalculus
algebra-precalculus
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
edited Dec 5 '18 at 13:10
Chinnapparaj R
5,2481828
5,2481828
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
asked Dec 5 '18 at 12:53
ALLearnerALLearner
62
62
1
math.stackexchange.com/questions/2047349/… math.stackexchange.com/questions/1274936/…
– lab bhattacharjee
Dec 5 '18 at 12:55
I think that, since you need to rationalize the denominator, both your solution and Symbolab answer are perfectly fine.
– rafa11111
Dec 5 '18 at 13:00
add a comment |
1
math.stackexchange.com/questions/2047349/… math.stackexchange.com/questions/1274936/…
– lab bhattacharjee
Dec 5 '18 at 12:55
I think that, since you need to rationalize the denominator, both your solution and Symbolab answer are perfectly fine.
– rafa11111
Dec 5 '18 at 13:00
1
1
math.stackexchange.com/questions/2047349/… math.stackexchange.com/questions/1274936/…
– lab bhattacharjee
Dec 5 '18 at 12:55
math.stackexchange.com/questions/2047349/… math.stackexchange.com/questions/1274936/…
– lab bhattacharjee
Dec 5 '18 at 12:55
I think that, since you need to rationalize the denominator, both your solution and Symbolab answer are perfectly fine.
– rafa11111
Dec 5 '18 at 13:00
I think that, since you need to rationalize the denominator, both your solution and Symbolab answer are perfectly fine.
– rafa11111
Dec 5 '18 at 13:00
add a comment |
1 Answer
1
active
oldest
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The book just wrote the answer in another form.
$$frac{sqrt{7}}{42}cdot(3-sqrt{3}) = frac{1}{42}cdotsqrt 7cdot(3-sqrt 3) = frac{1}{42}cdot(3sqrt 7-sqrt{21})$$
I think your answer is fine as well. You’ve rationalized the denominator, which is precisely what the question asked for. Whether you want to leave your answer as it is or “play around” with it is a matter of choice.
As another note, I think you started off by factoring $sqrt 7$ in the denominator while the solutions given probably involved rationalizing immediately. It makes no difference anyway and both are acceptable.
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
add a comment |
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The book just wrote the answer in another form.
$$frac{sqrt{7}}{42}cdot(3-sqrt{3}) = frac{1}{42}cdotsqrt 7cdot(3-sqrt 3) = frac{1}{42}cdot(3sqrt 7-sqrt{21})$$
I think your answer is fine as well. You’ve rationalized the denominator, which is precisely what the question asked for. Whether you want to leave your answer as it is or “play around” with it is a matter of choice.
As another note, I think you started off by factoring $sqrt 7$ in the denominator while the solutions given probably involved rationalizing immediately. It makes no difference anyway and both are acceptable.
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
add a comment |
The book just wrote the answer in another form.
$$frac{sqrt{7}}{42}cdot(3-sqrt{3}) = frac{1}{42}cdotsqrt 7cdot(3-sqrt 3) = frac{1}{42}cdot(3sqrt 7-sqrt{21})$$
I think your answer is fine as well. You’ve rationalized the denominator, which is precisely what the question asked for. Whether you want to leave your answer as it is or “play around” with it is a matter of choice.
As another note, I think you started off by factoring $sqrt 7$ in the denominator while the solutions given probably involved rationalizing immediately. It makes no difference anyway and both are acceptable.
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
add a comment |
The book just wrote the answer in another form.
$$frac{sqrt{7}}{42}cdot(3-sqrt{3}) = frac{1}{42}cdotsqrt 7cdot(3-sqrt 3) = frac{1}{42}cdot(3sqrt 7-sqrt{21})$$
I think your answer is fine as well. You’ve rationalized the denominator, which is precisely what the question asked for. Whether you want to leave your answer as it is or “play around” with it is a matter of choice.
As another note, I think you started off by factoring $sqrt 7$ in the denominator while the solutions given probably involved rationalizing immediately. It makes no difference anyway and both are acceptable.
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
The book just wrote the answer in another form.
$$frac{sqrt{7}}{42}cdot(3-sqrt{3}) = frac{1}{42}cdotsqrt 7cdot(3-sqrt 3) = frac{1}{42}cdot(3sqrt 7-sqrt{21})$$
I think your answer is fine as well. You’ve rationalized the denominator, which is precisely what the question asked for. Whether you want to leave your answer as it is or “play around” with it is a matter of choice.
As another note, I think you started off by factoring $sqrt 7$ in the denominator while the solutions given probably involved rationalizing immediately. It makes no difference anyway and both are acceptable.
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
answered Dec 5 '18 at 13:04
KM101KM101
5,7511423
5,7511423
add a comment |
add a comment |
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1
math.stackexchange.com/questions/2047349/… math.stackexchange.com/questions/1274936/…
– lab bhattacharjee
Dec 5 '18 at 12:55
I think that, since you need to rationalize the denominator, both your solution and Symbolab answer are perfectly fine.
– rafa11111
Dec 5 '18 at 13:00