Mixing
Please explain why 2x * /2 became 2/2x
$begingroup$
I'm studying a Grade 10 maths book and was surprised to see the correct answer for this question was:
(/2 - 3x)(/3 + 2x) - /5x
= /6 + 2/2x - 3/3x - 6x - /5x
I thought it should be:
= /6 + 2x/2 - 3x/3 - 6x - /5x
I'm using this '/' symbol for the radical sign
Can you please help me understand why the variable x is placed inside the radical sign?
Thank you.
quadratics radicals factoring
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
$endgroup$
|
show 4 more comments
$begingroup$
I'm studying a Grade 10 maths book and was surprised to see the correct answer for this question was:
(/2 - 3x)(/3 + 2x) - /5x
= /6 + 2/2x - 3/3x - 6x - /5x
I thought it should be:
= /6 + 2x/2 - 3x/3 - 6x - /5x
I'm using this '/' symbol for the radical sign
Can you please help me understand why the variable x is placed inside the radical sign?
Thank you.
quadratics radicals factoring
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
$endgroup$
$begingroup$
What you don't understand is that $2xsqrt2=2sqrt2 x$?
$endgroup$
– Martin Argerami
Jan 27 at 0:21
$begingroup$
correct. Not sure why that happens
$endgroup$
– Owen A Bowes
Jan 27 at 0:22
$begingroup$
$2xsqrt{2} = 2cdot x cdot sqrt{2}$. Now... remember that $acdot bcdot c = acdot ccdot b$ by the commutative property of multiplication... so $2cdot xcdot sqrt{2} = 2cdot sqrt{2}cdot x$
$endgroup$
– JMoravitz
Jan 27 at 0:22
$begingroup$
is it equally correct to say 2x/2 in that case?
$endgroup$
– Owen A Bowes
Jan 27 at 0:23
$begingroup$
By /5x, do you mean $sqrt{5x}$ or $sqrt 5cdot x$?
$endgroup$
– timtfj
Jan 27 at 0:24
|
show 4 more comments
$begingroup$
I'm studying a Grade 10 maths book and was surprised to see the correct answer for this question was:
(/2 - 3x)(/3 + 2x) - /5x
= /6 + 2/2x - 3/3x - 6x - /5x
I thought it should be:
= /6 + 2x/2 - 3x/3 - 6x - /5x
I'm using this '/' symbol for the radical sign
Can you please help me understand why the variable x is placed inside the radical sign?
Thank you.
quadratics radicals factoring
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
$endgroup$
I'm studying a Grade 10 maths book and was surprised to see the correct answer for this question was:
(/2 - 3x)(/3 + 2x) - /5x
= /6 + 2/2x - 3/3x - 6x - /5x
I thought it should be:
= /6 + 2x/2 - 3x/3 - 6x - /5x
I'm using this '/' symbol for the radical sign
Can you please help me understand why the variable x is placed inside the radical sign?
Thank you.
quadratics radicals factoring
quadratics radicals factoring
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
edited Jan 27 at 9:15
Owen A Bowes
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
asked Jan 27 at 0:19
Owen A BowesOwen A Bowes
154
154
$begingroup$
What you don't understand is that $2xsqrt2=2sqrt2 x$?
$endgroup$
– Martin Argerami
Jan 27 at 0:21
$begingroup$
correct. Not sure why that happens
$endgroup$
– Owen A Bowes
Jan 27 at 0:22
$begingroup$
$2xsqrt{2} = 2cdot x cdot sqrt{2}$. Now... remember that $acdot bcdot c = acdot ccdot b$ by the commutative property of multiplication... so $2cdot xcdot sqrt{2} = 2cdot sqrt{2}cdot x$
$endgroup$
– JMoravitz
Jan 27 at 0:22
$begingroup$
is it equally correct to say 2x/2 in that case?
$endgroup$
– Owen A Bowes
Jan 27 at 0:23
$begingroup$
By /5x, do you mean $sqrt{5x}$ or $sqrt 5cdot x$?
$endgroup$
– timtfj
Jan 27 at 0:24
|
show 4 more comments
$begingroup$
What you don't understand is that $2xsqrt2=2sqrt2 x$?
$endgroup$
– Martin Argerami
Jan 27 at 0:21
$begingroup$
correct. Not sure why that happens
$endgroup$
– Owen A Bowes
Jan 27 at 0:22
$begingroup$
$2xsqrt{2} = 2cdot x cdot sqrt{2}$. Now... remember that $acdot bcdot c = acdot ccdot b$ by the commutative property of multiplication... so $2cdot xcdot sqrt{2} = 2cdot sqrt{2}cdot x$
$endgroup$
– JMoravitz
Jan 27 at 0:22
$begingroup$
is it equally correct to say 2x/2 in that case?
$endgroup$
– Owen A Bowes
Jan 27 at 0:23
$begingroup$
By /5x, do you mean $sqrt{5x}$ or $sqrt 5cdot x$?
$endgroup$
– timtfj
Jan 27 at 0:24
$begingroup$
What you don't understand is that $2xsqrt2=2sqrt2 x$?
$endgroup$
– Martin Argerami
Jan 27 at 0:21
$begingroup$
What you don't understand is that $2xsqrt2=2sqrt2 x$?
$endgroup$
– Martin Argerami
Jan 27 at 0:21
$begingroup$
correct. Not sure why that happens
$endgroup$
– Owen A Bowes
Jan 27 at 0:22
$begingroup$
correct. Not sure why that happens
$endgroup$
– Owen A Bowes
Jan 27 at 0:22
$begingroup$
$2xsqrt{2} = 2cdot x cdot sqrt{2}$. Now... remember that $acdot bcdot c = acdot ccdot b$ by the commutative property of multiplication... so $2cdot xcdot sqrt{2} = 2cdot sqrt{2}cdot x$
$endgroup$
– JMoravitz
Jan 27 at 0:22
$begingroup$
$2xsqrt{2} = 2cdot x cdot sqrt{2}$. Now... remember that $acdot bcdot c = acdot ccdot b$ by the commutative property of multiplication... so $2cdot xcdot sqrt{2} = 2cdot sqrt{2}cdot x$
$endgroup$
– JMoravitz
Jan 27 at 0:22
$begingroup$
is it equally correct to say 2x/2 in that case?
$endgroup$
– Owen A Bowes
Jan 27 at 0:23
$begingroup$
is it equally correct to say 2x/2 in that case?
$endgroup$
– Owen A Bowes
Jan 27 at 0:23
$begingroup$
By /5x, do you mean $sqrt{5x}$ or $sqrt 5cdot x$?
$endgroup$
– timtfj
Jan 27 at 0:24
$begingroup$
By /5x, do you mean $sqrt{5x}$ or $sqrt 5cdot x$?
$endgroup$
– timtfj
Jan 27 at 0:24
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Summarizing comments above., by the commutative property of multiplication (which says that $atimes b = btimes a$ for any reals $a,b$) you have that:
$2times x times sqrt{2} = 2times sqrt{2} times x$
or, written with fewer multiplication signs but meaning the same thing:
$2xtimes sqrt{2} = 2sqrt{2} ~x$
"can you help me understand why $x$ is placed inside the radical sign?"
$x$ does not go inside of the radical sign. Note the difference in meaning of $sqrt{2x}=sqrt{(2x)}$ and $sqrt{2}~x=(sqrt{2})~x$. The length of the horizontal bar of the radical denotes which terms following are included inside of the radical versus outside of the radical.
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Summarizing comments above., by the commutative property of multiplication (which says that $atimes b = btimes a$ for any reals $a,b$) you have that:
$2times x times sqrt{2} = 2times sqrt{2} times x$
or, written with fewer multiplication signs but meaning the same thing:
$2xtimes sqrt{2} = 2sqrt{2} ~x$
"can you help me understand why $x$ is placed inside the radical sign?"
$x$ does not go inside of the radical sign. Note the difference in meaning of $sqrt{2x}=sqrt{(2x)}$ and $sqrt{2}~x=(sqrt{2})~x$. The length of the horizontal bar of the radical denotes which terms following are included inside of the radical versus outside of the radical.
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
$endgroup$
add a comment |
$begingroup$
Summarizing comments above., by the commutative property of multiplication (which says that $atimes b = btimes a$ for any reals $a,b$) you have that:
$2times x times sqrt{2} = 2times sqrt{2} times x$
or, written with fewer multiplication signs but meaning the same thing:
$2xtimes sqrt{2} = 2sqrt{2} ~x$
"can you help me understand why $x$ is placed inside the radical sign?"
$x$ does not go inside of the radical sign. Note the difference in meaning of $sqrt{2x}=sqrt{(2x)}$ and $sqrt{2}~x=(sqrt{2})~x$. The length of the horizontal bar of the radical denotes which terms following are included inside of the radical versus outside of the radical.
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
$endgroup$
add a comment |
$begingroup$
Summarizing comments above., by the commutative property of multiplication (which says that $atimes b = btimes a$ for any reals $a,b$) you have that:
$2times x times sqrt{2} = 2times sqrt{2} times x$
or, written with fewer multiplication signs but meaning the same thing:
$2xtimes sqrt{2} = 2sqrt{2} ~x$
"can you help me understand why $x$ is placed inside the radical sign?"
$x$ does not go inside of the radical sign. Note the difference in meaning of $sqrt{2x}=sqrt{(2x)}$ and $sqrt{2}~x=(sqrt{2})~x$. The length of the horizontal bar of the radical denotes which terms following are included inside of the radical versus outside of the radical.
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
$endgroup$
Summarizing comments above., by the commutative property of multiplication (which says that $atimes b = btimes a$ for any reals $a,b$) you have that:
$2times x times sqrt{2} = 2times sqrt{2} times x$
or, written with fewer multiplication signs but meaning the same thing:
$2xtimes sqrt{2} = 2sqrt{2} ~x$
"can you help me understand why $x$ is placed inside the radical sign?"
$x$ does not go inside of the radical sign. Note the difference in meaning of $sqrt{2x}=sqrt{(2x)}$ and $sqrt{2}~x=(sqrt{2})~x$. The length of the horizontal bar of the radical denotes which terms following are included inside of the radical versus outside of the radical.
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
answered Jan 27 at 0:32
JMoravitzJMoravitz
48.8k43988
48.8k43988
add a comment |
add a comment |
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$begingroup$
What you don't understand is that $2xsqrt2=2sqrt2 x$?
$endgroup$
– Martin Argerami
Jan 27 at 0:21
$begingroup$
correct. Not sure why that happens
$endgroup$
– Owen A Bowes
Jan 27 at 0:22
$begingroup$
$2xsqrt{2} = 2cdot x cdot sqrt{2}$. Now... remember that $acdot bcdot c = acdot ccdot b$ by the commutative property of multiplication... so $2cdot xcdot sqrt{2} = 2cdot sqrt{2}cdot x$
$endgroup$
– JMoravitz
Jan 27 at 0:22
$begingroup$
is it equally correct to say 2x/2 in that case?
$endgroup$
– Owen A Bowes
Jan 27 at 0:23
$begingroup$
By /5x, do you mean $sqrt{5x}$ or $sqrt 5cdot x$?
$endgroup$
– timtfj
Jan 27 at 0:24