Mixing
Trying to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ into $frac{-5sqrt{2}-6}{7}$
$begingroup$
I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:
$frac{2sqrt{2}-4}{4-sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$
This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?
algebra-precalculus radicals
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
$endgroup$
add a comment |
$begingroup$
I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:
$frac{2sqrt{2}-4}{4-sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$
This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?
algebra-precalculus radicals
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
$endgroup$
1
$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48
add a comment |
$begingroup$
I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:
$frac{2sqrt{2}-4}{4-sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$
This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?
algebra-precalculus radicals
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
$endgroup$
I'm asked to simplify $frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2}$ and am provided with the solution $frac{-5sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2sqrt{2}$ thus:
$frac{2sqrt{2}-4}{4-sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$frac{2sqrt{2}-4}{4-sqrt{2}}$ = $frac{2sqrt{2}-4}{4-sqrt{2}} * frac{4+sqrt{2}}{4+sqrt{2}}$ = $frac{(2sqrt{2}-4)(4+sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$frac{8sqrt{2}*2(sqrt{2}^2)-16-4sqrt{2}}{14}$ = $frac{8sqrt{2}*4-16-4sqrt{2}}{14}$ = $frac{32sqrt{2}-16-4sqrt{2}}{14}$ = $frac{28sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $sqrt{2}$
This is as far as I can get. I don't know if $frac{28sqrt{2}-16}{14}-sqrt{2}$ is still correct or close to the solution. How can I arrive at $frac{-5sqrt{2}-6}{7}$?
algebra-precalculus radicals
algebra-precalculus radicals
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
edited Jan 7 at 10:22
Martin Sleziak
44.7k9117272
44.7k9117272
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
asked Jan 7 at 2:44
Doug FirDoug Fir
3227
3227
1
$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48
add a comment |
1
$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48
1
1
$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48
$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You were doing fine until the place where you tried to expand
$(2sqrt2 - 4)(4 + sqrt2).$
There are mnemonic techniques for this but I think plain old distributive law works well enough:
begin{align}
(2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
&= (8sqrt2 - 16) + (4 - 4sqrt2) \
&= 4sqrt2 - 12.
end{align}
Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$frac{4sqrt2 - 12}{14}.$
And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.
answered Jan 7 at 3:20
David KDavid K
53.8k342116
$endgroup$
$begingroup$
Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
$endgroup$
– Doug Fir
Jan 7 at 3:38
1
$begingroup$
It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
$endgroup$
– David K
Jan 7 at 3:42
$begingroup$
Hi David, thanks for clarifying that, I understand it now
$endgroup$
– Doug Fir
Jan 7 at 4:01
add a comment |
$begingroup$
$$begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
&=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
&=frac{-10sqrt{2}-12}{14}\
&=frac{-5sqrt{2}-6}{7}
end{align}$$
answered Jan 7 at 3:01
LarryLarry
2,31431028
$endgroup$
$begingroup$
Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
$endgroup$
– Doug Fir
Jan 7 at 3:12
add a comment |
$begingroup$
begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
& = frac{4sqrt{2}-12}{14} - sqrt{2} \
& = frac{2sqrt{2}-6}{7} - sqrt{2} \
& = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
& = frac{-5sqrt{2} -6 }{7}
end{align}
answered Jan 7 at 9:08
kelalakakelalaka
3261212
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You were doing fine until the place where you tried to expand
$(2sqrt2 - 4)(4 + sqrt2).$
There are mnemonic techniques for this but I think plain old distributive law works well enough:
begin{align}
(2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
&= (8sqrt2 - 16) + (4 - 4sqrt2) \
&= 4sqrt2 - 12.
end{align}
Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$frac{4sqrt2 - 12}{14}.$
And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.
answered Jan 7 at 3:20
David KDavid K
53.8k342116
$endgroup$
$begingroup$
Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
$endgroup$
– Doug Fir
Jan 7 at 3:38
1
$begingroup$
It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
$endgroup$
– David K
Jan 7 at 3:42
$begingroup$
Hi David, thanks for clarifying that, I understand it now
$endgroup$
– Doug Fir
Jan 7 at 4:01
add a comment |
$begingroup$
You were doing fine until the place where you tried to expand
$(2sqrt2 - 4)(4 + sqrt2).$
There are mnemonic techniques for this but I think plain old distributive law works well enough:
begin{align}
(2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
&= (8sqrt2 - 16) + (4 - 4sqrt2) \
&= 4sqrt2 - 12.
end{align}
Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$frac{4sqrt2 - 12}{14}.$
And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.
answered Jan 7 at 3:20
David KDavid K
53.8k342116
$endgroup$
$begingroup$
Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
$endgroup$
– Doug Fir
Jan 7 at 3:38
1
$begingroup$
It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
$endgroup$
– David K
Jan 7 at 3:42
$begingroup$
Hi David, thanks for clarifying that, I understand it now
$endgroup$
– Doug Fir
Jan 7 at 4:01
add a comment |
$begingroup$
You were doing fine until the place where you tried to expand
$(2sqrt2 - 4)(4 + sqrt2).$
There are mnemonic techniques for this but I think plain old distributive law works well enough:
begin{align}
(2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
&= (8sqrt2 - 16) + (4 - 4sqrt2) \
&= 4sqrt2 - 12.
end{align}
Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$frac{4sqrt2 - 12}{14}.$
And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.
answered Jan 7 at 3:20
David KDavid K
53.8k342116
$endgroup$
You were doing fine until the place where you tried to expand
$(2sqrt2 - 4)(4 + sqrt2).$
There are mnemonic techniques for this but I think plain old distributive law works well enough:
begin{align}
(2sqrt2 - 4)(4 + sqrt2) &= (2sqrt2 - 4)4 + (2sqrt2 - 4)sqrt2 \
&= (8sqrt2 - 16) + (4 - 4sqrt2) \
&= 4sqrt2 - 12.
end{align}
Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$frac{4sqrt2 - 12}{14}.$
And finally you'll want to change the $-sqrt2$ so that you have two fractions with a common denominator and can finish.
answered Jan 7 at 3:20
David KDavid K
53.8k342116
answered Jan 7 at 3:20
David KDavid K
53.8k342116
answered Jan 7 at 3:20
David KDavid K
53.8k342116
answered Jan 7 at 3:20
David KDavid K
53.8k342116
53.8k342116
$begingroup$
Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
$endgroup$
– Doug Fir
Jan 7 at 3:38
1
$begingroup$
It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
$endgroup$
– David K
Jan 7 at 3:42
$begingroup$
Hi David, thanks for clarifying that, I understand it now
$endgroup$
– Doug Fir
Jan 7 at 4:01
add a comment |
$begingroup$
Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
$endgroup$
– Doug Fir
Jan 7 at 3:38
1
$begingroup$
It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
$endgroup$
– David K
Jan 7 at 3:42
$begingroup$
Hi David, thanks for clarifying that, I understand it now
$endgroup$
– Doug Fir
Jan 7 at 4:01
$begingroup$
Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
$endgroup$
– Doug Fir
Jan 7 at 3:38
$begingroup$
Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here?
$endgroup$
– Doug Fir
Jan 7 at 3:38
1
1
$begingroup$
It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
$endgroup$
– David K
Jan 7 at 3:42
$begingroup$
It's the same rule you would apply to simplify something like $frac37 - 2.$ The $2$ is equal to $frac21,$ which is equal to $frac{7cdot2}{7cdot1}.$ In your problem you have $sqrt2$ instead of $2$ but the principle is the same.
$endgroup$
– David K
Jan 7 at 3:42
$begingroup$
Hi David, thanks for clarifying that, I understand it now
$endgroup$
– Doug Fir
Jan 7 at 4:01
$begingroup$
Hi David, thanks for clarifying that, I understand it now
$endgroup$
– Doug Fir
Jan 7 at 4:01
add a comment |
$begingroup$
$$begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
&=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
&=frac{-10sqrt{2}-12}{14}\
&=frac{-5sqrt{2}-6}{7}
end{align}$$
answered Jan 7 at 3:01
LarryLarry
2,31431028
$endgroup$
$begingroup$
Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
$endgroup$
– Doug Fir
Jan 7 at 3:12
add a comment |
$begingroup$
$$begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
&=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
&=frac{-10sqrt{2}-12}{14}\
&=frac{-5sqrt{2}-6}{7}
end{align}$$
answered Jan 7 at 3:01
LarryLarry
2,31431028
$endgroup$
$begingroup$
Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
$endgroup$
– Doug Fir
Jan 7 at 3:12
add a comment |
$begingroup$
$$begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
&=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
&=frac{-10sqrt{2}-12}{14}\
&=frac{-5sqrt{2}-6}{7}
end{align}$$
answered Jan 7 at 3:01
LarryLarry
2,31431028
$endgroup$
$$begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}}-sqrt{2}&=frac{2sqrt{2}-4}{4-sqrt{2}}-sqrt{2}\
&=frac{2sqrt{2}-4}{4-sqrt{2}}-frac{4sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}\
&=frac{-2sqrt{2}-2}{4-sqrt{2}}~cdot~frac{4+sqrt{2}}{4+sqrt{2}}\
&=frac{-10sqrt{2}-12}{14}\
&=frac{-5sqrt{2}-6}{7}
end{align}$$
answered Jan 7 at 3:01
LarryLarry
2,31431028
answered Jan 7 at 3:01
LarryLarry
2,31431028
answered Jan 7 at 3:01
LarryLarry
2,31431028
answered Jan 7 at 3:01
LarryLarry
2,31431028
2,31431028
$begingroup$
Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
$endgroup$
– Doug Fir
Jan 7 at 3:12
add a comment |
$begingroup$
Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
$endgroup$
– Doug Fir
Jan 7 at 3:12
$begingroup$
Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
$endgroup$
– Doug Fir
Jan 7 at 3:12
$begingroup$
Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $frac{4sqrt{2}-2}{4sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-sqrt{2}$?
$endgroup$
– Doug Fir
Jan 7 at 3:12
add a comment |
$begingroup$
begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
& = frac{4sqrt{2}-12}{14} - sqrt{2} \
& = frac{2sqrt{2}-6}{7} - sqrt{2} \
& = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
& = frac{-5sqrt{2} -6 }{7}
end{align}
answered Jan 7 at 9:08
kelalakakelalaka
3261212
$endgroup$
add a comment |
$begingroup$
begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
& = frac{4sqrt{2}-12}{14} - sqrt{2} \
& = frac{2sqrt{2}-6}{7} - sqrt{2} \
& = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
& = frac{-5sqrt{2} -6 }{7}
end{align}
answered Jan 7 at 9:08
kelalakakelalaka
3261212
$endgroup$
add a comment |
$begingroup$
begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
& = frac{4sqrt{2}-12}{14} - sqrt{2} \
& = frac{2sqrt{2}-6}{7} - sqrt{2} \
& = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
& = frac{-5sqrt{2} -6 }{7}
end{align}
answered Jan 7 at 9:08
kelalakakelalaka
3261212
$endgroup$
begin{align}
frac{sqrt{8}-sqrt{16}}{4-sqrt{2}} - 2^{1/2} & = frac{2sqrt{2}-4}{4-sqrt{2}}cdot frac{4+sqrt{2}}{4+sqrt{2}} - sqrt{2} \
& = frac{4sqrt{2}-12}{14} - sqrt{2} \
& = frac{2sqrt{2}-6}{7} - sqrt{2} \
& = frac{2sqrt{2}-6 -7 sqrt{2}}{7}\
& = frac{-5sqrt{2} -6 }{7}
end{align}
answered Jan 7 at 9:08
kelalakakelalaka
3261212
answered Jan 7 at 9:08
kelalakakelalaka
3261212
answered Jan 7 at 9:08
kelalakakelalaka
3261212
answered Jan 7 at 9:08
kelalakakelalaka
3261212
3261212
add a comment |
add a comment |
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1
$begingroup$
Did you mean $8sqrt{2} ast 2 ast (sqrt{2})^2$? I think you should have two terms here
$endgroup$
– bounceback
Jan 7 at 2:48