Mixing
Simplify $sqrt[4]{frac{162x^6}{16x^4}}$ is $frac{3sqrt[4]{2x^2}}{2}$
$begingroup$
(I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)
I'm asked to simplify $sqrt[4]{frac{162x^6}{16x^4}}$ and am provided the text book solution $frac{3sqrt[4]{2x^2}}{2}$.
I arrived at $frac{3sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.
Here is my working:
$sqrt[4]{frac{162x^6}{16x^4}}$ = $frac{sqrt[4]{162x^6}}{sqrt[4]{16x^4}}$
Denominator: $sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16
Numerator: $sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3sqrt[4]{2}sqrt[4]{x^6}$ since:
$sqrt[4]{162x^6}$ = $sqrt[4]{81}$ * $sqrt[4]{2}$ * $sqrt[4]{x^6}$ = $3 * sqrt[4]{2} * sqrt[4]{x^6}$
Thus I got:
$frac{3sqrt[4]{2}sqrt[4]{x^6}}{2x^4}$ which I think is equal to $frac{3sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).
How ca I arrive at the provided solution $frac{3sqrt[4]{2x^2}}{2}$?
algebra-precalculus radicals
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
$endgroup$
add a comment |
$begingroup$
(I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)
I'm asked to simplify $sqrt[4]{frac{162x^6}{16x^4}}$ and am provided the text book solution $frac{3sqrt[4]{2x^2}}{2}$.
I arrived at $frac{3sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.
Here is my working:
$sqrt[4]{frac{162x^6}{16x^4}}$ = $frac{sqrt[4]{162x^6}}{sqrt[4]{16x^4}}$
Denominator: $sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16
Numerator: $sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3sqrt[4]{2}sqrt[4]{x^6}$ since:
$sqrt[4]{162x^6}$ = $sqrt[4]{81}$ * $sqrt[4]{2}$ * $sqrt[4]{x^6}$ = $3 * sqrt[4]{2} * sqrt[4]{x^6}$
Thus I got:
$frac{3sqrt[4]{2}sqrt[4]{x^6}}{2x^4}$ which I think is equal to $frac{3sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).
How ca I arrive at the provided solution $frac{3sqrt[4]{2x^2}}{2}$?
algebra-precalculus radicals
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
$endgroup$
$begingroup$
Hint, $sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$.
$endgroup$
– Gnumbertester
Jan 6 at 21:05
4
$begingroup$
On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one),
$endgroup$
– Ethan Bolker
Jan 6 at 21:13
add a comment |
$begingroup$
(I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)
I'm asked to simplify $sqrt[4]{frac{162x^6}{16x^4}}$ and am provided the text book solution $frac{3sqrt[4]{2x^2}}{2}$.
I arrived at $frac{3sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.
Here is my working:
$sqrt[4]{frac{162x^6}{16x^4}}$ = $frac{sqrt[4]{162x^6}}{sqrt[4]{16x^4}}$
Denominator: $sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16
Numerator: $sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3sqrt[4]{2}sqrt[4]{x^6}$ since:
$sqrt[4]{162x^6}$ = $sqrt[4]{81}$ * $sqrt[4]{2}$ * $sqrt[4]{x^6}$ = $3 * sqrt[4]{2} * sqrt[4]{x^6}$
Thus I got:
$frac{3sqrt[4]{2}sqrt[4]{x^6}}{2x^4}$ which I think is equal to $frac{3sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).
How ca I arrive at the provided solution $frac{3sqrt[4]{2x^2}}{2}$?
algebra-precalculus radicals
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
$endgroup$
(I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)
I'm asked to simplify $sqrt[4]{frac{162x^6}{16x^4}}$ and am provided the text book solution $frac{3sqrt[4]{2x^2}}{2}$.
I arrived at $frac{3sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.
Here is my working:
$sqrt[4]{frac{162x^6}{16x^4}}$ = $frac{sqrt[4]{162x^6}}{sqrt[4]{16x^4}}$
Denominator: $sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16
Numerator: $sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3sqrt[4]{2}sqrt[4]{x^6}$ since:
$sqrt[4]{162x^6}$ = $sqrt[4]{81}$ * $sqrt[4]{2}$ * $sqrt[4]{x^6}$ = $3 * sqrt[4]{2} * sqrt[4]{x^6}$
Thus I got:
$frac{3sqrt[4]{2}sqrt[4]{x^6}}{2x^4}$ which I think is equal to $frac{3sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).
How ca I arrive at the provided solution $frac{3sqrt[4]{2x^2}}{2}$?
algebra-precalculus radicals
algebra-precalculus radicals
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
edited Jan 6 at 21:11
Michael Rozenberg
99.4k1590189
99.4k1590189
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
asked Jan 6 at 21:00
Doug FirDoug Fir
3227
3227
$begingroup$
Hint, $sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$.
$endgroup$
– Gnumbertester
Jan 6 at 21:05
4
$begingroup$
On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one),
$endgroup$
– Ethan Bolker
Jan 6 at 21:13
add a comment |
$begingroup$
Hint, $sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$.
$endgroup$
– Gnumbertester
Jan 6 at 21:05
4
$begingroup$
On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one),
$endgroup$
– Ethan Bolker
Jan 6 at 21:13
$begingroup$
Hint, $sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$.
$endgroup$
– Gnumbertester
Jan 6 at 21:05
$begingroup$
Hint, $sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$.
$endgroup$
– Gnumbertester
Jan 6 at 21:05
4
4
$begingroup$
On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one),
$endgroup$
– Ethan Bolker
Jan 6 at 21:13
$begingroup$
On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one),
$endgroup$
– Ethan Bolker
Jan 6 at 21:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You made an error:
$$sqrt[4]{16x^4} = 2vert xvert$$
because $sqrt[4]{16x^4} = sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get
$$frac{3sqrt[4]{2x^6}}{2vert xvert} = frac{3sqrt[4]{2x^4x^2}}{2x} = frac{3vert xvertsqrt[4]{2x^2}}{2vert xvert} = frac{3sqrt[4]{2x^2}}{2}$$
As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).
$$sqrt[4]{frac{162x^6}{16x^4}} = sqrt[4]{frac{2cdot3^4x^2}{2^4}} = frac{3sqrt[4]{2x^2}}{2}$$
answered Jan 6 at 21:07
KM101KM101
5,9281524
$endgroup$
$begingroup$
In your second approach you have $sqrt[4]{frac{2cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there?
$endgroup$
– Doug Fir
Jan 6 at 21:24
1
$begingroup$
I simplified $frac{x^6}{x^4}$ first.
$endgroup$
– KM101
Jan 6 at 21:25
$begingroup$
In your first answer, your numerator goes from $3sqrt[4]{2x^4x^2}$ to $3vert xvertsqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute?
$endgroup$
– Doug Fir
Jan 6 at 21:40
$begingroup$
Sure. $sqrt[4]{x^4} = vert xvert$, like how $sqrt{x^2} = vert xvert$, $sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.)
$endgroup$
– KM101
Jan 6 at 21:47
1
$begingroup$
A good way of thinking about it: $$sqrt[4]{2x^6} = sqrt[4]{2x^2}cdotsqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $vert xvert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $sqrt[4]{2}$ would literally be... $sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.)
$endgroup$
– KM101
Jan 6 at 22:00
|
show 3 more comments
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$$sqrt[4]{frac{162x^6}{16x^4}}=sqrt[4]{frac{81cdot2x^2}{16}}=frac{3sqrt[4]{2x^2}}{2}.$$
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You made an error:
$$sqrt[4]{16x^4} = 2vert xvert$$
because $sqrt[4]{16x^4} = sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get
$$frac{3sqrt[4]{2x^6}}{2vert xvert} = frac{3sqrt[4]{2x^4x^2}}{2x} = frac{3vert xvertsqrt[4]{2x^2}}{2vert xvert} = frac{3sqrt[4]{2x^2}}{2}$$
As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).
$$sqrt[4]{frac{162x^6}{16x^4}} = sqrt[4]{frac{2cdot3^4x^2}{2^4}} = frac{3sqrt[4]{2x^2}}{2}$$
answered Jan 6 at 21:07
KM101KM101
5,9281524
$endgroup$
$begingroup$
In your second approach you have $sqrt[4]{frac{2cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there?
$endgroup$
– Doug Fir
Jan 6 at 21:24
1
$begingroup$
I simplified $frac{x^6}{x^4}$ first.
$endgroup$
– KM101
Jan 6 at 21:25
$begingroup$
In your first answer, your numerator goes from $3sqrt[4]{2x^4x^2}$ to $3vert xvertsqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute?
$endgroup$
– Doug Fir
Jan 6 at 21:40
$begingroup$
Sure. $sqrt[4]{x^4} = vert xvert$, like how $sqrt{x^2} = vert xvert$, $sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.)
$endgroup$
– KM101
Jan 6 at 21:47
1
$begingroup$
A good way of thinking about it: $$sqrt[4]{2x^6} = sqrt[4]{2x^2}cdotsqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $vert xvert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $sqrt[4]{2}$ would literally be... $sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.)
$endgroup$
– KM101
Jan 6 at 22:00
|
show 3 more comments
$begingroup$
You made an error:
$$sqrt[4]{16x^4} = 2vert xvert$$
because $sqrt[4]{16x^4} = sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get
$$frac{3sqrt[4]{2x^6}}{2vert xvert} = frac{3sqrt[4]{2x^4x^2}}{2x} = frac{3vert xvertsqrt[4]{2x^2}}{2vert xvert} = frac{3sqrt[4]{2x^2}}{2}$$
As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).
$$sqrt[4]{frac{162x^6}{16x^4}} = sqrt[4]{frac{2cdot3^4x^2}{2^4}} = frac{3sqrt[4]{2x^2}}{2}$$
answered Jan 6 at 21:07
KM101KM101
5,9281524
$endgroup$
$begingroup$
In your second approach you have $sqrt[4]{frac{2cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there?
$endgroup$
– Doug Fir
Jan 6 at 21:24
1
$begingroup$
I simplified $frac{x^6}{x^4}$ first.
$endgroup$
– KM101
Jan 6 at 21:25
$begingroup$
In your first answer, your numerator goes from $3sqrt[4]{2x^4x^2}$ to $3vert xvertsqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute?
$endgroup$
– Doug Fir
Jan 6 at 21:40
$begingroup$
Sure. $sqrt[4]{x^4} = vert xvert$, like how $sqrt{x^2} = vert xvert$, $sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.)
$endgroup$
– KM101
Jan 6 at 21:47
1
$begingroup$
A good way of thinking about it: $$sqrt[4]{2x^6} = sqrt[4]{2x^2}cdotsqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $vert xvert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $sqrt[4]{2}$ would literally be... $sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.)
$endgroup$
– KM101
Jan 6 at 22:00
|
show 3 more comments
$begingroup$
You made an error:
$$sqrt[4]{16x^4} = 2vert xvert$$
because $sqrt[4]{16x^4} = sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get
$$frac{3sqrt[4]{2x^6}}{2vert xvert} = frac{3sqrt[4]{2x^4x^2}}{2x} = frac{3vert xvertsqrt[4]{2x^2}}{2vert xvert} = frac{3sqrt[4]{2x^2}}{2}$$
As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).
$$sqrt[4]{frac{162x^6}{16x^4}} = sqrt[4]{frac{2cdot3^4x^2}{2^4}} = frac{3sqrt[4]{2x^2}}{2}$$
answered Jan 6 at 21:07
KM101KM101
5,9281524
$endgroup$
You made an error:
$$sqrt[4]{16x^4} = 2vert xvert$$
because $sqrt[4]{16x^4} = sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get
$$frac{3sqrt[4]{2x^6}}{2vert xvert} = frac{3sqrt[4]{2x^4x^2}}{2x} = frac{3vert xvertsqrt[4]{2x^2}}{2vert xvert} = frac{3sqrt[4]{2x^2}}{2}$$
As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).
$$sqrt[4]{frac{162x^6}{16x^4}} = sqrt[4]{frac{2cdot3^4x^2}{2^4}} = frac{3sqrt[4]{2x^2}}{2}$$
answered Jan 6 at 21:07
KM101KM101
5,9281524
edited Jan 6 at 21:15
answered Jan 6 at 21:07
KM101KM101
5,9281524
answered Jan 6 at 21:07
KM101KM101
5,9281524
answered Jan 6 at 21:07
KM101KM101
5,9281524
5,9281524
$begingroup$
In your second approach you have $sqrt[4]{frac{2cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there?
$endgroup$
– Doug Fir
Jan 6 at 21:24
1
$begingroup$
I simplified $frac{x^6}{x^4}$ first.
$endgroup$
– KM101
Jan 6 at 21:25
$begingroup$
In your first answer, your numerator goes from $3sqrt[4]{2x^4x^2}$ to $3vert xvertsqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute?
$endgroup$
– Doug Fir
Jan 6 at 21:40
$begingroup$
Sure. $sqrt[4]{x^4} = vert xvert$, like how $sqrt{x^2} = vert xvert$, $sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.)
$endgroup$
– KM101
Jan 6 at 21:47
1
$begingroup$
A good way of thinking about it: $$sqrt[4]{2x^6} = sqrt[4]{2x^2}cdotsqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $vert xvert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $sqrt[4]{2}$ would literally be... $sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.)
$endgroup$
– KM101
Jan 6 at 22:00
|
show 3 more comments
$begingroup$
In your second approach you have $sqrt[4]{frac{2cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there?
$endgroup$
– Doug Fir
Jan 6 at 21:24
1
$begingroup$
I simplified $frac{x^6}{x^4}$ first.
$endgroup$
– KM101
Jan 6 at 21:25
$begingroup$
In your first answer, your numerator goes from $3sqrt[4]{2x^4x^2}$ to $3vert xvertsqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute?
$endgroup$
– Doug Fir
Jan 6 at 21:40
$begingroup$
Sure. $sqrt[4]{x^4} = vert xvert$, like how $sqrt{x^2} = vert xvert$, $sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.)
$endgroup$
– KM101
Jan 6 at 21:47
1
$begingroup$
A good way of thinking about it: $$sqrt[4]{2x^6} = sqrt[4]{2x^2}cdotsqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $vert xvert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $sqrt[4]{2}$ would literally be... $sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.)
$endgroup$
– KM101
Jan 6 at 22:00
$begingroup$
In your second approach you have $sqrt[4]{frac{2cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there?
$endgroup$
– Doug Fir
Jan 6 at 21:24
$begingroup$
In your second approach you have $sqrt[4]{frac{2cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there?
$endgroup$
– Doug Fir
Jan 6 at 21:24
1
1
$begingroup$
I simplified $frac{x^6}{x^4}$ first.
$endgroup$
– KM101
Jan 6 at 21:25
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I simplified $frac{x^6}{x^4}$ first.
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– KM101
Jan 6 at 21:25
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In your first answer, your numerator goes from $3sqrt[4]{2x^4x^2}$ to $3vert xvertsqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute?
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– Doug Fir
Jan 6 at 21:40
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In your first answer, your numerator goes from $3sqrt[4]{2x^4x^2}$ to $3vert xvertsqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute?
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– Doug Fir
Jan 6 at 21:40
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Sure. $sqrt[4]{x^4} = vert xvert$, like how $sqrt{x^2} = vert xvert$, $sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.)
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– KM101
Jan 6 at 21:47
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Sure. $sqrt[4]{x^4} = vert xvert$, like how $sqrt{x^2} = vert xvert$, $sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.)
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– KM101
Jan 6 at 21:47
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A good way of thinking about it: $$sqrt[4]{2x^6} = sqrt[4]{2x^2}cdotsqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $vert xvert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $sqrt[4]{2}$ would literally be... $sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.)
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– KM101
Jan 6 at 22:00
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A good way of thinking about it: $$sqrt[4]{2x^6} = sqrt[4]{2x^2}cdotsqrt[4]{x^4}$$ The first quartic root can’t be simplified, and it is therefore left as it is. The second simplifies to $vert xvert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $sqrt[4]{2}$ would literally be... $sqrt[4]{2}$ again, so you don’t bring it out as it won’t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.)
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– KM101
Jan 6 at 22:00
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show 3 more comments
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$$sqrt[4]{frac{162x^6}{16x^4}}=sqrt[4]{frac{81cdot2x^2}{16}}=frac{3sqrt[4]{2x^2}}{2}.$$
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
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add a comment |
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$$sqrt[4]{frac{162x^6}{16x^4}}=sqrt[4]{frac{81cdot2x^2}{16}}=frac{3sqrt[4]{2x^2}}{2}.$$
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
$endgroup$
add a comment |
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$$sqrt[4]{frac{162x^6}{16x^4}}=sqrt[4]{frac{81cdot2x^2}{16}}=frac{3sqrt[4]{2x^2}}{2}.$$
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
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$$sqrt[4]{frac{162x^6}{16x^4}}=sqrt[4]{frac{81cdot2x^2}{16}}=frac{3sqrt[4]{2x^2}}{2}.$$
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
answered Jan 6 at 21:06
Michael RozenbergMichael Rozenberg
99.4k1590189
99.4k1590189
add a comment |
add a comment |
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Hint, $sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$.
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– Gnumbertester
Jan 6 at 21:05
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On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one),
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– Ethan Bolker
Jan 6 at 21:13