How do you simplify a radical equation? [closed]
2
$begingroup$
I am having trouble figuring out how this equation was simplified. I have looked everywhere on the internet. No luck. Thank you in advance.
algebra-precalculus
edited Dec 13 '18 at 11:52
amWhy
1
asked Dec 12 '18 at 23:59
user2984143user2984143
163
$endgroup$
closed as off-topic by JMoravitz, Jyrki Lahtonen, KReiser, Leucippus, Shailesh Dec 13 '18 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
2
$begingroup$
I am having trouble figuring out how this equation was simplified. I have looked everywhere on the internet. No luck. Thank you in advance.
algebra-precalculus
edited Dec 13 '18 at 11:52
amWhy
1
asked Dec 12 '18 at 23:59
user2984143user2984143
163
$endgroup$
closed as off-topic by JMoravitz, Jyrki Lahtonen, KReiser, Leucippus, Shailesh Dec 13 '18 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What equation? I don't see one.
$endgroup$
– Ross Millikan
Dec 13 '18 at 0:04
$begingroup$
Your original post had reference to the problem of presumably finding the real value(s) of $x$ such that $8=sqrt{3x}$. First, notice that $x$ must be positive (otherwise you are taking the square root of a negative number). Next, you can square each side, giving $(8)^2=left(sqrt{3x}right)^2$ or in other words $64 = 3x$. Continue from there.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:06
$begingroup$
Now that you have posted the image, we see that in fact you did not mean to have $sqrt{3x}$ (where both the $3$ AND the $x$ are under the radical) but rather you meant to have $sqrt{3} times x$ (where the $x$ is outside of the radical). You can continue exactly as you would if these were integers or rational numbers. $8=sqrt{3}~x$, you can divide both sides by $sqrt{3}$ to get $frac{8}{sqrt{3}} = x$. From here, if you so desire, you can "multiply the left side by $1$" to simplify, here by multiplying by $frac{sqrt{3}}{sqrt{3}}$ and continuing to simplify.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:08
$begingroup$
Hi, I got the same answer. 64=3x but then on the book is showing something else. I added a picture, please kindly see it. Thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:08
$begingroup$
They are just rationalizing the denominator by multiplying the top and bottom by $sqrt 3$
$endgroup$
– Andrew Li
Dec 13 '18 at 0:09
|
show 1 more comment
2
2
2
$begingroup$
I am having trouble figuring out how this equation was simplified. I have looked everywhere on the internet. No luck. Thank you in advance.
algebra-precalculus
edited Dec 13 '18 at 11:52
amWhy
1
asked Dec 12 '18 at 23:59
user2984143user2984143
163
$endgroup$
I am having trouble figuring out how this equation was simplified. I have looked everywhere on the internet. No luck. Thank you in advance.
algebra-precalculus
algebra-precalculus
edited Dec 13 '18 at 11:52
amWhy
1
asked Dec 12 '18 at 23:59
user2984143user2984143
163
edited Dec 13 '18 at 11:52
amWhy
1
asked Dec 12 '18 at 23:59
user2984143user2984143
163
edited Dec 13 '18 at 11:52
amWhy
1
edited Dec 13 '18 at 11:52
amWhy
1
edited Dec 13 '18 at 11:52
amWhy
1
1
asked Dec 12 '18 at 23:59
user2984143user2984143
163
asked Dec 12 '18 at 23:59
user2984143user2984143
163
asked Dec 12 '18 at 23:59
user2984143user2984143
163
163
closed as off-topic by JMoravitz, Jyrki Lahtonen, KReiser, Leucippus, Shailesh Dec 13 '18 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by JMoravitz, Jyrki Lahtonen, KReiser, Leucippus, Shailesh Dec 13 '18 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, KReiser, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What equation? I don't see one.
$endgroup$
– Ross Millikan
Dec 13 '18 at 0:04
$begingroup$
Your original post had reference to the problem of presumably finding the real value(s) of $x$ such that $8=sqrt{3x}$. First, notice that $x$ must be positive (otherwise you are taking the square root of a negative number). Next, you can square each side, giving $(8)^2=left(sqrt{3x}right)^2$ or in other words $64 = 3x$. Continue from there.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:06
$begingroup$
Now that you have posted the image, we see that in fact you did not mean to have $sqrt{3x}$ (where both the $3$ AND the $x$ are under the radical) but rather you meant to have $sqrt{3} times x$ (where the $x$ is outside of the radical). You can continue exactly as you would if these were integers or rational numbers. $8=sqrt{3}~x$, you can divide both sides by $sqrt{3}$ to get $frac{8}{sqrt{3}} = x$. From here, if you so desire, you can "multiply the left side by $1$" to simplify, here by multiplying by $frac{sqrt{3}}{sqrt{3}}$ and continuing to simplify.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:08
$begingroup$
Hi, I got the same answer. 64=3x but then on the book is showing something else. I added a picture, please kindly see it. Thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:08
$begingroup$
They are just rationalizing the denominator by multiplying the top and bottom by $sqrt 3$
$endgroup$
– Andrew Li
Dec 13 '18 at 0:09
|
show 1 more comment
$begingroup$
What equation? I don't see one.
$endgroup$
– Ross Millikan
Dec 13 '18 at 0:04
$begingroup$
Your original post had reference to the problem of presumably finding the real value(s) of $x$ such that $8=sqrt{3x}$. First, notice that $x$ must be positive (otherwise you are taking the square root of a negative number). Next, you can square each side, giving $(8)^2=left(sqrt{3x}right)^2$ or in other words $64 = 3x$. Continue from there.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:06
$begingroup$
Now that you have posted the image, we see that in fact you did not mean to have $sqrt{3x}$ (where both the $3$ AND the $x$ are under the radical) but rather you meant to have $sqrt{3} times x$ (where the $x$ is outside of the radical). You can continue exactly as you would if these were integers or rational numbers. $8=sqrt{3}~x$, you can divide both sides by $sqrt{3}$ to get $frac{8}{sqrt{3}} = x$. From here, if you so desire, you can "multiply the left side by $1$" to simplify, here by multiplying by $frac{sqrt{3}}{sqrt{3}}$ and continuing to simplify.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:08
$begingroup$
Hi, I got the same answer. 64=3x but then on the book is showing something else. I added a picture, please kindly see it. Thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:08
$begingroup$
They are just rationalizing the denominator by multiplying the top and bottom by $sqrt 3$
$endgroup$
– Andrew Li
Dec 13 '18 at 0:09
$begingroup$
What equation? I don't see one.
$endgroup$
– Ross Millikan
Dec 13 '18 at 0:04
$begingroup$
What equation? I don't see one.
$endgroup$
– Ross Millikan
Dec 13 '18 at 0:04
$begingroup$
Your original post had reference to the problem of presumably finding the real value(s) of $x$ such that $8=sqrt{3x}$. First, notice that $x$ must be positive (otherwise you are taking the square root of a negative number). Next, you can square each side, giving $(8)^2=left(sqrt{3x}right)^2$ or in other words $64 = 3x$. Continue from there.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:06
$begingroup$
Your original post had reference to the problem of presumably finding the real value(s) of $x$ such that $8=sqrt{3x}$. First, notice that $x$ must be positive (otherwise you are taking the square root of a negative number). Next, you can square each side, giving $(8)^2=left(sqrt{3x}right)^2$ or in other words $64 = 3x$. Continue from there.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:06
$begingroup$
Now that you have posted the image, we see that in fact you did not mean to have $sqrt{3x}$ (where both the $3$ AND the $x$ are under the radical) but rather you meant to have $sqrt{3} times x$ (where the $x$ is outside of the radical). You can continue exactly as you would if these were integers or rational numbers. $8=sqrt{3}~x$, you can divide both sides by $sqrt{3}$ to get $frac{8}{sqrt{3}} = x$. From here, if you so desire, you can "multiply the left side by $1$" to simplify, here by multiplying by $frac{sqrt{3}}{sqrt{3}}$ and continuing to simplify.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:08
$begingroup$
Now that you have posted the image, we see that in fact you did not mean to have $sqrt{3x}$ (where both the $3$ AND the $x$ are under the radical) but rather you meant to have $sqrt{3} times x$ (where the $x$ is outside of the radical). You can continue exactly as you would if these were integers or rational numbers. $8=sqrt{3}~x$, you can divide both sides by $sqrt{3}$ to get $frac{8}{sqrt{3}} = x$. From here, if you so desire, you can "multiply the left side by $1$" to simplify, here by multiplying by $frac{sqrt{3}}{sqrt{3}}$ and continuing to simplify.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:08
$begingroup$
Hi, I got the same answer. 64=3x but then on the book is showing something else. I added a picture, please kindly see it. Thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:08
$begingroup$
Hi, I got the same answer. 64=3x but then on the book is showing something else. I added a picture, please kindly see it. Thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:08
$begingroup$
They are just rationalizing the denominator by multiplying the top and bottom by $sqrt 3$
$endgroup$
– Andrew Li
Dec 13 '18 at 0:09
$begingroup$
They are just rationalizing the denominator by multiplying the top and bottom by $sqrt 3$
$endgroup$
– Andrew Li
Dec 13 '18 at 0:09
|
show 1 more comment
1 Answer
1
active
oldest
votes
1
$begingroup$
We have that
$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Perfect.! thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:13
$begingroup$
@user2984143 That's a standard manipulation. Another similar is for example $$frac1{sqrt 2+1}=frac1{sqrt 2+1}frac{sqrt 2-1}{sqrt 2-1}=frac{sqrt 2-1}{2-1}={sqrt 2-1}$$ using that $(A+B)(A-B)=A^2-B^2$.
$endgroup$
– gimusi
Dec 13 '18 at 0:15
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
We have that
$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Perfect.! thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:13
$begingroup$
@user2984143 That's a standard manipulation. Another similar is for example $$frac1{sqrt 2+1}=frac1{sqrt 2+1}frac{sqrt 2-1}{sqrt 2-1}=frac{sqrt 2-1}{2-1}={sqrt 2-1}$$ using that $(A+B)(A-B)=A^2-B^2$.
$endgroup$
– gimusi
Dec 13 '18 at 0:15
add a comment |
1
$begingroup$
We have that
$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Perfect.! thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:13
$begingroup$
@user2984143 That's a standard manipulation. Another similar is for example $$frac1{sqrt 2+1}=frac1{sqrt 2+1}frac{sqrt 2-1}{sqrt 2-1}=frac{sqrt 2-1}{2-1}={sqrt 2-1}$$ using that $(A+B)(A-B)=A^2-B^2$.
$endgroup$
– gimusi
Dec 13 '18 at 0:15
add a comment |
1
1
1
$begingroup$
We have that
$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
$endgroup$
We have that
$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
answered Dec 13 '18 at 0:08
gimusigimusi
92.9k84494
92.9k84494
$begingroup$
Perfect.! thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:13
$begingroup$
@user2984143 That's a standard manipulation. Another similar is for example $$frac1{sqrt 2+1}=frac1{sqrt 2+1}frac{sqrt 2-1}{sqrt 2-1}=frac{sqrt 2-1}{2-1}={sqrt 2-1}$$ using that $(A+B)(A-B)=A^2-B^2$.
$endgroup$
– gimusi
Dec 13 '18 at 0:15
add a comment |
$begingroup$
Perfect.! thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:13
$begingroup$
@user2984143 That's a standard manipulation. Another similar is for example $$frac1{sqrt 2+1}=frac1{sqrt 2+1}frac{sqrt 2-1}{sqrt 2-1}=frac{sqrt 2-1}{2-1}={sqrt 2-1}$$ using that $(A+B)(A-B)=A^2-B^2$.
$endgroup$
– gimusi
Dec 13 '18 at 0:15
$begingroup$
Perfect.! thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:13
$begingroup$
Perfect.! thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:13
$begingroup$
@user2984143 That's a standard manipulation. Another similar is for example $$frac1{sqrt 2+1}=frac1{sqrt 2+1}frac{sqrt 2-1}{sqrt 2-1}=frac{sqrt 2-1}{2-1}={sqrt 2-1}$$ using that $(A+B)(A-B)=A^2-B^2$.
$endgroup$
– gimusi
Dec 13 '18 at 0:15
$begingroup$
@user2984143 That's a standard manipulation. Another similar is for example $$frac1{sqrt 2+1}=frac1{sqrt 2+1}frac{sqrt 2-1}{sqrt 2-1}=frac{sqrt 2-1}{2-1}={sqrt 2-1}$$ using that $(A+B)(A-B)=A^2-B^2$.
$endgroup$
– gimusi
Dec 13 '18 at 0:15
add a comment |
$begingroup$
What equation? I don't see one.
$endgroup$
– Ross Millikan
Dec 13 '18 at 0:04
$begingroup$
Your original post had reference to the problem of presumably finding the real value(s) of $x$ such that $8=sqrt{3x}$. First, notice that $x$ must be positive (otherwise you are taking the square root of a negative number). Next, you can square each side, giving $(8)^2=left(sqrt{3x}right)^2$ or in other words $64 = 3x$. Continue from there.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:06
$begingroup$
Now that you have posted the image, we see that in fact you did not mean to have $sqrt{3x}$ (where both the $3$ AND the $x$ are under the radical) but rather you meant to have $sqrt{3} times x$ (where the $x$ is outside of the radical). You can continue exactly as you would if these were integers or rational numbers. $8=sqrt{3}~x$, you can divide both sides by $sqrt{3}$ to get $frac{8}{sqrt{3}} = x$. From here, if you so desire, you can "multiply the left side by $1$" to simplify, here by multiplying by $frac{sqrt{3}}{sqrt{3}}$ and continuing to simplify.
$endgroup$
– JMoravitz
Dec 13 '18 at 0:08
$begingroup$
Hi, I got the same answer. 64=3x but then on the book is showing something else. I added a picture, please kindly see it. Thank you
$endgroup$
– user2984143
Dec 13 '18 at 0:08
$begingroup$
They are just rationalizing the denominator by multiplying the top and bottom by $sqrt 3$
$endgroup$
– Andrew Li
Dec 13 '18 at 0:09