How do you simplify a radical equation? [closed]

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Multi tool use












2












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I am having trouble figuring out how this equation was simplified. I have looked everywhere on the internet. No luck. Thank you in advance.



enter image description here










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$endgroup$



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
















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.



enter image description here










share|cite|improve this question











$endgroup$



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














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.



enter image description here










share|cite|improve this question











$endgroup$




I am having trouble figuring out how this equation was simplified. I have looked everywhere on the internet. No luck. Thank you in advance.



enter image description here







algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 11:52









amWhy

1




1










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

We have that



$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

We have that



$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

We have that



$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

We have that



$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$






share|cite|improve this answer









$endgroup$



We have that



$$8=sqrt 3xiff x = frac{8}{sqrt 3}iff x = frac{8}{sqrt 3}frac{sqrt 3}{sqrt 3}=frac {8sqrt 3} 3 $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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





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