Trying to simplify $frac{sqrt{8}}{1-sqrt{3x}}$ to be $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$
$begingroup$
I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?
algebra-precalculus
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
$endgroup$
add a comment |
$begingroup$
I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?
algebra-precalculus
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
$endgroup$
$begingroup$
Please review your computations. There is a least 2 errors.
$endgroup$
– mathcounterexamples.net
Jan 5 at 18:10
add a comment |
$begingroup$
I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?
algebra-precalculus
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
$endgroup$
I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$
Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?
algebra-precalculus
algebra-precalculus
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
edited Jan 5 at 18:11
Doug Fir
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
asked Jan 5 at 18:07
Doug FirDoug Fir
3177
3177
$begingroup$
Please review your computations. There is a least 2 errors.
$endgroup$
– mathcounterexamples.net
Jan 5 at 18:10
add a comment |
$begingroup$
Please review your computations. There is a least 2 errors.
$endgroup$
– mathcounterexamples.net
Jan 5 at 18:10
$begingroup$
Please review your computations. There is a least 2 errors.
$endgroup$
– mathcounterexamples.net
Jan 5 at 18:10
$begingroup$
Please review your computations. There is a least 2 errors.
$endgroup$
– mathcounterexamples.net
Jan 5 at 18:10
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$
$$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.
answered Jan 5 at 18:27
KM101KM101
5,9261523
$endgroup$
add a comment |
$begingroup$
Observe that
$$
sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
$$ and
$$
sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
$$
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
$endgroup$
add a comment |
$begingroup$
Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.
Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
$endgroup$
add a comment |
$begingroup$
Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
$$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$
$sqrt24=2sqrt6$. (Why?)
$(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
$endgroup$
add a comment |
$begingroup$
You just made some mistakes in your arithmetic.
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
answered Jan 5 at 18:13
greeliousgreelious
19410
$endgroup$
add a comment |
$begingroup$
The second equality is where you mess up - note that
$$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
by the distributive property of real numbers
edited Jan 5 at 18:29
KM101
5,9261523
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
$endgroup$
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$
$$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.
answered Jan 5 at 18:27
KM101KM101
5,9261523
$endgroup$
add a comment |
$begingroup$
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$
$$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.
answered Jan 5 at 18:27
KM101KM101
5,9261523
$endgroup$
add a comment |
$begingroup$
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$
$$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.
answered Jan 5 at 18:27
KM101KM101
5,9261523
$endgroup$
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$
$$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.
answered Jan 5 at 18:27
KM101KM101
5,9261523
answered Jan 5 at 18:27
KM101KM101
5,9261523
answered Jan 5 at 18:27
KM101KM101
5,9261523
answered Jan 5 at 18:27
KM101KM101
5,9261523
5,9261523
add a comment |
add a comment |
$begingroup$
Observe that
$$
sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
$$ and
$$
sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
$$
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
$endgroup$
add a comment |
$begingroup$
Observe that
$$
sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
$$ and
$$
sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
$$
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
$endgroup$
add a comment |
$begingroup$
Observe that
$$
sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
$$ and
$$
sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
$$
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
$endgroup$
Observe that
$$
sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
$$ and
$$
sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
$$
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
answered Jan 5 at 18:10
Olivier OloaOlivier Oloa
108k17176293
108k17176293
add a comment |
add a comment |
$begingroup$
Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.
Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
$endgroup$
add a comment |
$begingroup$
Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.
Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
$endgroup$
add a comment |
$begingroup$
Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.
Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
$endgroup$
Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.
Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
answered Jan 5 at 18:11
GnumbertesterGnumbertester
1675
1675
add a comment |
add a comment |
$begingroup$
Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
$$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
$$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
$$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
$$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 5 at 18:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
73.6k42864
add a comment |
add a comment |
$begingroup$
Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$
$sqrt24=2sqrt6$. (Why?)
$(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
$endgroup$
add a comment |
$begingroup$
Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$
$sqrt24=2sqrt6$. (Why?)
$(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
$endgroup$
add a comment |
$begingroup$
Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$
$sqrt24=2sqrt6$. (Why?)
$(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
$endgroup$
Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$
$sqrt24=2sqrt6$. (Why?)
$(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
answered Jan 5 at 18:13
Thomas ShelbyThomas Shelby
2,102220
2,102220
add a comment |
add a comment |
$begingroup$
You just made some mistakes in your arithmetic.
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
answered Jan 5 at 18:13
greeliousgreelious
19410
$endgroup$
add a comment |
$begingroup$
You just made some mistakes in your arithmetic.
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
answered Jan 5 at 18:13
greeliousgreelious
19410
$endgroup$
add a comment |
$begingroup$
You just made some mistakes in your arithmetic.
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
answered Jan 5 at 18:13
greeliousgreelious
19410
$endgroup$
You just made some mistakes in your arithmetic.
$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$
answered Jan 5 at 18:13
greeliousgreelious
19410
answered Jan 5 at 18:13
greeliousgreelious
19410
answered Jan 5 at 18:13
greeliousgreelious
19410
answered Jan 5 at 18:13
greeliousgreelious
19410
19410
add a comment |
add a comment |
$begingroup$
The second equality is where you mess up - note that
$$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
by the distributive property of real numbers
edited Jan 5 at 18:29
KM101
5,9261523
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
$endgroup$
add a comment |
$begingroup$
The second equality is where you mess up - note that
$$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
by the distributive property of real numbers
edited Jan 5 at 18:29
KM101
5,9261523
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
$endgroup$
add a comment |
$begingroup$
The second equality is where you mess up - note that
$$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
by the distributive property of real numbers
edited Jan 5 at 18:29
KM101
5,9261523
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
$endgroup$
The second equality is where you mess up - note that
$$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
by the distributive property of real numbers
edited Jan 5 at 18:29
KM101
5,9261523
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
edited Jan 5 at 18:29
KM101
5,9261523
edited Jan 5 at 18:29
KM101
5,9261523
edited Jan 5 at 18:29
KM101
5,9261523
5,9261523
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
answered Jan 5 at 18:15
dromastyxdromastyx
2,2901517
2,2901517
add a comment |
add a comment |
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$begingroup$
Please review your computations. There is a least 2 errors.
$endgroup$
– mathcounterexamples.net
Jan 5 at 18:10