Mixing
$frac{sqrt{12x}}{2+2sqrt{3}}$ simplifies to $frac{3sqrt{x}-sqrt{3x}}{2}$ but I get $-12x + 6xsqrt{3}$
$begingroup$
I am asked to simplify $frac{sqrt{12x}}{2+2sqrt{3}}$ and the solution is provided as $frac{3sqrt{x}-sqrt{3x}}{2}$. I arrived at $-12x + 6xsqrt{3}$ and I'm not sure how to arrive at the text book solution.
My working:
$$frac{sqrt{12x}}{2+2sqrt{3}} = frac{sqrt{12x}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}} = frac{12x(2-sqrt{3})}{(2+2sqrt{3})(2-sqrt{3})} = frac{24x-12xsqrt{3}}{4+(2cdot(-3))}=frac{24x-12xsqrt{3}}{-2}$$
Then, multiplying out the denominator I get:
$-12x+6xsqrt{3}$
Is my thought process sound up to a point? Where did I go wrong and how can I arrive at $frac{3sqrt{x}-sqrt{3x}}{2}$?
algebra-precalculus
edited Jan 6 at 19:07
egreg
180k1485202
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
$endgroup$
add a comment |
$begingroup$
I am asked to simplify $frac{sqrt{12x}}{2+2sqrt{3}}$ and the solution is provided as $frac{3sqrt{x}-sqrt{3x}}{2}$. I arrived at $-12x + 6xsqrt{3}$ and I'm not sure how to arrive at the text book solution.
My working:
$$frac{sqrt{12x}}{2+2sqrt{3}} = frac{sqrt{12x}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}} = frac{12x(2-sqrt{3})}{(2+2sqrt{3})(2-sqrt{3})} = frac{24x-12xsqrt{3}}{4+(2cdot(-3))}=frac{24x-12xsqrt{3}}{-2}$$
Then, multiplying out the denominator I get:
$-12x+6xsqrt{3}$
Is my thought process sound up to a point? Where did I go wrong and how can I arrive at $frac{3sqrt{x}-sqrt{3x}}{2}$?
algebra-precalculus
edited Jan 6 at 19:07
egreg
180k1485202
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
$endgroup$
2
$begingroup$
Why do you get $sqrt {12x}(2-sqrt{3}) = 12x(2-sqrt 3)$? You arbitrarily just removed the radical sign over the $sqrt{12x}$. That was your error. Do it again without removing the radical sign.
$endgroup$
– fleablood
Jan 6 at 19:02
2
$begingroup$
Your second error is multiplying numerator and denominator by $2-sqrt 3$ instead of $2-2sqrt 3$.
$endgroup$
– TonyK
Jan 6 at 19:07
$begingroup$
And your third and fourth errors occur when you try to calculate $(2+2sqrt 3)(2-sqrt 3)$.
$endgroup$
– TonyK
Jan 6 at 19:08
$begingroup$
@TonyK good catch.
$endgroup$
– fleablood
Jan 6 at 19:08
add a comment |
$begingroup$
I am asked to simplify $frac{sqrt{12x}}{2+2sqrt{3}}$ and the solution is provided as $frac{3sqrt{x}-sqrt{3x}}{2}$. I arrived at $-12x + 6xsqrt{3}$ and I'm not sure how to arrive at the text book solution.
My working:
$$frac{sqrt{12x}}{2+2sqrt{3}} = frac{sqrt{12x}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}} = frac{12x(2-sqrt{3})}{(2+2sqrt{3})(2-sqrt{3})} = frac{24x-12xsqrt{3}}{4+(2cdot(-3))}=frac{24x-12xsqrt{3}}{-2}$$
Then, multiplying out the denominator I get:
$-12x+6xsqrt{3}$
Is my thought process sound up to a point? Where did I go wrong and how can I arrive at $frac{3sqrt{x}-sqrt{3x}}{2}$?
algebra-precalculus
edited Jan 6 at 19:07
egreg
180k1485202
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
$endgroup$
I am asked to simplify $frac{sqrt{12x}}{2+2sqrt{3}}$ and the solution is provided as $frac{3sqrt{x}-sqrt{3x}}{2}$. I arrived at $-12x + 6xsqrt{3}$ and I'm not sure how to arrive at the text book solution.
My working:
$$frac{sqrt{12x}}{2+2sqrt{3}} = frac{sqrt{12x}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}} = frac{12x(2-sqrt{3})}{(2+2sqrt{3})(2-sqrt{3})} = frac{24x-12xsqrt{3}}{4+(2cdot(-3))}=frac{24x-12xsqrt{3}}{-2}$$
Then, multiplying out the denominator I get:
$-12x+6xsqrt{3}$
Is my thought process sound up to a point? Where did I go wrong and how can I arrive at $frac{3sqrt{x}-sqrt{3x}}{2}$?
algebra-precalculus
algebra-precalculus
edited Jan 6 at 19:07
egreg
180k1485202
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
edited Jan 6 at 19:07
egreg
180k1485202
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
edited Jan 6 at 19:07
egreg
180k1485202
edited Jan 6 at 19:07
egreg
180k1485202
edited Jan 6 at 19:07
egreg
180k1485202
180k1485202
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
asked Jan 6 at 18:59
Doug FirDoug Fir
3227
3227
2
$begingroup$
Why do you get $sqrt {12x}(2-sqrt{3}) = 12x(2-sqrt 3)$? You arbitrarily just removed the radical sign over the $sqrt{12x}$. That was your error. Do it again without removing the radical sign.
$endgroup$
– fleablood
Jan 6 at 19:02
2
$begingroup$
Your second error is multiplying numerator and denominator by $2-sqrt 3$ instead of $2-2sqrt 3$.
$endgroup$
– TonyK
Jan 6 at 19:07
$begingroup$
And your third and fourth errors occur when you try to calculate $(2+2sqrt 3)(2-sqrt 3)$.
$endgroup$
– TonyK
Jan 6 at 19:08
$begingroup$
@TonyK good catch.
$endgroup$
– fleablood
Jan 6 at 19:08
add a comment |
2
$begingroup$
Why do you get $sqrt {12x}(2-sqrt{3}) = 12x(2-sqrt 3)$? You arbitrarily just removed the radical sign over the $sqrt{12x}$. That was your error. Do it again without removing the radical sign.
$endgroup$
– fleablood
Jan 6 at 19:02
2
$begingroup$
Your second error is multiplying numerator and denominator by $2-sqrt 3$ instead of $2-2sqrt 3$.
$endgroup$
– TonyK
Jan 6 at 19:07
$begingroup$
And your third and fourth errors occur when you try to calculate $(2+2sqrt 3)(2-sqrt 3)$.
$endgroup$
– TonyK
Jan 6 at 19:08
$begingroup$
@TonyK good catch.
$endgroup$
– fleablood
Jan 6 at 19:08
2
2
$begingroup$
Why do you get $sqrt {12x}(2-sqrt{3}) = 12x(2-sqrt 3)$? You arbitrarily just removed the radical sign over the $sqrt{12x}$. That was your error. Do it again without removing the radical sign.
$endgroup$
– fleablood
Jan 6 at 19:02
$begingroup$
Why do you get $sqrt {12x}(2-sqrt{3}) = 12x(2-sqrt 3)$? You arbitrarily just removed the radical sign over the $sqrt{12x}$. That was your error. Do it again without removing the radical sign.
$endgroup$
– fleablood
Jan 6 at 19:02
2
2
$begingroup$
Your second error is multiplying numerator and denominator by $2-sqrt 3$ instead of $2-2sqrt 3$.
$endgroup$
– TonyK
Jan 6 at 19:07
$begingroup$
Your second error is multiplying numerator and denominator by $2-sqrt 3$ instead of $2-2sqrt 3$.
$endgroup$
– TonyK
Jan 6 at 19:07
$begingroup$
And your third and fourth errors occur when you try to calculate $(2+2sqrt 3)(2-sqrt 3)$.
$endgroup$
– TonyK
Jan 6 at 19:08
$begingroup$
And your third and fourth errors occur when you try to calculate $(2+2sqrt 3)(2-sqrt 3)$.
$endgroup$
– TonyK
Jan 6 at 19:08
$begingroup$
@TonyK good catch.
$endgroup$
– fleablood
Jan 6 at 19:08
$begingroup$
@TonyK good catch.
$endgroup$
– fleablood
Jan 6 at 19:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First error:
$$
sqrt{12x}(2-sqrt{3})ne 12x(2-sqrt{3})
$$
Second error:
$$
(2+2sqrt{3})(2-sqrt{3})=4+4sqrt{3}-2sqrt{3}-2cdot3=2sqrt{3}-2ne-2
$$
Third error: in order to rationalize the denominator, you have to multiply by $2-2sqrt{3}$.
On the other hand, you can proceed more simply:
$$
frac{sqrt{12x}}{2+2sqrt{3}}=frac{2sqrt{3x}}{2(sqrt{3}+1)}
=frac{sqrt{3x}}{sqrt{3}+1}frac{sqrt{3}-1}{sqrt{3}-1}=
frac{sqrt{9x}-sqrt{3x}}{3-1}=frac{3sqrt{x}-sqrt{3x}}{2}
$$
answered Jan 6 at 19:13
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{color{green}{sqrt{12x}}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}}=frac{color{red}{12x}(2-sqrt{3})}{color{purple}{(2+2sqrt{3})(2-sqrt{3})}} = frac{24x-12xsqrt{3}}{color{orange}{4+(2cdot(-3))}}=frac{24x-12xsqrt{3}}{-2}$
You magically turned $color{green}{sqrt{12x}}$ into $color{red}{12x}$ for no reason whatsoever.
And you incorrectly calculated $color{purple}{(2+2sqrt{3})(2-sqrt{3})} = color{orange}{4+(2cdot(-3))}$.
You should have done:
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{sqrt{12x}}{2+2sqrt{3}}color{green}{frac{2 - 2sqrt{3}}{2 - 2sqrt{3}}}=frac{sqrt{12x}(2-2sqrt{3})}{color{purple}{(2+2sqrt{3})(2-2sqrt{3})}} = frac{...}{color{purple}{4-4*3}}=...$
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
$endgroup$
$begingroup$
Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here.
$endgroup$
– Doug Fir
Jan 6 at 19:19
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First error:
$$
sqrt{12x}(2-sqrt{3})ne 12x(2-sqrt{3})
$$
Second error:
$$
(2+2sqrt{3})(2-sqrt{3})=4+4sqrt{3}-2sqrt{3}-2cdot3=2sqrt{3}-2ne-2
$$
Third error: in order to rationalize the denominator, you have to multiply by $2-2sqrt{3}$.
On the other hand, you can proceed more simply:
$$
frac{sqrt{12x}}{2+2sqrt{3}}=frac{2sqrt{3x}}{2(sqrt{3}+1)}
=frac{sqrt{3x}}{sqrt{3}+1}frac{sqrt{3}-1}{sqrt{3}-1}=
frac{sqrt{9x}-sqrt{3x}}{3-1}=frac{3sqrt{x}-sqrt{3x}}{2}
$$
answered Jan 6 at 19:13
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
First error:
$$
sqrt{12x}(2-sqrt{3})ne 12x(2-sqrt{3})
$$
Second error:
$$
(2+2sqrt{3})(2-sqrt{3})=4+4sqrt{3}-2sqrt{3}-2cdot3=2sqrt{3}-2ne-2
$$
Third error: in order to rationalize the denominator, you have to multiply by $2-2sqrt{3}$.
On the other hand, you can proceed more simply:
$$
frac{sqrt{12x}}{2+2sqrt{3}}=frac{2sqrt{3x}}{2(sqrt{3}+1)}
=frac{sqrt{3x}}{sqrt{3}+1}frac{sqrt{3}-1}{sqrt{3}-1}=
frac{sqrt{9x}-sqrt{3x}}{3-1}=frac{3sqrt{x}-sqrt{3x}}{2}
$$
answered Jan 6 at 19:13
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
First error:
$$
sqrt{12x}(2-sqrt{3})ne 12x(2-sqrt{3})
$$
Second error:
$$
(2+2sqrt{3})(2-sqrt{3})=4+4sqrt{3}-2sqrt{3}-2cdot3=2sqrt{3}-2ne-2
$$
Third error: in order to rationalize the denominator, you have to multiply by $2-2sqrt{3}$.
On the other hand, you can proceed more simply:
$$
frac{sqrt{12x}}{2+2sqrt{3}}=frac{2sqrt{3x}}{2(sqrt{3}+1)}
=frac{sqrt{3x}}{sqrt{3}+1}frac{sqrt{3}-1}{sqrt{3}-1}=
frac{sqrt{9x}-sqrt{3x}}{3-1}=frac{3sqrt{x}-sqrt{3x}}{2}
$$
answered Jan 6 at 19:13
egregegreg
180k1485202
$endgroup$
First error:
$$
sqrt{12x}(2-sqrt{3})ne 12x(2-sqrt{3})
$$
Second error:
$$
(2+2sqrt{3})(2-sqrt{3})=4+4sqrt{3}-2sqrt{3}-2cdot3=2sqrt{3}-2ne-2
$$
Third error: in order to rationalize the denominator, you have to multiply by $2-2sqrt{3}$.
On the other hand, you can proceed more simply:
$$
frac{sqrt{12x}}{2+2sqrt{3}}=frac{2sqrt{3x}}{2(sqrt{3}+1)}
=frac{sqrt{3x}}{sqrt{3}+1}frac{sqrt{3}-1}{sqrt{3}-1}=
frac{sqrt{9x}-sqrt{3x}}{3-1}=frac{3sqrt{x}-sqrt{3x}}{2}
$$
answered Jan 6 at 19:13
egregegreg
180k1485202
answered Jan 6 at 19:13
egregegreg
180k1485202
answered Jan 6 at 19:13
egregegreg
180k1485202
answered Jan 6 at 19:13
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{color{green}{sqrt{12x}}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}}=frac{color{red}{12x}(2-sqrt{3})}{color{purple}{(2+2sqrt{3})(2-sqrt{3})}} = frac{24x-12xsqrt{3}}{color{orange}{4+(2cdot(-3))}}=frac{24x-12xsqrt{3}}{-2}$
You magically turned $color{green}{sqrt{12x}}$ into $color{red}{12x}$ for no reason whatsoever.
And you incorrectly calculated $color{purple}{(2+2sqrt{3})(2-sqrt{3})} = color{orange}{4+(2cdot(-3))}$.
You should have done:
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{sqrt{12x}}{2+2sqrt{3}}color{green}{frac{2 - 2sqrt{3}}{2 - 2sqrt{3}}}=frac{sqrt{12x}(2-2sqrt{3})}{color{purple}{(2+2sqrt{3})(2-2sqrt{3})}} = frac{...}{color{purple}{4-4*3}}=...$
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
$endgroup$
$begingroup$
Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here.
$endgroup$
– Doug Fir
Jan 6 at 19:19
add a comment |
$begingroup$
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{color{green}{sqrt{12x}}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}}=frac{color{red}{12x}(2-sqrt{3})}{color{purple}{(2+2sqrt{3})(2-sqrt{3})}} = frac{24x-12xsqrt{3}}{color{orange}{4+(2cdot(-3))}}=frac{24x-12xsqrt{3}}{-2}$
You magically turned $color{green}{sqrt{12x}}$ into $color{red}{12x}$ for no reason whatsoever.
And you incorrectly calculated $color{purple}{(2+2sqrt{3})(2-sqrt{3})} = color{orange}{4+(2cdot(-3))}$.
You should have done:
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{sqrt{12x}}{2+2sqrt{3}}color{green}{frac{2 - 2sqrt{3}}{2 - 2sqrt{3}}}=frac{sqrt{12x}(2-2sqrt{3})}{color{purple}{(2+2sqrt{3})(2-2sqrt{3})}} = frac{...}{color{purple}{4-4*3}}=...$
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
$endgroup$
$begingroup$
Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here.
$endgroup$
– Doug Fir
Jan 6 at 19:19
add a comment |
$begingroup$
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{color{green}{sqrt{12x}}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}}=frac{color{red}{12x}(2-sqrt{3})}{color{purple}{(2+2sqrt{3})(2-sqrt{3})}} = frac{24x-12xsqrt{3}}{color{orange}{4+(2cdot(-3))}}=frac{24x-12xsqrt{3}}{-2}$
You magically turned $color{green}{sqrt{12x}}$ into $color{red}{12x}$ for no reason whatsoever.
And you incorrectly calculated $color{purple}{(2+2sqrt{3})(2-sqrt{3})} = color{orange}{4+(2cdot(-3))}$.
You should have done:
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{sqrt{12x}}{2+2sqrt{3}}color{green}{frac{2 - 2sqrt{3}}{2 - 2sqrt{3}}}=frac{sqrt{12x}(2-2sqrt{3})}{color{purple}{(2+2sqrt{3})(2-2sqrt{3})}} = frac{...}{color{purple}{4-4*3}}=...$
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
$endgroup$
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{color{green}{sqrt{12x}}}{2+2sqrt{3}}frac{2 - sqrt{3}}{2 - sqrt{3}}=frac{color{red}{12x}(2-sqrt{3})}{color{purple}{(2+2sqrt{3})(2-sqrt{3})}} = frac{24x-12xsqrt{3}}{color{orange}{4+(2cdot(-3))}}=frac{24x-12xsqrt{3}}{-2}$
You magically turned $color{green}{sqrt{12x}}$ into $color{red}{12x}$ for no reason whatsoever.
And you incorrectly calculated $color{purple}{(2+2sqrt{3})(2-sqrt{3})} = color{orange}{4+(2cdot(-3))}$.
You should have done:
$frac{sqrt{12x}}{2+2sqrt{3}}=frac{sqrt{12x}}{2+2sqrt{3}}color{green}{frac{2 - 2sqrt{3}}{2 - 2sqrt{3}}}=frac{sqrt{12x}(2-2sqrt{3})}{color{purple}{(2+2sqrt{3})(2-2sqrt{3})}} = frac{...}{color{purple}{4-4*3}}=...$
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
edited Jan 6 at 19:17
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
answered Jan 6 at 19:06
fleabloodfleablood
69.4k22685
69.4k22685
$begingroup$
Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here.
$endgroup$
– Doug Fir
Jan 6 at 19:19
add a comment |
$begingroup$
Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here.
$endgroup$
– Doug Fir
Jan 6 at 19:19
$begingroup$
Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here.
$endgroup$
– Doug Fir
Jan 6 at 19:19
$begingroup$
Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here.
$endgroup$
– Doug Fir
Jan 6 at 19:19
add a comment |
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2
$begingroup$
Why do you get $sqrt {12x}(2-sqrt{3}) = 12x(2-sqrt 3)$? You arbitrarily just removed the radical sign over the $sqrt{12x}$. That was your error. Do it again without removing the radical sign.
$endgroup$
– fleablood
Jan 6 at 19:02
2
$begingroup$
Your second error is multiplying numerator and denominator by $2-sqrt 3$ instead of $2-2sqrt 3$.
$endgroup$
– TonyK
Jan 6 at 19:07
$begingroup$
And your third and fourth errors occur when you try to calculate $(2+2sqrt 3)(2-sqrt 3)$.
$endgroup$
– TonyK
Jan 6 at 19:08
$begingroup$
@TonyK good catch.
$endgroup$
– fleablood
Jan 6 at 19:08