Rationalizing denominator of $frac{7}{2+sqrt{3}}$. Cannot match textbook solution
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I am given this expression and asked to simplify by rationalizing the denominator:
$$frac{7}{2+sqrt{3}}$$
The solution is provided:
$14 - 7sqrt{3}$
I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:
$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$
=
$$frac{14 - 7sqrt{3}}{4}$$
Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?
algebra-precalculus rationalising-denominator
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add a comment |
$begingroup$
I am given this expression and asked to simplify by rationalizing the denominator:
$$frac{7}{2+sqrt{3}}$$
The solution is provided:
$14 - 7sqrt{3}$
I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:
$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$
=
$$frac{14 - 7sqrt{3}}{4}$$
Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?
algebra-precalculus rationalising-denominator
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1
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How did you compute $4$ in the denominator?
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– Martin R
Jan 1 at 18:15
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@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
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– Doug Fir
Jan 1 at 18:18
1
$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22
add a comment |
$begingroup$
I am given this expression and asked to simplify by rationalizing the denominator:
$$frac{7}{2+sqrt{3}}$$
The solution is provided:
$14 - 7sqrt{3}$
I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:
$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$
=
$$frac{14 - 7sqrt{3}}{4}$$
Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?
algebra-precalculus rationalising-denominator
$endgroup$
I am given this expression and asked to simplify by rationalizing the denominator:
$$frac{7}{2+sqrt{3}}$$
The solution is provided:
$14 - 7sqrt{3}$
I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:
$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$
=
$$frac{14 - 7sqrt{3}}{4}$$
Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?
algebra-precalculus rationalising-denominator
algebra-precalculus rationalising-denominator
edited Jan 1 at 18:27
José Carlos Santos
165k22132235
165k22132235
asked Jan 1 at 17:53
Doug FirDoug Fir
4098
4098
1
$begingroup$
How did you compute $4$ in the denominator?
$endgroup$
– Martin R
Jan 1 at 18:15
$begingroup$
@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
$endgroup$
– Doug Fir
Jan 1 at 18:18
1
$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22
add a comment |
1
$begingroup$
How did you compute $4$ in the denominator?
$endgroup$
– Martin R
Jan 1 at 18:15
$begingroup$
@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
$endgroup$
– Doug Fir
Jan 1 at 18:18
1
$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22
1
1
$begingroup$
How did you compute $4$ in the denominator?
$endgroup$
– Martin R
Jan 1 at 18:15
$begingroup$
How did you compute $4$ in the denominator?
$endgroup$
– Martin R
Jan 1 at 18:15
$begingroup$
@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
$endgroup$
– Doug Fir
Jan 1 at 18:18
$begingroup$
@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
$endgroup$
– Doug Fir
Jan 1 at 18:18
1
1
$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22
$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.
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$begingroup$
Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
$endgroup$
– Doug Fir
Jan 1 at 17:57
4
$begingroup$
@DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:58
add a comment |
$begingroup$
note that:
$$(a+b)(c+d)=ac+ad+bc+bd$$
so in some cases it simplifies to:
$$(a+b)(a-b)=a^2-b^2$$
for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$
$endgroup$
add a comment |
$begingroup$
Trick to remember forever and use again and again.
$(a + b)(a-b) = a(a-b) + b(a-b)=$
$a^2 - ab + ab - b^2 = a^2 - b^2$.
So
1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$
and
2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.
So
3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.
So:
$frac {7}{2 + sqrt 3} = $
$frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$
$frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$
$frac {7(2-sqrt 3)}{4-3} =$
$frac {7(2-sqrt 3)}{1} =$
$7(2-sqrt 3)$.
Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.
You will be using it for the REST OF YOUR LIFE!
$endgroup$
$begingroup$
Thank you for the tip!
$endgroup$
– Doug Fir
Jan 1 at 20:32
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.
$endgroup$
$begingroup$
Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
$endgroup$
– Doug Fir
Jan 1 at 17:57
4
$begingroup$
@DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:58
add a comment |
$begingroup$
Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.
$endgroup$
$begingroup$
Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
$endgroup$
– Doug Fir
Jan 1 at 17:57
4
$begingroup$
@DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:58
add a comment |
$begingroup$
Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.
$endgroup$
Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.
answered Jan 1 at 17:54
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
$begingroup$
Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
$endgroup$
– Doug Fir
Jan 1 at 17:57
4
$begingroup$
@DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:58
add a comment |
$begingroup$
Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
$endgroup$
– Doug Fir
Jan 1 at 17:57
4
$begingroup$
@DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:58
$begingroup$
Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
$endgroup$
– Doug Fir
Jan 1 at 17:57
$begingroup$
Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
$endgroup$
– Doug Fir
Jan 1 at 17:57
4
4
$begingroup$
@DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:58
$begingroup$
@DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:58
add a comment |
$begingroup$
note that:
$$(a+b)(c+d)=ac+ad+bc+bd$$
so in some cases it simplifies to:
$$(a+b)(a-b)=a^2-b^2$$
for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$
$endgroup$
add a comment |
$begingroup$
note that:
$$(a+b)(c+d)=ac+ad+bc+bd$$
so in some cases it simplifies to:
$$(a+b)(a-b)=a^2-b^2$$
for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$
$endgroup$
add a comment |
$begingroup$
note that:
$$(a+b)(c+d)=ac+ad+bc+bd$$
so in some cases it simplifies to:
$$(a+b)(a-b)=a^2-b^2$$
for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$
$endgroup$
note that:
$$(a+b)(c+d)=ac+ad+bc+bd$$
so in some cases it simplifies to:
$$(a+b)(a-b)=a^2-b^2$$
for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$
answered Jan 1 at 18:31
Henry LeeHenry Lee
2,054219
2,054219
add a comment |
add a comment |
$begingroup$
Trick to remember forever and use again and again.
$(a + b)(a-b) = a(a-b) + b(a-b)=$
$a^2 - ab + ab - b^2 = a^2 - b^2$.
So
1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$
and
2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.
So
3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.
So:
$frac {7}{2 + sqrt 3} = $
$frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$
$frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$
$frac {7(2-sqrt 3)}{4-3} =$
$frac {7(2-sqrt 3)}{1} =$
$7(2-sqrt 3)$.
Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.
You will be using it for the REST OF YOUR LIFE!
$endgroup$
$begingroup$
Thank you for the tip!
$endgroup$
– Doug Fir
Jan 1 at 20:32
add a comment |
$begingroup$
Trick to remember forever and use again and again.
$(a + b)(a-b) = a(a-b) + b(a-b)=$
$a^2 - ab + ab - b^2 = a^2 - b^2$.
So
1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$
and
2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.
So
3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.
So:
$frac {7}{2 + sqrt 3} = $
$frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$
$frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$
$frac {7(2-sqrt 3)}{4-3} =$
$frac {7(2-sqrt 3)}{1} =$
$7(2-sqrt 3)$.
Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.
You will be using it for the REST OF YOUR LIFE!
$endgroup$
$begingroup$
Thank you for the tip!
$endgroup$
– Doug Fir
Jan 1 at 20:32
add a comment |
$begingroup$
Trick to remember forever and use again and again.
$(a + b)(a-b) = a(a-b) + b(a-b)=$
$a^2 - ab + ab - b^2 = a^2 - b^2$.
So
1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$
and
2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.
So
3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.
So:
$frac {7}{2 + sqrt 3} = $
$frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$
$frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$
$frac {7(2-sqrt 3)}{4-3} =$
$frac {7(2-sqrt 3)}{1} =$
$7(2-sqrt 3)$.
Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.
You will be using it for the REST OF YOUR LIFE!
$endgroup$
Trick to remember forever and use again and again.
$(a + b)(a-b) = a(a-b) + b(a-b)=$
$a^2 - ab + ab - b^2 = a^2 - b^2$.
So
1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$
and
2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.
So
3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.
So:
$frac {7}{2 + sqrt 3} = $
$frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$
$frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$
$frac {7(2-sqrt 3)}{4-3} =$
$frac {7(2-sqrt 3)}{1} =$
$7(2-sqrt 3)$.
Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.
You will be using it for the REST OF YOUR LIFE!
answered Jan 1 at 18:45
fleabloodfleablood
71.7k22686
71.7k22686
$begingroup$
Thank you for the tip!
$endgroup$
– Doug Fir
Jan 1 at 20:32
add a comment |
$begingroup$
Thank you for the tip!
$endgroup$
– Doug Fir
Jan 1 at 20:32
$begingroup$
Thank you for the tip!
$endgroup$
– Doug Fir
Jan 1 at 20:32
$begingroup$
Thank you for the tip!
$endgroup$
– Doug Fir
Jan 1 at 20:32
add a comment |
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1
$begingroup$
How did you compute $4$ in the denominator?
$endgroup$
– Martin R
Jan 1 at 18:15
$begingroup$
@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
$endgroup$
– Doug Fir
Jan 1 at 18:18
1
$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22