Dividing one equation by another [closed]
0
$begingroup$
This is from Higher Algebra by Hall and Knight. 
I don't understand how this is done. Can you explain?
algebra-precalculus
edited Dec 26 '18 at 17:07
Zacky
7,73511061
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
$endgroup$
closed as off-topic by RRL, Holo, Did, Saad, Abcd Jan 6 at 13:10
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." – RRL, Holo, Did, Saad, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
This is from Higher Algebra by Hall and Knight.
I don't understand how this is done. Can you explain?
algebra-precalculus
edited Dec 26 '18 at 17:07
Zacky
7,73511061
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
$endgroup$
closed as off-topic by RRL, Holo, Did, Saad, Abcd Jan 6 at 13:10
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." – RRL, Holo, Did, Saad, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
6
$begingroup$
difference of two squares
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 17:05
$begingroup$
Thanks. I got it.
$endgroup$
– Md Masood
Dec 26 '18 at 17:06
2
$begingroup$
To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
$endgroup$
– littleO
Dec 26 '18 at 17:10
$begingroup$
I first thought so.
$endgroup$
– Md Masood
Dec 26 '18 at 17:53
$begingroup$
Now try to derive the cartesian equation for an ellipse. :-)
$endgroup$
– John Joy
Dec 28 '18 at 0:48
add a comment |
0
0
0
$begingroup$
This is from Higher Algebra by Hall and Knight.
I don't understand how this is done. Can you explain?
algebra-precalculus
edited Dec 26 '18 at 17:07
Zacky
7,73511061
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
$endgroup$
This is from Higher Algebra by Hall and Knight.
I don't understand how this is done. Can you explain?
algebra-precalculus
algebra-precalculus
edited Dec 26 '18 at 17:07
Zacky
7,73511061
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
edited Dec 26 '18 at 17:07
Zacky
7,73511061
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
edited Dec 26 '18 at 17:07
Zacky
7,73511061
edited Dec 26 '18 at 17:07
Zacky
7,73511061
edited Dec 26 '18 at 17:07
Zacky
7,73511061
7,73511061
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
asked Dec 26 '18 at 17:03
Md MasoodMd Masood
508
508
closed as off-topic by RRL, Holo, Did, Saad, Abcd Jan 6 at 13:10
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." – RRL, Holo, Did, Saad, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Holo, Did, Saad, Abcd Jan 6 at 13:10
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." – RRL, Holo, Did, Saad, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
6
$begingroup$
difference of two squares
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 17:05
$begingroup$
Thanks. I got it.
$endgroup$
– Md Masood
Dec 26 '18 at 17:06
2
$begingroup$
To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
$endgroup$
– littleO
Dec 26 '18 at 17:10
$begingroup$
I first thought so.
$endgroup$
– Md Masood
Dec 26 '18 at 17:53
$begingroup$
Now try to derive the cartesian equation for an ellipse. :-)
$endgroup$
– John Joy
Dec 28 '18 at 0:48
add a comment |
6
$begingroup$
difference of two squares
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 17:05
$begingroup$
Thanks. I got it.
$endgroup$
– Md Masood
Dec 26 '18 at 17:06
2
$begingroup$
To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
$endgroup$
– littleO
Dec 26 '18 at 17:10
$begingroup$
I first thought so.
$endgroup$
– Md Masood
Dec 26 '18 at 17:53
$begingroup$
Now try to derive the cartesian equation for an ellipse. :-)
$endgroup$
– John Joy
Dec 28 '18 at 0:48
6
6
$begingroup$
difference of two squares
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 17:05
$begingroup$
difference of two squares
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 17:05
$begingroup$
Thanks. I got it.
$endgroup$
– Md Masood
Dec 26 '18 at 17:06
$begingroup$
Thanks. I got it.
$endgroup$
– Md Masood
Dec 26 '18 at 17:06
2
2
$begingroup$
To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
$endgroup$
– littleO
Dec 26 '18 at 17:10
$begingroup$
To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
$endgroup$
– littleO
Dec 26 '18 at 17:10
$begingroup$
I first thought so.
$endgroup$
– Md Masood
Dec 26 '18 at 17:53
$begingroup$
I first thought so.
$endgroup$
– Md Masood
Dec 26 '18 at 17:53
$begingroup$
Now try to derive the cartesian equation for an ellipse. :-)
$endgroup$
– John Joy
Dec 28 '18 at 0:48
$begingroup$
Now try to derive the cartesian equation for an ellipse. :-)
$endgroup$
– John Joy
Dec 28 '18 at 0:48
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
Note that (2) could be factored as
$$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$
Now dividing by $(1)$ results in $(3)$
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
4
$begingroup$
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Note that (2) could be factored as
$$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$
Now dividing by $(1)$ results in $(3)$
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
2
$begingroup$
Note that (2) could be factored as
$$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$
Now dividing by $(1)$ results in $(3)$
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
2
2
2
$begingroup$
Note that (2) could be factored as
$$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$
Now dividing by $(1)$ results in $(3)$
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
Note that (2) could be factored as
$$ (3x^2-4x+34)-(3x^2-4x-11) = (sqrt {3x^2-4x+34} -sqrt {3x^2-4x-11})(sqrt {3x^2-4x+34} +sqrt {3x^2-4x-11})=45$$
Now dividing by $(1)$ results in $(3)$
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 26 '18 at 17:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
4
$begingroup$
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
$endgroup$
add a comment |
4
$begingroup$
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
$endgroup$
add a comment |
4
4
4
$begingroup$
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
$endgroup$
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
answered Dec 26 '18 at 17:08
Thomas ShelbyThomas Shelby
4,0342625
4,0342625
add a comment |
add a comment |
6
$begingroup$
difference of two squares
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 17:05
$begingroup$
Thanks. I got it.
$endgroup$
– Md Masood
Dec 26 '18 at 17:06
2
$begingroup$
To me the explanation given in the book seems rather unclear. It really should have mentioned that we are using the difference of two squares factorization. Otherwise it's not clear why dividing each member of (2) by the corresponding member of (1) would be a valid thing to do. I'm a bit afraid that other students who read this (not you) might think that if $a+b = x$ and $c+d= y$ then $a/c + b/d = x/y$, but of course that's usually not correct.
$endgroup$
– littleO
Dec 26 '18 at 17:10
$begingroup$
I first thought so.
$endgroup$
– Md Masood
Dec 26 '18 at 17:53
$begingroup$
Now try to derive the cartesian equation for an ellipse. :-)
$endgroup$
– John Joy
Dec 28 '18 at 0:48