Dividing one equation by another
This is from Higher Algebra by Hall and Knight.
I don't understand how this is done. Can you explain?
algebra-precalculus
add a comment |
This is from Higher Algebra by Hall and Knight.
I don't understand how this is done. Can you explain?
algebra-precalculus
6
difference of two squares
– Lord Shark the Unknown
11 hours ago
Thanks. I got it.
– Md Masood
11 hours ago
2
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.
– littleO
11 hours ago
I first thought so.
– Md Masood
10 hours ago
add a comment |
This is from Higher Algebra by Hall and Knight.
I don't understand how this is done. Can you explain?
algebra-precalculus
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 11 hours ago
Zacky
4,1351546
4,1351546
asked 11 hours ago
Md Masood
528
528
6
difference of two squares
– Lord Shark the Unknown
11 hours ago
Thanks. I got it.
– Md Masood
11 hours ago
2
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.
– littleO
11 hours ago
I first thought so.
– Md Masood
10 hours ago
add a comment |
6
difference of two squares
– Lord Shark the Unknown
11 hours ago
Thanks. I got it.
– Md Masood
11 hours ago
2
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.
– littleO
11 hours ago
I first thought so.
– Md Masood
10 hours ago
6
6
difference of two squares
– Lord Shark the Unknown
11 hours ago
difference of two squares
– Lord Shark the Unknown
11 hours ago
Thanks. I got it.
– Md Masood
11 hours ago
Thanks. I got it.
– Md Masood
11 hours ago
2
2
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.
– littleO
11 hours ago
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.
– littleO
11 hours ago
I first thought so.
– Md Masood
10 hours ago
I first thought so.
– Md Masood
10 hours ago
add a comment |
2 Answers
2
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oldest
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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 $(3)$
add a comment |
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $(3)$
add a comment |
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 $(3)$
add a comment |
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 $(3)$
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 11 hours ago
Mohammad Riazi-Kermani
40.7k42058
40.7k42058
add a comment |
add a comment |
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
add a comment |
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
add a comment |
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
Hint: Use the identity
$$a^2-b^2=(a-b)(a+b) $$
answered 11 hours ago
Thomas Shelby
1,242116
1,242116
add a comment |
add a comment |
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6
difference of two squares
– Lord Shark the Unknown
11 hours ago
Thanks. I got it.
– Md Masood
11 hours ago
2
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.
– littleO
11 hours ago
I first thought so.
– Md Masood
10 hours ago