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How to know when to complete the square
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up vote
2
down vote
favorite
Question is: integration of $1over x^2+8x+20$
Why can I not just solve for: $A/(x+2) +B/(x+10)$ and integrate it this way?
The answer on symbolab shows I need to complete the square of the denominator first but I don't know hen to do that or when to factor it out.
Any help would be great!
integration
asked Aug 11 at 14:15
Shauna
286
add a comment |Â
up vote
2
down vote
favorite
Question is: integration of $1over x^2+8x+20$
Why can I not just solve for: $A/(x+2) +B/(x+10)$ and integrate it this way?
The answer on symbolab shows I need to complete the square of the denominator first but I don't know hen to do that or when to factor it out.
Any help would be great!
integration
asked Aug 11 at 14:15
Shauna
286
8
$(x+2)(x+10)=x^2+12x+20$, not $x^2+8x+20$. You can indeed solve it by your method, but you will need to use the appropriate factors $x^2+8x+20=(x+4-2i)(x+4+2i)$.
– user583012
Aug 11 at 14:18
You can use this: en.wikipedia.org/wiki/… after you integrated to get back to a "real looking" form.
– Zacky
Aug 11 at 14:34
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Question is: integration of $1over x^2+8x+20$
Why can I not just solve for: $A/(x+2) +B/(x+10)$ and integrate it this way?
The answer on symbolab shows I need to complete the square of the denominator first but I don't know hen to do that or when to factor it out.
Any help would be great!
integration
asked Aug 11 at 14:15
Shauna
286
Question is: integration of $1over x^2+8x+20$
Why can I not just solve for: $A/(x+2) +B/(x+10)$ and integrate it this way?
The answer on symbolab shows I need to complete the square of the denominator first but I don't know hen to do that or when to factor it out.
Any help would be great!
integration
asked Aug 11 at 14:15
Shauna
286
edited Aug 11 at 14:20
user582949
asked Aug 11 at 14:15
Shauna
286
asked Aug 11 at 14:15
Shauna
286
asked Aug 11 at 14:15
Shauna
286
286
8
$(x+2)(x+10)=x^2+12x+20$, not $x^2+8x+20$. You can indeed solve it by your method, but you will need to use the appropriate factors $x^2+8x+20=(x+4-2i)(x+4+2i)$.
– user583012
Aug 11 at 14:18
You can use this: en.wikipedia.org/wiki/… after you integrated to get back to a "real looking" form.
– Zacky
Aug 11 at 14:34
add a comment |Â
8
$(x+2)(x+10)=x^2+12x+20$, not $x^2+8x+20$. You can indeed solve it by your method, but you will need to use the appropriate factors $x^2+8x+20=(x+4-2i)(x+4+2i)$.
– user583012
Aug 11 at 14:18
You can use this: en.wikipedia.org/wiki/… after you integrated to get back to a "real looking" form.
– Zacky
Aug 11 at 14:34
8
8
$(x+2)(x+10)=x^2+12x+20$, not $x^2+8x+20$. You can indeed solve it by your method, but you will need to use the appropriate factors $x^2+8x+20=(x+4-2i)(x+4+2i)$.
– user583012
Aug 11 at 14:18
$(x+2)(x+10)=x^2+12x+20$, not $x^2+8x+20$. You can indeed solve it by your method, but you will need to use the appropriate factors $x^2+8x+20=(x+4-2i)(x+4+2i)$.
– user583012
Aug 11 at 14:18
You can use this: en.wikipedia.org/wiki/… after you integrated to get back to a "real looking" form.
– Zacky
Aug 11 at 14:34
You can use this: en.wikipedia.org/wiki/… after you integrated to get back to a "real looking" form.
– Zacky
Aug 11 at 14:34
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
8
down vote
If the roots are real, you can factor in binomials and convert to simple fractions.
But if they are complex, it may be better to just complete the square in order to stay in the reals.
E.g., it is easier to deal with
$$intfracdxx^2+1$$ than with
$$intfracdx(x-i)(x+i).$$
answered Aug 11 at 14:23
Yves Daoust
113k665207
Is it?$$
– tomasz
Aug 12 at 15:54
@tomasz: It is.
– Yves Daoust
Aug 12 at 16:05
add a comment |Â
up vote
6
down vote
Beacuse $(x+2)(x+10)= x^2+12x+20$ and not $x^2+8x+20$
Of course!! I can't believe how stupid that question was thanks guys.
– Shauna
Aug 11 at 14:19
add a comment |Â
up vote
4
down vote
Because it is not true that $x^2+8x+20=(x+2)(x+10)$. Completing the square is a natural choice whenever (as in this case) the quadratic has no real roots.
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
add a comment |Â
up vote
3
down vote
You're factoring wrong, the right factorization is $x^2+8x+20=(x+4-2i)(x+4+2i)$
I think completing a square is a natural choice if quadratic doesn't have real roots.
$$x^2+8x+20=x^2+8x+16+4=(x+4)^2+2^2$$
$$int frac1(x+4)^2+2^2,dx = frac12arctanleft(fracx+42right)+C $$
answered Aug 11 at 14:20
Deepesh Meena
2,688721
-1 If the question is "Why do I have to complete the square to solve the integral", an answer showing you completing the square and solving the integral is pointless. The OP is specifically asking why the other method doesn't work, he's not interested in how to solve the integral with this method.
– Ant
Aug 12 at 11:31
but OP was doing wrong factorization of the polynomial
– Deepesh Meena
Aug 12 at 11:35
$x^2+8x+20=(x+4-2i)(x+4+2i)$ I think completing a square is a natural choice if quadratic doesn't have real roots
– Deepesh Meena
Aug 12 at 11:39
Sure, I agree. But then the answer (that was given) is "You're factoring wrong, if you do it correctly it will work too". But I think that showing how the square completion work is just unhelpful. If you instead add this to your ansswer, show how factorization works, and then make a comparison with the "completing the square method", elaborating on why you think it's easier / more natural, then I would upvote it. But as it is you're just ignoring the OP question and just showing that you can do that integral, which is unhelpful
– Ant
Aug 12 at 13:22
sure I have edited the answer
– Deepesh Meena
Aug 12 at 13:59
add a comment |Â
up vote
2
down vote
A more general answer to the question "when should I complete the square?" is:
- If you are fundamentally concerned with the roots of the quadratic, you should factorise (since factorising gives you the roots for free).
- If you are more concerned with the curve as a whole, then completing the square can often help, because then you get the transformations of $y=x^2$ that produce the quadratic.
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
If the roots are real, you can factor in binomials and convert to simple fractions.
But if they are complex, it may be better to just complete the square in order to stay in the reals.
E.g., it is easier to deal with
$$intfracdxx^2+1$$ than with
$$intfracdx(x-i)(x+i).$$
answered Aug 11 at 14:23
Yves Daoust
113k665207
Is it?$$
– tomasz
Aug 12 at 15:54
@tomasz: It is.
– Yves Daoust
Aug 12 at 16:05
add a comment |Â
up vote
8
down vote
If the roots are real, you can factor in binomials and convert to simple fractions.
But if they are complex, it may be better to just complete the square in order to stay in the reals.
E.g., it is easier to deal with
$$intfracdxx^2+1$$ than with
$$intfracdx(x-i)(x+i).$$
answered Aug 11 at 14:23
Yves Daoust
113k665207
Is it?$$
– tomasz
Aug 12 at 15:54
@tomasz: It is.
– Yves Daoust
Aug 12 at 16:05
add a comment |Â
up vote
8
down vote
up vote
8
down vote
If the roots are real, you can factor in binomials and convert to simple fractions.
But if they are complex, it may be better to just complete the square in order to stay in the reals.
E.g., it is easier to deal with
$$intfracdxx^2+1$$ than with
$$intfracdx(x-i)(x+i).$$
answered Aug 11 at 14:23
Yves Daoust
113k665207
If the roots are real, you can factor in binomials and convert to simple fractions.
But if they are complex, it may be better to just complete the square in order to stay in the reals.
E.g., it is easier to deal with
$$intfracdxx^2+1$$ than with
$$intfracdx(x-i)(x+i).$$
answered Aug 11 at 14:23
Yves Daoust
113k665207
answered Aug 11 at 14:23
Yves Daoust
113k665207
answered Aug 11 at 14:23
Yves Daoust
113k665207
answered Aug 11 at 14:23
Yves Daoust
113k665207
113k665207
Is it?$$
– tomasz
Aug 12 at 15:54
@tomasz: It is.
– Yves Daoust
Aug 12 at 16:05
add a comment |Â
Is it?$$
– tomasz
Aug 12 at 15:54
@tomasz: It is.
– Yves Daoust
Aug 12 at 16:05
Is it?$$
– tomasz
Aug 12 at 15:54
Is it?$$
– tomasz
Aug 12 at 15:54
@tomasz: It is.
– Yves Daoust
Aug 12 at 16:05
@tomasz: It is.
– Yves Daoust
Aug 12 at 16:05
add a comment |Â
up vote
6
down vote
Beacuse $(x+2)(x+10)= x^2+12x+20$ and not $x^2+8x+20$
Of course!! I can't believe how stupid that question was thanks guys.
– Shauna
Aug 11 at 14:19
add a comment |Â
up vote
6
down vote
Beacuse $(x+2)(x+10)= x^2+12x+20$ and not $x^2+8x+20$
Of course!! I can't believe how stupid that question was thanks guys.
– Shauna
Aug 11 at 14:19
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Beacuse $(x+2)(x+10)= x^2+12x+20$ and not $x^2+8x+20$
Beacuse $(x+2)(x+10)= x^2+12x+20$ and not $x^2+8x+20$
answered Aug 11 at 14:17
user582949
Of course!! I can't believe how stupid that question was thanks guys.
– Shauna
Aug 11 at 14:19
add a comment |Â
Of course!! I can't believe how stupid that question was thanks guys.
– Shauna
Aug 11 at 14:19
Of course!! I can't believe how stupid that question was thanks guys.
– Shauna
Aug 11 at 14:19
Of course!! I can't believe how stupid that question was thanks guys.
– Shauna
Aug 11 at 14:19
add a comment |Â
up vote
4
down vote
Because it is not true that $x^2+8x+20=(x+2)(x+10)$. Completing the square is a natural choice whenever (as in this case) the quadratic has no real roots.
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
add a comment |Â
up vote
4
down vote
Because it is not true that $x^2+8x+20=(x+2)(x+10)$. Completing the square is a natural choice whenever (as in this case) the quadratic has no real roots.
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Because it is not true that $x^2+8x+20=(x+2)(x+10)$. Completing the square is a natural choice whenever (as in this case) the quadratic has no real roots.
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
Because it is not true that $x^2+8x+20=(x+2)(x+10)$. Completing the square is a natural choice whenever (as in this case) the quadratic has no real roots.
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
edited Aug 11 at 14:28
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
answered Aug 11 at 14:19
José Carlos Santos
119k16101182
119k16101182
add a comment |Â
add a comment |Â
up vote
3
down vote
You're factoring wrong, the right factorization is $x^2+8x+20=(x+4-2i)(x+4+2i)$
I think completing a square is a natural choice if quadratic doesn't have real roots.
$$x^2+8x+20=x^2+8x+16+4=(x+4)^2+2^2$$
$$int frac1(x+4)^2+2^2,dx = frac12arctanleft(fracx+42right)+C $$
answered Aug 11 at 14:20
Deepesh Meena
2,688721
-1 If the question is "Why do I have to complete the square to solve the integral", an answer showing you completing the square and solving the integral is pointless. The OP is specifically asking why the other method doesn't work, he's not interested in how to solve the integral with this method.
– Ant
Aug 12 at 11:31
but OP was doing wrong factorization of the polynomial
– Deepesh Meena
Aug 12 at 11:35
$x^2+8x+20=(x+4-2i)(x+4+2i)$ I think completing a square is a natural choice if quadratic doesn't have real roots
– Deepesh Meena
Aug 12 at 11:39
Sure, I agree. But then the answer (that was given) is "You're factoring wrong, if you do it correctly it will work too". But I think that showing how the square completion work is just unhelpful. If you instead add this to your ansswer, show how factorization works, and then make a comparison with the "completing the square method", elaborating on why you think it's easier / more natural, then I would upvote it. But as it is you're just ignoring the OP question and just showing that you can do that integral, which is unhelpful
– Ant
Aug 12 at 13:22
sure I have edited the answer
– Deepesh Meena
Aug 12 at 13:59
add a comment |Â
up vote
3
down vote
You're factoring wrong, the right factorization is $x^2+8x+20=(x+4-2i)(x+4+2i)$
I think completing a square is a natural choice if quadratic doesn't have real roots.
$$x^2+8x+20=x^2+8x+16+4=(x+4)^2+2^2$$
$$int frac1(x+4)^2+2^2,dx = frac12arctanleft(fracx+42right)+C $$
answered Aug 11 at 14:20
Deepesh Meena
2,688721
-1 If the question is "Why do I have to complete the square to solve the integral", an answer showing you completing the square and solving the integral is pointless. The OP is specifically asking why the other method doesn't work, he's not interested in how to solve the integral with this method.
– Ant
Aug 12 at 11:31
but OP was doing wrong factorization of the polynomial
– Deepesh Meena
Aug 12 at 11:35
$x^2+8x+20=(x+4-2i)(x+4+2i)$ I think completing a square is a natural choice if quadratic doesn't have real roots
– Deepesh Meena
Aug 12 at 11:39
Sure, I agree. But then the answer (that was given) is "You're factoring wrong, if you do it correctly it will work too". But I think that showing how the square completion work is just unhelpful. If you instead add this to your ansswer, show how factorization works, and then make a comparison with the "completing the square method", elaborating on why you think it's easier / more natural, then I would upvote it. But as it is you're just ignoring the OP question and just showing that you can do that integral, which is unhelpful
– Ant
Aug 12 at 13:22
sure I have edited the answer
– Deepesh Meena
Aug 12 at 13:59
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You're factoring wrong, the right factorization is $x^2+8x+20=(x+4-2i)(x+4+2i)$
I think completing a square is a natural choice if quadratic doesn't have real roots.
$$x^2+8x+20=x^2+8x+16+4=(x+4)^2+2^2$$
$$int frac1(x+4)^2+2^2,dx = frac12arctanleft(fracx+42right)+C $$
answered Aug 11 at 14:20
Deepesh Meena
2,688721
You're factoring wrong, the right factorization is $x^2+8x+20=(x+4-2i)(x+4+2i)$
I think completing a square is a natural choice if quadratic doesn't have real roots.
$$x^2+8x+20=x^2+8x+16+4=(x+4)^2+2^2$$
$$int frac1(x+4)^2+2^2,dx = frac12arctanleft(fracx+42right)+C $$
answered Aug 11 at 14:20
Deepesh Meena
2,688721
edited Aug 12 at 13:59
answered Aug 11 at 14:20
Deepesh Meena
2,688721
answered Aug 11 at 14:20
Deepesh Meena
2,688721
answered Aug 11 at 14:20
Deepesh Meena
2,688721
2,688721
-1 If the question is "Why do I have to complete the square to solve the integral", an answer showing you completing the square and solving the integral is pointless. The OP is specifically asking why the other method doesn't work, he's not interested in how to solve the integral with this method.
– Ant
Aug 12 at 11:31
but OP was doing wrong factorization of the polynomial
– Deepesh Meena
Aug 12 at 11:35
$x^2+8x+20=(x+4-2i)(x+4+2i)$ I think completing a square is a natural choice if quadratic doesn't have real roots
– Deepesh Meena
Aug 12 at 11:39
Sure, I agree. But then the answer (that was given) is "You're factoring wrong, if you do it correctly it will work too". But I think that showing how the square completion work is just unhelpful. If you instead add this to your ansswer, show how factorization works, and then make a comparison with the "completing the square method", elaborating on why you think it's easier / more natural, then I would upvote it. But as it is you're just ignoring the OP question and just showing that you can do that integral, which is unhelpful
– Ant
Aug 12 at 13:22
sure I have edited the answer
– Deepesh Meena
Aug 12 at 13:59
add a comment |Â
-1 If the question is "Why do I have to complete the square to solve the integral", an answer showing you completing the square and solving the integral is pointless. The OP is specifically asking why the other method doesn't work, he's not interested in how to solve the integral with this method.
– Ant
Aug 12 at 11:31
but OP was doing wrong factorization of the polynomial
– Deepesh Meena
Aug 12 at 11:35
$x^2+8x+20=(x+4-2i)(x+4+2i)$ I think completing a square is a natural choice if quadratic doesn't have real roots
– Deepesh Meena
Aug 12 at 11:39
Sure, I agree. But then the answer (that was given) is "You're factoring wrong, if you do it correctly it will work too". But I think that showing how the square completion work is just unhelpful. If you instead add this to your ansswer, show how factorization works, and then make a comparison with the "completing the square method", elaborating on why you think it's easier / more natural, then I would upvote it. But as it is you're just ignoring the OP question and just showing that you can do that integral, which is unhelpful
– Ant
Aug 12 at 13:22
sure I have edited the answer
– Deepesh Meena
Aug 12 at 13:59
-1 If the question is "Why do I have to complete the square to solve the integral", an answer showing you completing the square and solving the integral is pointless. The OP is specifically asking why the other method doesn't work, he's not interested in how to solve the integral with this method.
– Ant
Aug 12 at 11:31
-1 If the question is "Why do I have to complete the square to solve the integral", an answer showing you completing the square and solving the integral is pointless. The OP is specifically asking why the other method doesn't work, he's not interested in how to solve the integral with this method.
– Ant
Aug 12 at 11:31
but OP was doing wrong factorization of the polynomial
– Deepesh Meena
Aug 12 at 11:35
but OP was doing wrong factorization of the polynomial
– Deepesh Meena
Aug 12 at 11:35
$x^2+8x+20=(x+4-2i)(x+4+2i)$ I think completing a square is a natural choice if quadratic doesn't have real roots
– Deepesh Meena
Aug 12 at 11:39
$x^2+8x+20=(x+4-2i)(x+4+2i)$ I think completing a square is a natural choice if quadratic doesn't have real roots
– Deepesh Meena
Aug 12 at 11:39
Sure, I agree. But then the answer (that was given) is "You're factoring wrong, if you do it correctly it will work too". But I think that showing how the square completion work is just unhelpful. If you instead add this to your ansswer, show how factorization works, and then make a comparison with the "completing the square method", elaborating on why you think it's easier / more natural, then I would upvote it. But as it is you're just ignoring the OP question and just showing that you can do that integral, which is unhelpful
– Ant
Aug 12 at 13:22
Sure, I agree. But then the answer (that was given) is "You're factoring wrong, if you do it correctly it will work too". But I think that showing how the square completion work is just unhelpful. If you instead add this to your ansswer, show how factorization works, and then make a comparison with the "completing the square method", elaborating on why you think it's easier / more natural, then I would upvote it. But as it is you're just ignoring the OP question and just showing that you can do that integral, which is unhelpful
– Ant
Aug 12 at 13:22
sure I have edited the answer
– Deepesh Meena
Aug 12 at 13:59
sure I have edited the answer
– Deepesh Meena
Aug 12 at 13:59
add a comment |Â
up vote
2
down vote
A more general answer to the question "when should I complete the square?" is:
- If you are fundamentally concerned with the roots of the quadratic, you should factorise (since factorising gives you the roots for free).
- If you are more concerned with the curve as a whole, then completing the square can often help, because then you get the transformations of $y=x^2$ that produce the quadratic.
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
add a comment |Â
up vote
2
down vote
A more general answer to the question "when should I complete the square?" is:
- If you are fundamentally concerned with the roots of the quadratic, you should factorise (since factorising gives you the roots for free).
- If you are more concerned with the curve as a whole, then completing the square can often help, because then you get the transformations of $y=x^2$ that produce the quadratic.
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A more general answer to the question "when should I complete the square?" is:
- If you are fundamentally concerned with the roots of the quadratic, you should factorise (since factorising gives you the roots for free).
- If you are more concerned with the curve as a whole, then completing the square can often help, because then you get the transformations of $y=x^2$ that produce the quadratic.
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
A more general answer to the question "when should I complete the square?" is:
- If you are fundamentally concerned with the roots of the quadratic, you should factorise (since factorising gives you the roots for free).
- If you are more concerned with the curve as a whole, then completing the square can often help, because then you get the transformations of $y=x^2$ that produce the quadratic.
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
answered Aug 11 at 19:46
Patrick Stevens
27.1k52769
27.1k52769
add a comment |Â
add a comment |Â
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8
$(x+2)(x+10)=x^2+12x+20$, not $x^2+8x+20$. You can indeed solve it by your method, but you will need to use the appropriate factors $x^2+8x+20=(x+4-2i)(x+4+2i)$.
– user583012
Aug 11 at 14:18
You can use this: en.wikipedia.org/wiki/… after you integrated to get back to a "real looking" form.
– Zacky
Aug 11 at 14:34