Mixing
Complex quadratic equation always comes out as wrong
$begingroup$
For some reason I always get the wrong answer and I don't understand why:
$$4z^2-12z+19=0$$
I got $z= frac{12 pm sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 pm sqrt{10}i}{2} $ where is my mistake?
algebra-precalculus complex-numbers quadratics
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
asked Jan 17 at 14:11
ythhtrgythhtrg
193
$endgroup$
add a comment |
$begingroup$
For some reason I always get the wrong answer and I don't understand why:
$$4z^2-12z+19=0$$
I got $z= frac{12 pm sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 pm sqrt{10}i}{2} $ where is my mistake?
algebra-precalculus complex-numbers quadratics
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
asked Jan 17 at 14:11
ythhtrgythhtrg
193
$endgroup$
2
$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
Jan 17 at 14:14
$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset assqrt{260}.
$endgroup$
– Martin R
Jan 17 at 14:26
add a comment |
$begingroup$
For some reason I always get the wrong answer and I don't understand why:
$$4z^2-12z+19=0$$
I got $z= frac{12 pm sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 pm sqrt{10}i}{2} $ where is my mistake?
algebra-precalculus complex-numbers quadratics
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
asked Jan 17 at 14:11
ythhtrgythhtrg
193
$endgroup$
For some reason I always get the wrong answer and I don't understand why:
$$4z^2-12z+19=0$$
I got $z= frac{12 pm sqrt{260}i}{8} $ and the answer is supposed to be $z=frac{3 pm sqrt{10}i}{2} $ where is my mistake?
algebra-precalculus complex-numbers quadratics
algebra-precalculus complex-numbers quadratics
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
asked Jan 17 at 14:11
ythhtrgythhtrg
193
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
asked Jan 17 at 14:11
ythhtrgythhtrg
193
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
edited Jan 17 at 20:53
J. W. Tanner
2,4311117
2,4311117
asked Jan 17 at 14:11
ythhtrgythhtrg
193
asked Jan 17 at 14:11
ythhtrgythhtrg
193
asked Jan 17 at 14:11
ythhtrgythhtrg
193
193
2
$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
Jan 17 at 14:14
$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset assqrt{260}.
$endgroup$
– Martin R
Jan 17 at 14:26
add a comment |
2
$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
Jan 17 at 14:14
$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset assqrt{260}.
$endgroup$
– Martin R
Jan 17 at 14:26
2
2
$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
Jan 17 at 14:14
$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
Jan 17 at 14:14
$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset as
sqrt{260} .$endgroup$
– Martin R
Jan 17 at 14:26
$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset as
sqrt{260} .$endgroup$
– Martin R
Jan 17 at 14:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your mistake is that the discriminant should be $12^2 - 4times 4times 19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.
edited Jan 17 at 18:32
Mutantoe
612513
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
$endgroup$
1
$begingroup$
In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process.
$endgroup$
– markbruns
Jan 18 at 3:49
add a comment |
$begingroup$
When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$
Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.
Then the roots are $$frac{-b pm sqrt{E}}{a}.$$
For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
$endgroup$
2
$begingroup$
When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one.
$endgroup$
– A. Howells
Jan 17 at 18:38
1
$begingroup$
@A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient.
$endgroup$
– coffeemath
Jan 17 at 18:49
1
$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
Jan 17 at 19:51
1
$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
Jan 18 at 0:20
1
$begingroup$
@coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$.
$endgroup$
– JiK
Jan 18 at 1:53
|
show 1 more 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
$begingroup$
Your mistake is that the discriminant should be $12^2 - 4times 4times 19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.
edited Jan 17 at 18:32
Mutantoe
612513
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
$endgroup$
1
$begingroup$
In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process.
$endgroup$
– markbruns
Jan 18 at 3:49
add a comment |
$begingroup$
Your mistake is that the discriminant should be $12^2 - 4times 4times 19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.
edited Jan 17 at 18:32
Mutantoe
612513
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
$endgroup$
1
$begingroup$
In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process.
$endgroup$
– markbruns
Jan 18 at 3:49
add a comment |
$begingroup$
Your mistake is that the discriminant should be $12^2 - 4times 4times 19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.
edited Jan 17 at 18:32
Mutantoe
612513
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
$endgroup$
Your mistake is that the discriminant should be $12^2 - 4times 4times 19 = -160$, not $-260$, and then dividing top and bottom by 4 gives the correct answer.
edited Jan 17 at 18:32
Mutantoe
612513
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
edited Jan 17 at 18:32
Mutantoe
612513
edited Jan 17 at 18:32
Mutantoe
612513
edited Jan 17 at 18:32
Mutantoe
612513
612513
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
answered Jan 17 at 14:22
J. W. TannerJ. W. Tanner
2,4311117
2,4311117
1
$begingroup$
In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process.
$endgroup$
– markbruns
Jan 18 at 3:49
add a comment |
1
$begingroup$
In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process.
$endgroup$
– markbruns
Jan 18 at 3:49
1
1
$begingroup$
In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process.
$endgroup$
– markbruns
Jan 18 at 3:49
$begingroup$
In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process.
$endgroup$
– markbruns
Jan 18 at 3:49
add a comment |
$begingroup$
When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$
Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.
Then the roots are $$frac{-b pm sqrt{E}}{a}.$$
For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
$endgroup$
2
$begingroup$
When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one.
$endgroup$
– A. Howells
Jan 17 at 18:38
1
$begingroup$
@A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient.
$endgroup$
– coffeemath
Jan 17 at 18:49
1
$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
Jan 17 at 19:51
1
$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
Jan 18 at 0:20
1
$begingroup$
@coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$.
$endgroup$
– JiK
Jan 18 at 1:53
|
show 1 more comment
$begingroup$
When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$
Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.
Then the roots are $$frac{-b pm sqrt{E}}{a}.$$
For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
$endgroup$
2
$begingroup$
When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one.
$endgroup$
– A. Howells
Jan 17 at 18:38
1
$begingroup$
@A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient.
$endgroup$
– coffeemath
Jan 17 at 18:49
1
$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
Jan 17 at 19:51
1
$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
Jan 18 at 0:20
1
$begingroup$
@coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$.
$endgroup$
– JiK
Jan 18 at 1:53
|
show 1 more comment
$begingroup$
When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$
Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.
Then the roots are $$frac{-b pm sqrt{E}}{a}.$$
For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
$endgroup$
When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$
Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when squareroot of discriminant is taken.
Then the roots are $$frac{-b pm sqrt{E}}{a}.$$
For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4cdot 19=36-4cdot 19=36-76=-40.$ Then roots are $(6 pm sqrt{-40})/12.$ This simplifies again in this case.
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
answered Jan 17 at 14:38
coffeemathcoffeemath
2,8471415
2,8471415
2
$begingroup$
When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one.
$endgroup$
– A. Howells
Jan 17 at 18:38
1
$begingroup$
@A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient.
$endgroup$
– coffeemath
Jan 17 at 18:49
1
$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
Jan 17 at 19:51
1
$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
Jan 18 at 0:20
1
$begingroup$
@coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$.
$endgroup$
– JiK
Jan 18 at 1:53
|
show 1 more comment
2
$begingroup$
When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one.
$endgroup$
– A. Howells
Jan 17 at 18:38
1
$begingroup$
@A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient.
$endgroup$
– coffeemath
Jan 17 at 18:49
1
$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
Jan 17 at 19:51
1
$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
Jan 18 at 0:20
1
$begingroup$
@coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$.
$endgroup$
– JiK
Jan 18 at 1:53
2
2
$begingroup$
When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one.
$endgroup$
– A. Howells
Jan 17 at 18:38
$begingroup$
When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one.
$endgroup$
– A. Howells
Jan 17 at 18:38
1
1
$begingroup$
@A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient.
$endgroup$
– coffeemath
Jan 17 at 18:49
$begingroup$
@A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient.
$endgroup$
– coffeemath
Jan 17 at 18:49
1
1
$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
Jan 17 at 19:51
$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
Jan 17 at 19:51
1
1
$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
Jan 18 at 0:20
$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
Jan 18 at 0:20
1
1
$begingroup$
@coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$.
$endgroup$
– JiK
Jan 18 at 1:53
$begingroup$
@coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$.
$endgroup$
– JiK
Jan 18 at 1:53
|
show 1 more comment
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2
$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
Jan 17 at 14:14
$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset as
sqrt{260}.$endgroup$
– Martin R
Jan 17 at 14:26