Complex quadratic equation always comes out as wrong












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    17 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    17 hours ago
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    17 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    17 hours ago














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago









J. W. Tanner

2169




2169










asked 17 hours ago









ythhtrgythhtrg

193




193








  • 2




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    17 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    17 hours ago














  • 2




    $begingroup$
    They are the same thing, only a factor of $4$ was cancelled.
    $endgroup$
    – Matti P.
    17 hours ago










  • $begingroup$
    Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
    $endgroup$
    – Martin R
    17 hours ago








2




2




$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
17 hours ago




$begingroup$
They are the same thing, only a factor of $4$ was cancelled.
$endgroup$
– Matti P.
17 hours ago












$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
$endgroup$
– Martin R
17 hours ago




$begingroup$
Note that $ sqrt{260} $ (is that is what you meant) is typeset as sqrt{260} .
$endgroup$
– Martin R
17 hours ago










2 Answers
2






active

oldest

votes


















8












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    3 hours ago





















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    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
    12 hours ago










  • $begingroup$
    @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
    12 hours ago










  • $begingroup$
    @coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
    $endgroup$
    – JiK
    11 hours ago










  • $begingroup$
    @JiK And that, divided by $4,$ is $b^2-ac.$
    $endgroup$
    – coffeemath
    7 hours ago










  • $begingroup$
    @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
    5 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    3 hours ago


















8












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    3 hours ago
















8












8








8





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 13 hours ago









Mutantoe

602512




602512










answered 17 hours ago









J. W. TannerJ. W. Tanner

2169




2169












  • $begingroup$
    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
    3 hours ago




















  • $begingroup$
    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
    3 hours ago


















$begingroup$
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
3 hours ago






$begingroup$
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
3 hours ago













3












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    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
    12 hours ago










  • $begingroup$
    @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
    12 hours ago










  • $begingroup$
    @coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
    $endgroup$
    – JiK
    11 hours ago










  • $begingroup$
    @JiK And that, divided by $4,$ is $b^2-ac.$
    $endgroup$
    – coffeemath
    7 hours ago










  • $begingroup$
    @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
    5 hours ago
















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    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
    12 hours ago










  • $begingroup$
    @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
    12 hours ago










  • $begingroup$
    @coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
    $endgroup$
    – JiK
    11 hours ago










  • $begingroup$
    @JiK And that, divided by $4,$ is $b^2-ac.$
    $endgroup$
    – coffeemath
    7 hours ago










  • $begingroup$
    @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
    5 hours ago














3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 16 hours ago









coffeemathcoffeemath

2,6881413




2,6881413








  • 1




    $begingroup$
    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
    12 hours ago










  • $begingroup$
    @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
    12 hours ago










  • $begingroup$
    @coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
    $endgroup$
    – JiK
    11 hours ago










  • $begingroup$
    @JiK And that, divided by $4,$ is $b^2-ac.$
    $endgroup$
    – coffeemath
    7 hours ago










  • $begingroup$
    @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
    5 hours ago














  • 1




    $begingroup$
    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
    12 hours ago










  • $begingroup$
    @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
    12 hours ago










  • $begingroup$
    @coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
    $endgroup$
    – JiK
    11 hours ago










  • $begingroup$
    @JiK And that, divided by $4,$ is $b^2-ac.$
    $endgroup$
    – coffeemath
    7 hours ago










  • $begingroup$
    @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
    5 hours ago








1




1




$begingroup$
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
12 hours ago




$begingroup$
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
12 hours ago












$begingroup$
@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
12 hours ago




$begingroup$
@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
12 hours ago












$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
11 hours ago




$begingroup$
@coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$.
$endgroup$
– JiK
11 hours ago












$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
7 hours ago




$begingroup$
@JiK And that, divided by $4,$ is $b^2-ac.$
$endgroup$
– coffeemath
7 hours ago












$begingroup$
@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
5 hours ago




$begingroup$
@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
5 hours ago


















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ASUS Zenbook UX433/UX333 — Configure Touchpad-embedded numpad on Linux

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0 1 I have an ASUS Zenbook UX433,on which ASUS decided to innovate by embedding a Numpad inside the touchpad (see picture 1). Toggling between classic touchpad mode and numpad (also turns the touchpad LEDs on) can be done through the golden button in the top-right corner of the touchpad. However, on my Manjaro arch Linux config, this behavior does not work by default and I can only use the touchpad, no numpad. Has anyone been able to configure the numpad to work on Linux, or have any idea how this could be achieved? EDIT1: I have had a look at the official Windows drivers for the UX433FA TouchPad, but was not able to make anything of it. I tried installing them on Linux using tools like NDISWrapper, but they do not detect the TouchPad/Numpad hardware, and my understanding of drivers is too limited. ...
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