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If the quadratic formula is complete formula for solving the quadratic equation why is i= [(5±i√7)/2]+i one of the infinitely many solutions of x^2+3x+5=0? If that is so are the formulas for cubic or quartic equations complete? Why for example does -8+3i and 17+2i fit to be roots of the same equation is it a coincidence or is it that the quadratic formula is incomplete? If the quadratic formula is incomplete what is the guarantee that the cubic or even equation formulas are complete? I realize that the roots I sent could not solve the problem. I was investigating a condition in which if: x^2+a_1 x+a_0=0 ---------- 1 If we permit a complex number, x=c+di ---------2 Then 〖(c+di)〗^2+a_1 (c+di)+a_0=0 ----------- 3 Expanding and simplifying: c^2-d^2+a_1 c+a_0+i(2cd+a_1 d)=0 ------------ 4 Equation the real terms of the expansion to zero: c^2-d^2+a_1 c+a_0=0 ---------------- 5 Equating the imaginary terms of the expansion to zero: 2cd+a_1 d=0 ---------------- 6 Adding together 5 and 6: c^2+c(a_1+2d)+a_1 d+a_0=0 ------------- 7 Solving 7: c=(a_1+2d±√(2&((a_1+2d)^2-4(a_1 d+a_0-d^2)))/2 ------------- 8 Substituting 8 into 2: x=(a_1+2d±√(2&((a_1+2d)^2-4(a_1 d+a_0-d^2)))/2+di ------- 9 When d = 0 the quadratic equation is recovered. This was a surprising result to me and I was wondering if equation 9 could be used to produce results using any random number d. I discovered it couldn’t. I realized that the only way for the imaginary coefficient to be zero Is for d to be equal to zero, in which case the quadratic formula as it is formulated is complete.

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closed as unclear what you're asking by The Chaz 2.0, azimut, Davide Giraudo, Hagen von Eitzen, Daniel Fischer Oct 16 '13 at 18:26

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What did you actually mean to write between "why is" and "one of the..." ?? –  DonAntonio Oct 16 '13 at 16:38
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$(-8 + 3 i)^2 + 3 (-8 + 3 i) + 5 = 36-39 i \neq 0$. None of your other "roots" work either. Why do you think they do? –  Macavity Oct 16 '13 at 17:29
    
Samuel, please edit your question and include a demonstration that $-8 +3i$ "fits to be a root". Also, wrap your math equations in dollar signs ($) –  The Chaz 2.0 Oct 16 '13 at 17:33
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2 Answers 2

The solutions to a quadratic equation written in the form $ax^2 + bx + c = 0$ are given by $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Hence the solutions to $$x^2 + 3x + 5 = 0$$ are given by $$x = \dfrac{ -3 \pm i\sqrt {11}}{2}$$ There are exactly $2$ solutions to this equation, not infinitely many!

Yes, the quadratic formula is "complete" given any quadratic, we can always find solutions using the quadratic formula. Sometimes, a root may not be real, as in the case above.

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Why is it that 1+2i and 6+2i fit also as roots of the question. –  Samuel Bonaya Buya Oct 16 '13 at 17:11
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$$(1 + 2i)^2 + 3(1 +2i) +5 = 5 + 10i$$ –  The Chaz 2.0 Oct 16 '13 at 17:30
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... and $5 + 10i \neq 0$. What made you think that those are solutions to the quadratic?!? –  The Chaz 2.0 Oct 16 '13 at 17:31
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@Samuel We cannot mix up the coefficients $a, b, c$ of a quadratic equation as we please. With respect to the quadratic formula for $ax^2 + bx + c = 0$, as given in the answer, $a$ must be the coefficient for $x^2$, $b$ the coefficient of $x$, and $c$ the coefficient of $x^0$. In this case, $a = 1, b = 3, c = 5$. We can not assign to the $a$ position any number other than $1$, $b$ no other than $3$, and $c$ no other than $5$. –  amWhy Oct 16 '13 at 17:36
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The quadratic formula is $$ x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}. $$

Substituting a = 1, b = 3 and c = 5 into this gives $$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(5)} }{2(1)} = \frac{-3 \pm \sqrt{9 - 20} }{2} = \frac{-3 \pm \sqrt{-11} }{2}= \frac{-3 \pm \sqrt{11}i }{2}$$

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why is it then when we substitute the the two roots the into the they fit as roots? –  Samuel Bonaya Buya Oct 16 '13 at 17:04
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I think you hit the "the" key a couple of times by mistake. –  Eric Tressler Oct 16 '13 at 17:27
    
How? the fundamental theory of algebra requires two roots of a quadratic equation at least under a given basis? In each case here we have two roots. What if we have a pair root involving π could it still be a coincidence? –  Samuel Bonaya Buya Oct 16 '13 at 17:37
    
If by 'the two roots' you mean 1+2i and 6+2i (do you?), The Chaz has demonstrated above that 1 + 2i isn't a root. You can check that 1 + 6i isn't using the same method. To answer your question 'how?', the reason $\frac{-3 + \sqrt{11}i }{2}$ is a root is because if we plug it into the equation $x^2 + 3x + 5$ we get $$( \frac{-3 + \sqrt{11}i }{2})^2 + 3( \frac{-3 + \sqrt{11}i }{2} ) + 5 = 0. \\ $$ The fundamental theorem of algebra does indeed say that a polynomial of degree n has n roots. Can what be a coincidence? @Samuel –  George Tomlinson Oct 16 '13 at 18:42
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