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I'm reading Sawyer's Prelude to Mathematics, here:


Book's excerpt


I can't understand what's the meaning and application of "condition" here. Also when he gives the example on the cubic equation, stating that the condition is: $$(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$$

I can understand that it is $b^2-4ac=0$ (I hope I'm right with this), I just have no idea on where is the order of the variables inside the parentheses coming from.


1: I noticed that the $b^2-4ac$ can be found here:

$$-b\pm \frac{\sqrt{b^2-4ac}}{2a}$$

Which could be found by solving a general form quadratic equation:

$$ax^2+bx+c=0$$

Then I thought about searching it on the solutions for cubic equations with some help of Mathematica, but I got nothing that was similar to:

$$(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$$

or:

$$a^2d^2-6abcd+4b^3d+4ac^3-3b^2c^2=0$$

With no success. You can see it here:

enter image description here

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    He writes $ax^2+2bx+c=0$ which completing the square gives $a(x+b/a)^2+c-ab^2/a^2=0$ for this to be a perfect square we need $c-b^2/a=0$ or simply $ac-b^2=0$. I guess you understand this part already. I'll have to think about the other part.2012-09-01
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    I'm reading the condition $(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$ as the condition for a cubic to have a three-peated root. You say "quadratic" I think this is a typo.2012-09-01
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    @James: No. If you look at the example $x^3-x^2$, which has two (not three) repeated roots, you will find that the expression is zero. (And I fixed the "quadratic" typo.)2012-09-01
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    Read the few words just before the condition. This tells where the constants $a$, $b$, $c$, and $d$ come from. As the few sentences before that describe, the author starts by talking about a quadratic equation then changes to talking about a cubic equation.2012-09-01
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    @JamesS.Cook I think the "typo" actually points out where the OP is confused. He might have missed that the second condition is for a cubic equation rather than a quadratic one.2012-09-01
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    @Code-Guru: You haven't answered the OP's question -- where does that expression come from? It's not completely obvious (not to me or the OP, anyway).2012-09-01
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    @TonyK The OP asked "I just have no idea on where are the variables inside the paretheses coming from". I don't see any question about where the expression itself comes from. I certainly agree that it isn't obvious. It probably isn't even trivial to derive.2012-09-01
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    Consider $ax^3+3bx^2+3cx+d=0$. We wish to write it as a perfect cube. Dividing by $a$ yields: $$ x^3+(3b/a)x^2+(3c/a)x+d/a = (x-m)^3$$ We need to find $m$. Note $$ x^3+(3b/a)x^2+(3c/a)x+d/a = x^3-3mx^2+3xm^2-m^3$$ Equating coefficients gives: $$ b/a=-m, \ \ c/a=m^2, \ \ d/a=-m^3 $$ and eliminating $m$ would reveal the necessary condtions on $a,b,c,d$ for the triple root. Although, clearly this was not asked :(2012-09-01
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    @TonyK I read the OP as confusion about the jump between the example of quadratic equations to a discussion about applying the same concept to cubic equations. This is just my interpretation of what I read. I may be wrong.2012-09-01
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    @Code-Guru: Yes, the OP is not as clear as I thought on first reading. But if there _is_ an interesting question in there, it has to be: Where does the expression $(bc-ad)^2-...$ come from?2012-09-01
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    With three roots $x_1, x_2, x_3$, the coefficients of the polynomial are symmetric expressions in these, e.g. $\frac{-3b}a=x_1+x_2+x_3$. Two roots are equal iff $(x_1-x_2)(x_1-x_3)(x_2-x_3)=0$, the square of which turns out to be symmetric in the $x_i$ and by a general theorem such can be expressed using $x_1+x_2+x_3$, $x_1x_2+x_1x_3+x_2x_3$ and $x_1x_2x_3$, i.e. essentially the polynomial coefficients. That's how one obtains that weird expression in the coefficients. (That's also how one might arrive at $b^2-(4)ac=0$ because the left hand is in fact $(x_1-x_2)^2$ (up to a factor of $a^2$))2012-09-01
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    @TonyK I agree that this is an interesting question. I'm just not inclined to tackle the derivation =p2012-09-01
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    It has just come to my attention that you were unhappy with the change I made to the title of your question, and I wanted to apologize. I am sorry.2012-12-14
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    @MJD No problem, I just wanted to express the exactly as it was in my mind, this is the the most plausible way to me.2012-12-14

4 Answers 4