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I would like to know if it is possible to select R1, R2, R3, C1, and C2 such that the quadratic equation yields complex roots.

$s_{1,2}=-\frac{b\pm \sqrt{b^2-4ac}}{2a}$


$a = R_{1}R_{2}R_{3}C_{1}C_{2}$

$b = R_{1}C_{1}(R_{2}+R_{3})+R_{3}C_{2}(R_{1}+R_{2})$

$c = R_{1}+R_{2}+R_{3}$

Obviously this depends solely on $b^2-4ac$. With 6 variables, even with maximum factoring and simplification, it is difficult to see whether 4a can be made greater than $b^2$. Is there a rule/theory/theorem which I can turn to in order to prove whether there exists any combination of the six variables that would have $4ac>b^2$?



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As Jason specified in commenting on Bill Cook's answer, he wants all variables real and positive. The answer is no, there are no such $R_1,R_2,R_3,C_1,C_2$ that make $b^2 - 4 a c < 0$.

In fact, we can write $b^2 - 4 a c = p C_2^2 + q C_2 + r$ with $p = {R_{{3}}}^{2} \left( R_{{1}}+R_{{2}} \right) ^{2}$, $q = -2\,C_{{1}}R_{{1}}R_{{3}} \left( -R_{{1}}R_{{3}}+R_{{2}}R_{{3}}+R_{{2} }R_{{1}}+{R_{{2}}}^{2} \right)$, and $r = {R_{{1}}}^{2}{C_{{1}}}^{2} \left( R_{{2}}+R_{{3}} \right) ^{2}$. Now $q^2 - 4 p r = -16\,{C_{{1}}}^{2}{R_{{1}}}^{3}{R_{{3}}}^{3}R_{{2}} \left( R_{{1}}+R_{ {2}}+R_{{3}} \right) < 0$ when all variables are positive, implying that $b^2 - 4 a c$ can't change sign.

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Using Maple code..

a := R[1]*R[2]*R[3]*C[1]*C[2];

b := R[1]C[1](R[2]+R[3])+R[3]C[2](R[1]+R[2]);

c := R[1]+R[2]+R[3];

roll := rand(-100..100):

R[1] := roll();

R[2] := roll();

R[3] := roll();

C[1] := roll();

C[2] := roll();

C[3] := roll();


After a few tries...

R[1] = -84

R[2] = 97

R[3] = -49

C[1] = -20

C[2] = 61

C[3] = 36


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Sorry, I should have specified the constraint that all R's and C's must be real and postive. – Jason Dugas Nov 20 '11 at 3:50
If you want the variables to be all positive, there is no solution according to Maple's Global Optimization Toolbox. – Robert Israel Nov 20 '11 at 4:16
What do you mean "no solution?" – Jason Dugas Nov 20 '11 at 4:54
I mean that for all $R_1, R_2, R_3, C_1, C_2 > 0$, $b^2 - 4 a c \ge 0$. – Robert Israel Nov 20 '11 at 6:49

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