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Because p and q are both positive, we know that the ratio q /p is positive Its positive and negative square roots are therefore both real numbers We can take the square root of both sides of the above equation, getting x = [ 1(q /p)]1/2 = ( 1)1/2 [ (q /p)1/2] = j[(q /p)1/2] The roots are therefore x = j[(q /p)1/2] or x = j[(q /p)1/2]

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Question 23-7

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Is it possible for the roots of a quadratic equation to be pure imaginary but not additive inverses If so, provide an example of such an equation If not, explain why not

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Answer 23-7

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A quadratic equation can have roots that are pure imaginary but not additive inverses Here is an example of such an equation in binomial factor form: (x + j2)(x + j3) = 0 The roots of this equation are x = j2 or x = j3, as we can verify by plugging them in They are not additive inverses

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Question 23-8

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We have learned in the last several chapters (but not explicitly stated in full, until now!), that if the polynomial standard form of a quadratic equation has real coefficients and a real constant, then one of these things must be true: There are two different real roots There is a single real root with multiplicity 2 There are two different pure imaginary roots, and they are additive inverses There are two different complex roots, and they are conjugates

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In Answer 23-7, we found a quadratic equation that has two pure imaginary roots that are not additive inverses How is this possible

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Answer 23-8

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The coefficients and constant in the polynomial standard form of this equation are not all real numbers To see that, we can multiply the product of binomials out: (x + j2)(x + j3) = x 2 + j3x + j2x + (j2j3) = x 2 + j5x 6

Part Three 509

In this case, the coefficient of x is imaginary The coefficient of x 2, as well as the stand-alone constant, are real The complete polynomial quadratic equation is x 2 + j5x 6 = 0

Question 23-9

Is it possible for one root of a quadratic equation to be pure real and the other pure imaginary If so, provide an example of such an equation in polynomial standard form If not, explain why not

Answer 23-9

Yes, this is possible Here is an example of such an equation in binomial factor form: (x + 2)(x + j3) = 0 The roots of this equation are x = 2 or x = j3, as we can verify by plugging them in To convert this to polynomial standard form, we multiply the product of binomials out: (x + 2)(x + j3) = x 2 + j3x + 2x + j6 = x 2 + (2 + j3)x + j6 Here, the coefficient of x is complex but not pure imaginary, and the stand-alone constant is pure imaginary The coefficient of x 2 is real The complete polynomial quadratic equation is x 2 + (2 + j3)x + j6 = 0

Question 23-10

Is it possible for a quadratic equation to have two nonconjugate complex roots, neither or which is pure imaginary If so, provide an example of such an equation in polynomial standard form If this sort of situation is impossible, explain why

Answer 23-10

This, too, is possible! Suppose the roots are 1 + j and 2 + j These are non-conjugate complex numbers, and neither of them is pure imaginary We can construct a binomial factor quadratic with these numbers as roots by subtracting the roots from x, like this: [x (1 + j )][x (2 + j )] = 0 When we multiply the left side of this equation out to obtain a polynomial, taking extra precautions to be sure that we don t mess up with the signs, we obtain [x (1 + j)][x (2 + j)] = [x + ( 1) + ( j)][x + ( 2) + ( j)] = x 2 + ( 2x) + ( jx) + ( x) + 2 + j + ( jx) + j2 + ( 1) = x 2 + ( 3x) + ( j2x) + j3 + 1 = x 2 + ( 3 j2)x + (1 + j3)