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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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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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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