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Complex Roots by Formula
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and b 2 = 32 =9 Therefore, d = b 2 4ac = 9 90 = 81 This tells us that the roots of the quadratic are not real numbers To find the roots, we can plug in the value 81 for d in the abbreviated discriminant form of the quadratic formula: x = [ b j (|d |1/2)] / (2a) = [ 3 j (| 81|)1/2] / [2 (45/2)] = [ 3 j (811/2)] / 45 = ( 3 j 9) / 45 = 3/45 j (9/45) = 1/15 j (1/5) The roots are x = 1/15 + j (1/5) or x = 1/15 j (1/5) These are complex conjugates As things work out, the roots are always complex conjugates in a quadratic where d < 0, even when b 0 The solution set X in this example is X = { 1/15 + j (1/5), 1/15 j (1/5)}
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For complementary credit, plug the roots we ve just found into the original quadratic to be sure that they work You re on your own Here s a hint: This is a messy process, but if you re careful and patient, all the garbage will drop out in the end
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The above derivations are abstract, but it s important that you follow through them so you understand the reasoning behind each step Here are the results, wrapped up into two statements In any quadratic: If the discriminant is a negative real number and the coefficient of x is 0, then the roots are pure imaginary, and are additive inverses If the discriminant is a negative real number and the coefficient of x is a nonzero real number, then the roots are not pure imaginary, but are complex conjugates
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386 Quadratic Equations with Complex Roots
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Consider once again the general polynomial equation ax 2 + bx + c = 0 The discriminant is d = b 2 4ac Suppose that d < 0 and b 0 We can use the abbreviated discriminant version of the quadratic formula for complex roots: x = [ b j (|d |1/2)] / (2a) Applying the right-hand distributive rule for addition and subtraction in reverse to the right side, we can split that expression into a sum or difference of two ratios with a common denominator, like this: x = b /(2a) j (|d |1/2)/(2a) Therefore, we can write the two roots as x = b /(2a) + j (|d |1/2)/(2a) or x = b /(2a) j (|d |1/2)/(2a) The fact that these are complex conjugates is obscure because the expressions are messy To bring things into focus, we can change a couple of names Let p = b /(2a) and q = (|d |1/2)/(2a) Now we can rewrite the roots as x = p + jq or x = p jq These are complex conjugates (but not pure imaginary numbers) for all nonzero real numbers p and q You might ask, Are p and q really both nonzero The answer is Yes We can be sure that p 0 because b 0 and a 0, so b /(2a) can t be 0 We can be sure that q 0 because d 0 and a 0, so (|d |1/2)/(2a) can t be 0
Imaginary Roots in Factors
Imaginary Roots in Factors
Now that we ve found a way to solve quadratics that have real coefficients, a real constant, and a negative real discriminant, let s see what the factors of such equations look like We ll start with situations where the coefficient of x is 0, so the roots are pure imaginary
A specific case Consider again the quadratic equation we solved earlier, in which the roots are pure imaginary and additive inverses:
3x 2 + 75 = 0 We simplified this equation in the second version of the solution when we divided both sides by 3, obtaining x 2 + 25 = 0 We found that the roots are j 5 and j 5 Knowing these roots, it s reasonable to think that we should be able to figure out what this equation looks like in binomial factor form In one case, we have x = j 5 If we take that equation and subtract j 5 from each side, we get x j5 = 0 In the other case, we have x = j 5 We can add j 5 to each side of that, getting x + j5 = 0 This suggests that the binomial factor form of the quadratic is (x j 5)(x + j 5) = 0 The left side of this equation is 0 if and only if x = j 5 or x = j 5 Let s multiply it out and see what we get To avoid confusion with the signs, we can rewrite the first term as a sum: [x + ( j 5)](x + j 5) = 0 Applying the product of sums rule gives us x 2 + x j 5 + ( j 5x) + ( j 5)( j 5) = 0 Using the commutative law for multiplication in the third and fourth terms, we obtain x 2 + x j 5 + ( x j 5) + ( j j ) 25 = 0 Note that j j = 1 Also, the second and third terms of the polynomial add up to 0 We can therefore simplify to get x 2 + 25 = 0