Quadratic Craze
The quadratic equation above has complex roots. Let be one of the roots. If the harmonic mean of the roots is and , determine the sum of all values of in degrees for . Add 360 to your sum and submit.
The answer is 1080.
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1 solution
If ( a + b i ) and ( a − b i ) are the roots, then a 2 + b 2 = 1 since ∣ z ∣ = a 2 + b 2 = 1 and a 2 + b 2 = 2 a since the harmonic mean is 2 . As a result, 2 a = 1 so a = 0 . 5 . In a Quadratic Equation with complex roots, x 2 − ( 2 a ) x + ( a 2 + b 2 ) = 0 by Vieta's Formula. Since a = 0 . 5 and a 2 + b 2 = 1 , the equation becomes x 2 − x + 1 = 0 . By substitution:
Next, by substitution, cos ( 2 β ) − sin ( β ) = a 2 + b 2 = 1 . cos ( 2 β ) = 1 − 2 sin ( β ) 2 , so 1 − 2 sin ( β ) 2 − sin ( β ) = 1 . By solving the Quadratic, sin ( β ) = − 0 . 5 and sin ( β ) = 0 . Thus, β = 3 3 0 , 2 1 0 , 1 8 0 since 0 < β < 3 6 0 . As a result, 3 3 0 + 2 1 0 + 1 8 0 = 7 2 0 , and 7 2 0 + 3 6 0 = 1 0 8 0 which is the final answer.