Complex Roots

Algebra Level 2

Let a + b i a+bi and c + d i c+di be the two roots of the quadratic equation i x 2 + ( 6 − 7 i ) x − 21 − 9 i = 0 , ix^2+(6-7i)x-21-9i=0, where a < c a<c . What is the value of a d + b c ? ad+bc?

i i is the imaginary number that satisfies i 2 = − 1. i^2 = -1.

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Mahindra Jain by Mahindra Jain

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1 solution

Tom Engelsman
Nov 7, 2020

The roots can be computed via the Quadratic Formula:

x = ( − 6 + 7 i ) ± ( 6 − 7 i ) 2 − 4 ( i ) ( − 21 − 9 i ) 2 i = ( − 6 + 7 i ) ± 36 − 49 − 84 i + 84 i − 36 2 i = ( − 6 + 7 i ) ± − 49 2 i = ( − 6 + 7 i ) ± 7 i 2 i = 3 i , 7 + 3 i . x = \frac{(-6+7i) \pm \sqrt{(6-7i)^2-4(i)(-21-9i)}}{2i} = \frac{(-6+7i) \pm \sqrt{36-49-84i + 84i - 36}}{2i} = \frac{(-6+7i) \pm \sqrt{-49}}{2i} = \frac{(-6+7i) \pm 7i}{2i} = 3i, 7+3i.

Thus a = 0 , b = d = 3 , c = 7 ⇒ a d + b c = ( 0 ) ( 3 ) + ( 3 ) ( 7 ) = 21 . a =0, b =d = 3, c = 7 \Rightarrow ad+bc = (0)(3) + (3)(7) = \boxed{21}.

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