Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.1 Complex Numbers - 8.1 Exercises - Page 365: 120


Since -3 + 4i satisfies the equation $x^2+6x+25=0$, it is a solution of the equation.

Work Step by Step

For -3 + 4i to be a solution of the equation $x^2+6x+25=0$, -3 + 4i should satisfy the equation. Substitute $x=−3+4i$ into $x^2+6x+25$, we have $(−3+4i)^2+6(−3+4i)+25$ = $9−24i+16i^2−18+24i+25$ = $16+16i^2$ = $16−16$ $(i^2=−1)$ = 0 = R.H.S Therefore, -3 + 4i is a solution of the equation $x^2+6x+25=0$.
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