How do I solve \((3+i)(3-i)\)?

( 3 + i ) ( 3 − i ) = 3 ( 3 − i ) + i ( 3 − i ) = 9 − 3 i + 3 i − i 2 = 9 − ( − 1 ) = = 9 + 1 = 10

Answered by kate200468 | 2024-06-10

(3+I)(3-1) -i^2+9=0 -I^2=-9 I^2=9 i= + or - 3

Answered by kayt91 | 2024-06-10

To solve ( 3 + i ) ( 3 − i ) , you can use the FOIL method, multiplying the parts of each complex number. The calculation leads to 9 + 1 = 10 because the imaginary parts cancel each other. Therefore, the answer is 10 .
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Answered by kate200468 | 2025-01-24