x^2 - 8x +32=0? | fedest.com

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x^2 - 8x +32=0?

the 2 following the x is an exponont. I need to find the square square root for this problem.

2007-07-11 06:14:14 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

x² - 8x + 32 = 0
x² - 8x = -32
x² - 8x + 16 = -32 + 16
(x - 4)² = -16
x - 4 = ±√-16
x - 4 = ±4i
x = 4 ± 4i

2007-07-11 06:18:42 · answer #1 · answered by Philo 7 · 1⤊ 0⤋

Complete the square

x^2 -8x = -32
x^2 -8x (+ 16) = -32 (+16)
(x - 4) ^2 = -16

x-4 = sqrt(-16)

x = 4 +- 4i

2007-07-11 13:20:20 · answer #2 · answered by gfulton57 4 · 0⤊ 0⤋

x^2-8x+32=0

x^2-8x+16= -16

(x-4)^2= -16 take sqrt of both sides

x-4= =/-4i

x= 4+/-4i

2007-07-11 13:44:27 · answer #3 · answered by Curious 2 · 0⤊ 0⤋