2007-05-30 09:56:43 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2 -10x = -29
(5+2i)^2 - 10(5+2i) = -29
25 +20i - 4 - 50 - 20i = -29
-29 = -29
2007-05-30 10:00:54 · answer #1 · answered by McFate 7 · 0⤊ 0⤋
So, by substituting, you get:
(5+2i)^2-10*(5+2i) = (5+2i)(5+2i) - 50 - 20i =
25+4i^2 + 10i + 10i -50 -20i = 25 + (4)(-1) -50 =
(Note that i^2 = -1)
25 - 4 - 50 = -29
2007-05-30 17:02:38 · answer #2 · answered by RG 3 · 0⤊ 0⤋
(5+2i)^2 = 25 + 20i - 4 = 21 + 20i
-10*(5+2i) = -50 - 20i
adding them, we get
21 - 50 = -29
Can you show that 5 - 2i is also a solution?
2007-05-30 17:15:52 · answer #3 · answered by Dr D 7 · 0⤊ 0⤋
x^2-10x=-29
(5+2i)^2-10(5+2i)=-29
reduce & solve...(i'm assuming that i is an invisible number)
25+20i+4-50-20i=-29
-29=-29
2007-05-30 17:00:31 · answer #4 · answered by Sum Girl 4 · 0⤊ 0⤋