(X^2) +25 = 0
how do i solve it?
2007-12-08 03:23:44 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
+-5i
2007-12-08 03:28:23 · answer #1 · answered by someone else 7 · 0⤊ 0⤋
Difference of perfect squares.
You take the square root of the first term plus the square root and the second and the square root of the first term minus the square root of the second.
(x+5)(x-5) = 0
Now set each term equal to zero.
x + 5 = 0
x = -5
x - 5 = 0
x = 5
Your answer is x = 5 and -5
2007-12-08 03:28:02 · answer #2 · answered by npontello12 2 · 0⤊ 1⤋
take the 25 to other side by subtraction
x^2 = -25
take the square root
x = +/- sqrt (-25)
x = 5i or -5i
2007-12-08 03:27:03 · answer #3 · answered by Linda K 5 · 0⤊ 2⤋
(x^2) +25 = (x+5i)(x-5i) = 0
Solve x+5i = 0 for x,
x = -5i
Solve x-5i = 0 for x,
x = 5i
2007-12-08 03:28:49 · answer #4 · answered by sahsjing 7 · 0⤊ 1⤋
Difference of two perfect squares
(x + 5i)(x - 5i) = 0
x = ±5i
2007-12-08 03:42:06 · answer #5 · answered by krazykyngekorny 4 · 0⤊ 1⤋
-7<( 9-2x)/ 9 <= 5 -63< 9 -2x <= 45 (multiply all sides of inequality by 9) -72 < -2x <= 36 (subtract 9 from all sides of inequality) 36 > x >= -18 (divide all sides of inequality by -2) (note the inequality changes from less than to greater than because of the division by negative number)
2016-05-22 04:08:33 · answer #6 · answered by ? 3 · 0⤊ 0⤋
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2007-12-08 03:52:05 · answer #7 · answered by Anonymous · 0⤊ 1⤋
x^2 = -25
x = +/- 5i
2007-12-08 03:27:21 · answer #8 · answered by norman 7 · 0⤊ 2⤋