x^2 - 4x +13 = 0
I got x = 2 (+-) sq root of 7
2007-05-07 08:58:33 · 5 answers · asked by confused 1 in Science & Mathematics ➔ Mathematics
Completing the square
x² - 4x + 13 = 0
x² - 4x + 13 - 13 = 0 - 13
x² - 4x = - 13
x² - 4x +______ = - 13 +______
x² - 4x + 4 = - 13 + 4
(x - 2)(x - 2) = - 9
(x - 2)² = - 9
(√x - 2)² = ± √- 9
x - 2 = ± i√9
x - 2 = ± 3i
x - 2 + 2 = 2 ± 3i
x = 2 ± 3i
- - - - - - - - - s-
2007-05-07 09:47:17 · answer #1 · answered by SAMUEL D 7 · 1⤊ 0⤋
ax²+bx+c=0 Here, a = 1, b = â4, and c = 13.
The roots are x = (âb ± â(b²â4ac))÷(2a). Thus the two values of x = [4±â(16-52)]÷2 = [4±â(â36)]÷2 = [4±6i]÷2 = 2 ± 3i, a set of complex roots.
2007-05-07 16:31:16 · answer #2 · answered by engineer01 5 · 0⤊ 0⤋
x^2-4x+13 is the same as x^2 - 4x +4 = -9
which can be factored into (x-2)^2 = -9
therefore x-2 = 3i, x = 3i+2
2007-05-07 16:13:51 · answer #3 · answered by Brad K 3 · 0⤊ 0⤋
If your equation is correctly typed, it only has complex roots,
x^2 -4x+13 = (x-2)^2+9
thus (x-2)^2=-9, so x=2 (+-) sqrt(-9) = 2(+-) 3i
2007-05-07 16:08:26 · answer #4 · answered by ulfsnilsson 2 · 1⤊ 0⤋
I'll tell you that the result is imaginary and that the answer of x is (4+/-6i)/2, 6i is sqrt of -36
2007-05-07 16:09:14 · answer #5 · answered by UnknownD 6 · 1⤊ 0⤋