2007-01-02 08:13:10 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
you'd do this by the quadratic forumla...
First, put it in standard form
x^2 - 2x - 5 = 0
so a = 1, b = -2, c = -5
Put this into the quadratic formula
x = (-(-2))+/- sqrt ((-2^2)-(4x1x(-5)))/(2x1) =
(2 +/- sqrt (4 - 20))/2 =
(2 +/- 4i)/2 = 1 +/- 2i, that is, 1 + 2i or 1 - 2i
2007-01-02 08:18:40 · answer #1 · answered by Joni DaNerd 6 · 1⤊ 0⤋
first subtract 2x: x²-2x+7=0 then use the quadratic formula: x = (2+/-?(4-28))/2 =(2+/-?(-24))/2 =(2+/-2?(-6))/2 =a million +/- ?(-6) =a million +/- ?(6)?(-a million) the roots are a million+?(6)i and a million-?(6)i. desire this helps.
2016-11-25 23:10:28 · answer #2 · answered by immanuel 4 · 0⤊ 0⤋
x^2 - 2x + 5 =0
a=2 , b=-2 , c=5
use the quadratic formula
there will be a negative number under the root sign so u have to use the imaginary root i=(-1)^1/2
then the last answer will be
x=0.5 + 1.5i
x=0.5 - 1.5i
where i is the imaginary square root of (-1)
2007-01-02 08:27:17 · answer #3 · answered by freddy 1 · 0⤊ 0⤋
x^2-2x+5=0
x=-(-2)^2+ or - square root ((-2)^2-4(1)(5))/2*1
=(2+ or -square root-16)/2
=(2+ or -4i)/2
x1+2i or x=1-2i
2007-01-02 08:32:57 · answer #4 · answered by eissa 3 · 0⤊ 0⤋
x^2 + 5=2x subtract 2x from each side
x^2-2x+5=0
x=(2+/-√(2^2-4*5))/2
x=1+/-√(-16)/2
x=1+/-2i
x=1+2i, 1-2i
2007-01-02 08:26:05 · answer #5 · answered by yupchagee 7 · 0⤊ 0⤋