Solve using the Quadratic Formula:?
回答 (7)
2008-06-30 1:58 pm
✔ 最佳答案
x = [-b ± √(b² - 4ac)] / (2a)x = [ -(6) ± √(6² - 4(1)(-3)] / (2(1))
x = [ -6 ± √(36 + 12)] / 2
x = [ -6 ± √(48)] / 2
x = [ -6 ± √(16 * 3)] / 2
x = [ -6 ± 4√(3)] / 2
x = -3 ± 2√3
2008-06-30 9:00 pm
You have to foil it.
but this one has lotsa decimals.
but like
x^2+6x+5=0
(x+3)(x+2)
x+3=0
x+2=0
so x = -2 or -3
but this one has lotsa decimals.
but like
x^2+6x+5=0
(x+3)(x+2)
x+3=0
x+2=0
so x = -2 or -3
參考: algebra last year.
2008-06-30 9:30 pm
x^2 + 6x - 3 = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 1
b = 6
c = -3
x = [-6 屉(36 + 12)]/2
x = [-6 屉48]/2
x = [-6 ±6.92]/2 (approx.)
x = [-6 + 6.92]/2
x = 0.92/2
x = 0.46
x = [-6 - 6.92]/2
x = -12.92/2
x = -6.46
â´ x = -6.46 , 0.46
x = [-b 屉(b^2 - 4ac)]/2a
a = 1
b = 6
c = -3
x = [-6 屉(36 + 12)]/2
x = [-6 屉48]/2
x = [-6 ±6.92]/2 (approx.)
x = [-6 + 6.92]/2
x = 0.92/2
x = 0.46
x = [-6 - 6.92]/2
x = -12.92/2
x = -6.46
â´ x = -6.46 , 0.46
參考: Square Root Calculator
http://www.math.com/students/calculators/source/square-root.htm
2008-06-30 9:01 pm
Ax^2 + Bx + C
-B +/- (B^2 - 4AC)^0.5 all divided by 2A
-3 +/- sqroot(6) will give you the two roots
-B +/- (B^2 - 4AC)^0.5 all divided by 2A
-3 +/- sqroot(6) will give you the two roots
2008-06-30 8:59 pm
x1,2=-6+-sqrt(36+12)/2=-6+-4sqrt(3)/2=-3+-2sqrt(3)
2008-06-30 8:58 pm
Quadratic Formula:
-b +/- sqrt(b² - 4ac)
----------------------------
2a
So...
-6 +/- sqrt(36 - 4(1)(-3))
---------------------------------
2
-3 +/- sqrt(36 + 12) / 2
-3 +/- sqrt(48) / 2
-3 +/- 4 * sqrt(3) / 2
-3 +/- (2 * sqrt(3))
x = -3 + (2 * sqrt(3))
and
x = -3 - (2 * sqrt(3))
-b +/- sqrt(b² - 4ac)
----------------------------
2a
So...
-6 +/- sqrt(36 - 4(1)(-3))
---------------------------------
2
-3 +/- sqrt(36 + 12) / 2
-3 +/- sqrt(48) / 2
-3 +/- 4 * sqrt(3) / 2
-3 +/- (2 * sqrt(3))
x = -3 + (2 * sqrt(3))
and
x = -3 - (2 * sqrt(3))
2008-06-30 8:56 pm
收錄日期: 2021-05-01 10:44:34
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