How do you solve this quadratic equation: r ^2 − 2r − 4 = 0?
2008-07-18 8:29 am
r ^2 − 2r − 4 = 0 I'm getting a different answer then the book. Your help is greatly appreciated!
回答 (4)
2008-07-18 8:33 am
✔ 最佳答案
r ^2 − 2r − 4 = 0 (r-1)^2-4-1=0
(r-1)^2-5=0
(r-1)^2=5
r-1=± √5
r=1± √5
2008-07-18 9:26 am
r^2 - 2r - 4 = 0
r = [-b ±√(b^2 - 4ac)]/2a (Quadratic Discriminant)
a = 1
b = -2
c = -4
r = [2 ±√(4 + 16)]/2
r = [2 ±√20]/2
r = [2 ±4.47]/2 (approx.)
r = [2 + 4.47]/2
r = 6.47/2
r = 3.235
r = [2 - 4.47]/2
r = -2.47/2
r = -1.235
∴ r = -1.235 , 3.235
r = [-b ±√(b^2 - 4ac)]/2a (Quadratic Discriminant)
a = 1
b = -2
c = -4
r = [2 ±√(4 + 16)]/2
r = [2 ±√20]/2
r = [2 ±4.47]/2 (approx.)
r = [2 + 4.47]/2
r = 6.47/2
r = 3.235
r = [2 - 4.47]/2
r = -2.47/2
r = -1.235
∴ r = -1.235 , 3.235
參考: Square Root Calculator
http://www.math.com/students/calculators/source/square-root.htm
2008-07-18 9:10 am
r = [ 2 ± √ (4 + 16) ] / 2
r = [ 2 ± √ (20) ] / 2
r = [ 2 ± 2√5 ] / 2
r = 1 ± √5
r = [ 2 ± √ (20) ] / 2
r = [ 2 ± 2√5 ] / 2
r = 1 ± √5
2008-07-18 8:33 am
Why simply complete the square of course....
(r^2 - 2r + 1) - 4 - 1 = 0
(r-1)^2 = 5
r = + sqrt (5) + 1 and r = - sqrt (5) + 1
eh?
(r^2 - 2r + 1) - 4 - 1 = 0
(r-1)^2 = 5
r = + sqrt (5) + 1 and r = - sqrt (5) + 1
eh?
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