(leave the answer in surd form if necessary, and identify those equations that have no real roots.)
a)6(x + 3)^2 - 25 = 2(x + 3)^2
b) 4.8x = 2.7+2.1x^2
c) (x+2 )^2 + (x+1)^2 =0
更新1:
d) 2(5-x)=(5-x)(3x+4) 詳細解釋
更新2:
d) 2(5-x)=(5-x)(3x+4) 詳細解釋
MATHS...quadratic equations in one unknown
2007-12-01 11:28 pm
1.solve each of the following equations using any method.
(leave the answer in surd form if necessary, and identify those equations that have no real roots.)
a)6(x + 3)^2 - 25 = 2(x + 3)^2
b) 4.8x = 2.7+2.1x^2
c) (x+2 )^2 + (x+1)^2 =0
(leave the answer in surd form if necessary, and identify those equations that have no real roots.)
a)6(x + 3)^2 - 25 = 2(x + 3)^2
b) 4.8x = 2.7+2.1x^2
c) (x+2 )^2 + (x+1)^2 =0
回答 (2)
2007-12-01 11:37 pm
✔ 最佳答案
a)(x+3)^2 = 25/4
x+3 = 5/2 or x+3 = -5/2
x = -1/2 or x = -11/2
b)
7x^2 - 16x + 9 = 0
(7x - 9)(x - 1) = 0
x = 9/7 or x = 1
c)
2x^2 + 6x + 5 =0
there is no real root......
參考: me
2007-12-08 3:39 am
c)
(x+2)^2 + (x+1)^2 =0
x^2 + 4x + 4 + x^2 +2x + 1 = 0
2x^2 + 6x + 5 = 0
delta = [-b +/- root (4ac)] / 2a
= [-6 +/- root (4*2*5)]/ (2*2)
= (-6 +/- root40) /4
= (-3 + root10) /2 or (-3 - root10) /2
d)
2(5-x) = (5-x) (3x+4)
10-2x = -3x^2 +11x+20
-3x^2 -13x -10 = 0
delta = [-b +/- root (4ac)] / 2a
= [13 +/- root (4*3*-10)] / (2*3)
= [13 +/- root (-120)] / 6
because root(-120) is not real number, so the equation has no real root.
(x+2)^2 + (x+1)^2 =0
x^2 + 4x + 4 + x^2 +2x + 1 = 0
2x^2 + 6x + 5 = 0
delta = [-b +/- root (4ac)] / 2a
= [-6 +/- root (4*2*5)]/ (2*2)
= (-6 +/- root40) /4
= (-3 + root10) /2 or (-3 - root10) /2
d)
2(5-x) = (5-x) (3x+4)
10-2x = -3x^2 +11x+20
-3x^2 -13x -10 = 0
delta = [-b +/- root (4ac)] / 2a
= [13 +/- root (4*3*-10)] / (2*3)
= [13 +/- root (-120)] / 6
because root(-120) is not real number, so the equation has no real root.
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