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addmaths 问题
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我今年form 4但是我已经学了addmaths两个chapter
有几个问题我不太明白所以希望你们帮我解答
1) The quadratic equation (x+m)^2 = kx has the roots 1 and 16. Find the values of m and the corresponding values of k
2) The roots of quadratic equation 3x^2 + px + 2 = 0 are in the ratio 2:3
(A) find the value of p if p>0
(B) hence,find the roots of the equation
3) One of the roots of the equation 9x^2 - 9(k+2)x + 9k^2 + 18k + 11 = 0 is two times the other root. Find the values of k. |
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楼主 |
发表于 3-1-2009 10:27 PM
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还有这两题，谢谢！
1) Find the range of values of p if each of the following quadratic equation does not have real roots
(a) 2(x^2 + 1) = 2x - p
(b) 5x^2 + p = 2x
2) Find the values of m if each of the following quadratic equation has two real and equal roots
(a) x^2 - 2mx +3m +4=0
(b) x^2 + 2(x+2) = m(x^2+4)
(c) 5x^2-4x+2=m(4x^2-2x-1) |
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发表于 4-1-2009 12:58 AM
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1) The quadratic equation (x+m)² = kx has the roots 1 and 16. Find the values of m and the corresponding values of k.
(x+m)² = kx
x² + 2mx + m² = kx
x² + (2m-k)x + m² = 0
1 + 16 = -(2m-k) —— (1)
1 × 16 = m² —— (2)
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发表于 4-1-2009 01:06 AM
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2) The roots of quadratic equation 3x² + px + 2 = 0 are in the ratio 2:3
(A) find the value of p if p>0
(B) hence,find the roots of the equation
设两个根分别为2α, 3α.
2α + 3α = -p/3 —— (1)
2α · 3α = 2/3 —— (2)
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发表于 4-1-2009 01:11 AM
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3) One of the roots of the equation 9x² - 9(k+2)x + 9k² + 18k + 11 = 0 is two times the other root. Find the values of k.
设两个根分别为α, 2α.
α + 2α = -[-9(k+2)]/9
3α = k+2 —— (1)
α · 2α = (9k² + 18k + 11)/9
2α² = k² + 2k + 11/9 —— (2)
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发表于 5-1-2009 09:38 AM
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1) Find the range of values of p if each of the following quadratic equation does not have real roots
(a) 2(x² + 1) = 2x - p
(b) 5x² + p = 2x
(a) 2(x² + 1) = 2x - p
2x² - 2x + p + 2 = 0
△ < 0
(-2)² - 4×2×(p+2) < 0
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(b) 5x² + p = 2x
5x² - 2x + p = 0
△ < 0
(-2)² - 4×5×p < 0
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发表于 5-1-2009 10:00 AM
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2) Find the values of m if each of the following quadratic equation has two real and equal roots
(a) x² - 2mx +3m +4=0
(b) x² + 2(x+2) = m(x²+4)
(c) 5x²-4x+2=m(4x²-2x-1)
(a) x² - 2mx + 3m + 4 = 0
△ = 0
(-2m)² - 4×1×(3m+4) = 0
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(b) x² + 2(x+2) = m(x²+4)
(1-m)x² + 2x + 4 - 4m = 0
△ = 0
2² - 4×(1-m)×(4-4m) = 0
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(c) 5x² - 4x + 2 = m(4x² - 2x - 1)
(5-4m)x² + (2m-4)x + 2 + m = 0
△ = 0
(2m-4)² - 4×(5-4m)×(2+m) = 0
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发表于 5-1-2009 01:38 PM
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顺便解释一下，因为国中没学过这个：
△ 是判别式(Discriminant)的符号。 |
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发表于 5-1-2009 07:50 PM
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发表于 20-1-2009 10:29 AM
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D,Discriminant = b²-4ac
is because x = [-b ±√(b²-4ac)]2a
but when b²-4ac > 0, x is not real number
if b²-4ac ≥ 0 , x is real number |
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发表于 20-1-2009 08:34 PM
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回复 11# TKCboy 的帖子
應該是
x = [-b ±√(b² - 4ac)]/2a
還有
when b² - 4ac < 0, x is not a real number.
[ 本帖最后由 Ivanlsy 于 20-1-2009 08:37 PM 编辑 ] |
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发表于 21-1-2009 12:15 PM
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