標題:
(a+1)(b+2)=4 ,show that a= -2k
發問:
最佳解答:
a+b = k+2 ab = k (a+1)(b+2)=4 ab+ (a+b) + a + 2 = 4 k + k+2 + a + 2 = 4 2k+ a = 0 a = -2k Hence x = -2k is one of the roots of the equation. i.e. (-2k)^2 - (k+2)(-2k) + k = 0 4k^2 + 2k^2 + 4k + k = 0 k(6k+5) = 0 k = 0 or -5/6
其他解答:
given that a+b = k+2 and ab=k (a+1)(b+2) = 4 Step 1: a+b = k+2 b = k+2-a Step 2: ab+2a+b+2 = 4 ab+2a+b = 4-2 As ab=k and b = k+2-a k + 2a+ (k+2-a) = 2 2k+a+2 = 2 2k+a+(2-2) = 0 2k+a = 0 a = -2k Hence a = -2k 2007-11-03 20:35:55 補充： Step 3 find the two values of k:(a,b) = (x,y) == a=x= -2kx^2 -(k 2)x k=0(-2k)^2 -(k 2)(-2k) k=04k^2 -((-2k)^2-4k) k =04k^2 2k^2 4k k = 0k(6k 5) = 0k = 0 or (6k 5) = 0 == k=-5/6Hence the two values of k are k = 0 or -5/6
(a+1)(b+2)=4 ,show that a= -2k
發問:
此文章來自奇摩知識+如有不便請留言告知
given that a+b = k+2 and ab=k If (a+1)(b+2)=4 ,show that a= -2k. Hence find the twp values of k. P.S. a,b are the roots of x^2 -(k+2)x + k=0 Plz shpw the steps最佳解答:
a+b = k+2 ab = k (a+1)(b+2)=4 ab+ (a+b) + a + 2 = 4 k + k+2 + a + 2 = 4 2k+ a = 0 a = -2k Hence x = -2k is one of the roots of the equation. i.e. (-2k)^2 - (k+2)(-2k) + k = 0 4k^2 + 2k^2 + 4k + k = 0 k(6k+5) = 0 k = 0 or -5/6
其他解答:
given that a+b = k+2 and ab=k (a+1)(b+2) = 4 Step 1: a+b = k+2 b = k+2-a Step 2: ab+2a+b+2 = 4 ab+2a+b = 4-2 As ab=k and b = k+2-a k + 2a+ (k+2-a) = 2 2k+a+2 = 2 2k+a+(2-2) = 0 2k+a = 0 a = -2k Hence a = -2k 2007-11-03 20:35:55 補充： Step 3 find the two values of k:(a,b) = (x,y) == a=x= -2kx^2 -(k 2)x k=0(-2k)^2 -(k 2)(-2k) k=04k^2 -((-2k)^2-4k) k =04k^2 2k^2 4k k = 0k(6k 5) = 0k = 0 or (6k 5) = 0 == k=-5/6Hence the two values of k are k = 0 or -5/6
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