高一數學 絕對值方程式

方程式 |ax-b|=7 相當於 (ax-b)^2 = 49
x = -2 及 x = 5 代入, 得
(-2a-b)^2 = 49 <==> 4a^2+4ab+b^2 = 49
(5a-b)^2 = 49 <==> 25a^2-10ab+b^2 = 49

所以 21a^2 - 14ab = 0.
a≠0, 故 21a = 14b, 故 a = 2b/3.
代入第一式, 得 (49/9)b^2 = 49, 所以 b = ± 3, a = ± 2.

即 (a,b) = (2,3) 或 (-2,-3).



檢查:
(a,b) = (2,3), 則原方程式成為 |2x-3| = 7.

故 2x = 3±7 = -4 或 10, 所以 x = -2 或 5.


(a,b) = (-2,-3) 只是絕對值內正負完全倒過來, 但取絕對值不受影響.

| ax-b |=7
ax - b = 7 or ax - b = -7

so
- 2a - b = 7
5a - b = - 7
(a = - 2 , b = - 3)

or
- 2a - b = - 7
5a - b = 7
(a = 2 , b = 3)