F.4 數學問題 (一元二次方程)(2)

1.
Let y = hypotenuse of triangle.
x + (2x - 1) + y = 40
y = 41 - 3x .............(1)
By Pythagoras thm.,
x^2 + (2x - 1)^2 = y^2........(2)
So x^2 + (2x - 1)^2 = (41 - 3x)^2
x^2 + 4x^2 + 1 - 4x = 1681 + 9x^2 - 246x
4x^2 - 242x + 1680 = 0
2x^2 - 121x + 840 = 0
(x - 8)(2x - 105) = 0
x = 8 or 52.5 (rej because perimeter is 40 only.)





2013-08-03 08:18:35 補充：
2. Let speed of 慧敏 = x m/s. So speed of the other one = (x + 2) m/s.
After 6 sec., distance from the bus station = 27m. So 6 sec. EARLIER, distance of 慧敏 from the station = (27 + 6x) m. Distance of the other one from the station = 27 + 6(x + 2) = (6x + 39) m.

2013-08-03 08:27:27 補充：
By Pythagoras thm., (6x + 27)^2 + (6x + 39)^2 = 60^2. That is 4x^2 + 44x - 75 = 0. Solving equation we get x = 3/2 or - 25/2 ( rej because speed cannot be negative). So speed of 慧敏 is 1.5 m/s, speed of the other one is 3.5 m/s.

2013-08-03 08:36:07 補充：
To be exact, the answer of - 25/2 is correct if 慧敏 is standing in the west and the other one is standing in the north at the beginning.

2013-08-03 15:00:40 補充：
3. Obviously, one of the answer is x = a. Expanding LHS : x^2 - 7x + 12 - (a - 3)(a - 4) = 0.
Sum of roots = 7, one of the roots is a, so the other root = 7 - a. Answer is D.