F.4 Quadratic Function

1. For reciprocal roots, the two roots are in the form k and 1/k.
Product of roots = k(1/k) = 1
(2a^2-a)/6 = 1
2a^2 - a - 6 = 0
(2a+3)(a-2) = 0
a = -3/2 or 2
When a = -3/2, the equation is 6x^2 + 15x + 6 = 0, i.e. 2x^2 +5x+2 = 0
(2x+1)(x+2) = 0
x = -1/2 or -2

When a = 2, the equation is 6x^2 - 13x + 6 = 0
(2x-3)(3x-2) = 0
x = 3/2 or 2/3

2. f(x) = 2x^2 - 12x + p = 2(x^2 -6x + (-6/2)^2) - 2(-6/2)^2 + p
= 2(x-3)^2 -18+p
The minimum value is p-18 = 5
p = 23

1)
let m & 1/m be the roots of the equation
(m)(1/m) = (2a^2-a)/(6)
1= (2a^2-a)/(6)
6 = 2a^2-a
2a^2 - a - 6 = 0
(2a+3)(a-2) = 0
a = 2 or a = -3/2
when a = 2,
6x^2-(8(2)-3)x+(2(2)^2-2) = 0
6x^2-13x+6 = 0
[you may use calculator to solve the equation]
when a = -3/2,
6x^2-(8(-3/2)-3)x+(2(-3/2)^2-(-3/2)) = 0
6x^2+15x+3 = 0
2x^2+5x+1 = 0
[you may use calculator to solve the equation]


2)
f(x)= 2x^2-12x+p
= 2(x^2-6x)+p
= 2(x^2-6x +3^2-3^2)+p
= 2(x-3)^2-2(9)+p
= 2(x-3)^2-(18-p)
because the minimum value of the function f(x)=2x^2-12x+p is 5,
ie (18-p) = 5
p = 13

2007-10-21 14:07:54 補充：
1)when a = 2,6x^2-(8(2)-3)x+(2(2)^2-2) = 06x^2-13x+6 = 0(3x-2)(2x-3) = 0x = 3/2 or 2/3when a = -3/2,6x^2-(8(-3/2)-3)x+(2(-3/2)^2-(-3/2)) = 06x^2+15x+3 = 02x^2+5x+1 = 0 x = (-5+(13)^1/2)/4 or (-5-(13)^1/2)/4