中四MATHS題

Since the new roots = 3 * 1/(root of old equation)
Put new roots be yi, for i = 1,2 and old roots be xi
yi = 3/xi => xi = 3/yi
Put x = 3/y into the equation x(12x+1) = 1
(3/y)(12*3/y + 1) = 1
3(36/y + 1) = y
108 + 3y = y^2
The equation required : y^2 - 3y - 108 = 0

Alternative method:
x(12x+1) = 1 => 12x^2 + x - 1 = 0
Sum of old roots = -1/12
Product of old roots = -1/12
Sum of new roots
= 3/x1 + 3/x2
= 3(x1+x2)/x1x2
= 3(-1/12)/(-1/12)
= 3
Product of new roots
= 9/x1x2
= 9/(-1/12)
= -108
Therefore,
equation required: x^2 - 3x - 108 = 0


2011-09-28 18:44:07 補充：
108/x^2+3/x=1 ->>>
How to convert to ->>>
108+3x=108x^2 ?

Form aquadratic equation in x whose roots are three times the reciprocals of theroots of
x(12x+1)=1,and write the equation in general form .
Sol
x(12x+1)=1
12x^2+x=1
12(3/x)^2+(3/x)=1
108/x^2+3/x=1
108+3x=108x^2
108x^2－3x－108=0