SUM / PRODUCT OF THE ROOT

Actually for this question, you obviously cannot find the roots directly by
using quadratic formula because there is an unknown a in the equation.

So, the first step to solve this problem is letting the roots be x and 1/x
respectively.

Then, the only method you can use now is the sum of roots and product of
roots.

Product of roots: x(1/x) = ( 2 a^2 - a ) / 6
6 = 2 a^2 - a
0 = (2a+3)(a-2)
a = -3/2 or 2

Then you can put the two values of a into the equation respectively and use
the quadratic formula ( as the value of a is known now) to find out the roots
of the equation for the two values of a.

2008-05-07 23:22:52 補充：
Actually the most common type of question that requires you to use quadratic formula is like that:

Given a quadratic equation x^2-3ax-(a+1)=0 has real roots A and B, find a quadratic equation with real roots A^2 and B^2 .

You can try it out!

2008-05-07 23:23:58 補充：
The answer is in terms of a, just like the given equation

2008-05-09 21:45:29 補充：
yes
the one who comments on me is correct
I have typed the wrong word and you should use sum/product of roots

答問題果位人兄最底果條問題
唔係應該用sum/product of root的嗎?

A+B = 3a
AB = -(a+1)

A^2 + B^2 = (A+B)^2 - 2AB = 9a^2+2(a+1)
(A^2)(B^2) = (AB)^2 = (a+1)^2

required quadratic equation:
x^2-(9a^2+2a+2)x+(a+1)^2 = 0