1. ## Roots of quadratic functions

If a and b are roots of the quadratic equation 3x^2-6x+2=0.
(i) Find a+b
(ii) Find ab

Anyone can help?

2. Originally Posted by acc100jt
If a and b are roots of the quadratic equation 3x^2-6x+2=0.
(i) Find a+b
(ii) Find ab

Anyone can help?
If a and b are the roots of 3x^2-6x+2, then:

3(x-a)(x-b)=3x^2-6x+2

Expand the left hand side and equate corfficients of x^2, and x, and the constant terms:

3x^2 - 3(a+b)x + 3ab = 3x^2 - 6x + 2

So (a+b)= 2, and ab=2/3.

RonL

3. Originally Posted by acc100jt
If a and b are roots of the quadratic equation 3x^2-6x+2=0.
(i) Find a+b
(ii) Find ab

Anyone can help?
CaptainBlack's method is perfectly acceptable, but there is a general theorem for quadratics you can use to tackle this. (You can actually get theorems for this for higher order polynomials, but I forget the name of the method.)

If you have two roots of a quadratic equation $ax^2 + bx + c = 0$, $r_1~\text{and}~r_2$, then

$r_1 + r_2 = -\frac{b}{a}$
and
$r_1 \cdot r_2 = \frac{c}{a}$

(You can easily get this result by noting that
$r_{1, 2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ and doing the relevant sum or product of $r_1~\text{and}~r_2$.)

In your particular case, we have a = 3, b = -6, and c = 2. Thus
$r_1 + r_2 = -\frac{-6}{3} = 2$
and
$r1 \cdot r_2 = \frac{2}{3}$
just as CaptainBlack derived.

-Dan

4. It is called "Viete's Theorem".