f.4 maths5知點做??勁趕 quadratic formula!!

The curve passes through (-1,0) and (4,0)
So, put (x,y) = (-1,0) and (4,0)
0 = p(-1)^2 + q(-1) -2 and 0 = p(4)^2 + q(4) - 2
p - q = 2----------(1) and 16p + 4q = 2---------------------(2)
(1)*4 + (2),
20p = 10
p = 0.5
q = -1.5

係咪 p(x)sqaure?

(-1,0)同(4,0) 係條quadratic formula o既solutions,
所以你可以substitute返呢兩個root入條式
即係 : (^2 係square)
0 = p(-1)^2+q(-1)-2 ... (1)
{
0 = p(4)^2+q(4)-2 ... (2)

有simultaneous linear equation in two unknowns,
可以計返p同q

i.e.,
From 1: p-q-2 = 0 ... (3)
From 2 : 16p+4q-2 = 0
8p+2q-1 = 0 ... (4)

(3)*2 : 2p-2q-4 = 0 ... (5)
(4)+(5) : 10p-5 = 0
p = 1/2 ... (6)

sub. (6) into (3),
(1/2)-q-2 = 0
q = -(3/2)

-1 and 4 are the root of the equation px² + qx - 2 = 0
Product of the roots = 4 x -1 = -4
-4 = -2 / p
p = 0.5

sum of roots = -1 + 4 = 3
-q/p = 3
-q/0.5 = 3
q = -1.5

So p = 0.5 and q = -1.5