2題F.4maths 數學題唔識做!!高手!!

1
y=3x2+bx+c passes through A(1,3)and B(2,3).
3=3(1)2+b(1)+c
3=3+b+c
b+c=0___________________(1)
3=3(2)2+b(2)+c
3=12+2b+c
2b+c=-9_________________(2)
(2)-(1)
b=-9
Sub b=-9 into (2)
2(-9)+c=-9
c=9
2
(2x2+3x)2-7(2x2+3x)-18=0
[(2x2+3x)-9][(2x2+3x)+2]=0
(2x2+3x-9)(2x2+3x+2)=0
2x2+3x-9=0 or 2x2+3x+2=0
x=3/2 or -3

我淨識1a渣....

at point A(1,3)
3=3(1)^2+b(1)+c
3=3+b+c
b+c=0
b=-c ===== (1)

at point B(2,3)
3=3(2)^2+b(2)+c
3=3(4)+2b+c
3-12=2b+c
2b+c=-9 ====== (2)

Combining (1) and (2)
2(-c)+c=-9
-c=-9
c=9
then b=-c=-9

1)

a)

y = 3x^2 + bx + c

3 = 3 + b + c ...... (1)

3 = 12 + 2b + c ...... (2)

(2) - (1) , b + 9 = 0

b = - 9

c = 9

b)

Let the point be (x , 0)

y = 0

3x^2 - 9x + 9 = 0

x^2 - 3x + 3 = 0 , no real root.

So , no x - intercept in the graph.

2)

Let 2x^2 + 3x = y ,

y^2 - 7y - 18 = 0

(y - 9)(y + 2) = 0

y = 9 or y = - 2

So ,

2x^2 + 3x - 9 = 0 or 2x^2 + 3x + 2 = 0

(2x - 3)(x + 3) = 0 or no real root

x = 3 / 2 or x = - 3

1)
y=3x^2+bx+c
a) When x=1 y=3
3=3(1)+b+c
b+c=0
When x=2 y=3
3=3(4)+2b+c
-9=b+b+c
-9=b
sub b=-9 to b+c=0
c=9
b)y=3x^2-9x+9
When y=0
0=3x^2-9x+9
D=81-4x9x3
<0
so no x-intercept(s) of the graph.
2)
(2x^2+3x)^2-7(2x^2+3x)-18=0
(2x^2+3x-9)(2x^2+3x+2)=0
so (2x^2+3x-9)=0 or (2x^2+3x+2)=0 (no solution)
so (2x-3)(x+3)= 0
x=3/2 or x=-3

1a) 3=3*1^2 + b*1+ c => b + c=0 ... (1)
3=3*2^2+ b*2+ c => 2b+ c=-9 ... (2)
(2) - (1) gives b=-9 => c= 9
1b) y=3x^2-9x+ 9
If y=0, discriminant = (-9)^2-4*3*9 = -27 < 0
So there is no real solution of x for y=0
=> the number of x-intercept(s) is 0.

2) (2x^2 +3x)^2-7(2x^2 +3x)-18=0
2x^2 +3x = (7 +(7^2-4*(-18))^0.5)/2 or (7-(7^2-4*(-18))^0.5)/2 = 9 or -2
2x^2 +3x - 9 = 0 or 2x^2 +3x 2 = 0 (no real solution)
x = (-3 +(3^2-4*2*(-9))^0.5)/(2*2) or (-3-(3^2-4*2*(-9))^0.5)/(2*2)
x = 1.5 or -3