中四 aｍａｔｈｓ　一問

1) 3x^2+5x+3

= 3(x^2 + 5x/3 ) + 3

= 3(x^2+5x/3+25/36) + 3 - 3(25/36)

= 3 ( x + 5 / 6 )^2 + 11 / 12 > 0 [ Completing square ]

So 3x^2+5x+3>0 for all real value of x.

2a) y = ax^2 + bx + 8

= a ( x^2 + bx / a + b^2 / 4a^2 ) + 8 - a ( b^2 / 4a^2 )

= a ( x + b / 2a )^2 + 8 - b^2 / 4a

So,

-b / 2a = -2

b = 4a --- ( 1 )

8 - b^2 / 4a = -4

b^2 / 4a = 12 --- ( 2 )

Put ( 1 ) into ( 2 ),

( 4a )^2 / 4a = 12

4a = 12

a = 3

Put a = 3 into ( 1 ),

b = 4 ( 3 ) = 12

So a = 3, b = 12.

3) Let the two parts be x cm and ( 8 - x ) cm and so the length of each side of the squares are x / 4 cm and ( 8 - x ) / 4 cm respectively.

A = ( x / 4 )^2 + ( 8 - x )^2 / 16

= x^2 / 16 + ( 8 - x )^2 / 16

= ( x^2 + 64 - 16x + x^2 ) / 16

= ( 2x^2 - 16x + 64 ) / 16

= ( x^2 - 8x + 32 ) / 8

= [( x^2 - 8x + 16 ) + 16 ] / 8

= [ ( x - 4 )^2 + 16 ] / 8

So x = 4 when A is a min.

Hence each side of the squares:

4 / 4 = 1cm

and

( 8 - 4 ) / 4 = 1cm