maths

1. Using Pythagoras' Theorem, we have
x^2 + (x + 2)^2 = (x + 4)^2
x^2 + x^2 + 4x + 4 = x^2 + 8x + 16
x^2 - 4x - 12 = 0
(x - 6)(x + 2) = 0
x = 6 or x = -2(rejected)
2. Let x m be the length of the rectangle.
Then the width is (153 / x) m
We have 2[x + (153 / x)] = 52
x + (153 / x) = 26
x[x + (153 / x)] = 26x
x^2 + 153 = 26x
x^2 - 26x + 153 = 0
(x - 9)(x - 17) = 0
x = 9(rejected as length < width) or x = 17
The length is 17m
3. Put x = 3 into the equation. Then
5(3)^2 + a(3) - 12 = 0
45 + 3a - 12 = 0
3a = 12 - 45 = -33
a = -11
The equation becomes 5x^2 - 11x - 12 = 0
(x - 3)(5x + 4) = 0
x = 3 or -4 / 5
The other root is -4 / 5.

1. x^2 + (x+2)^2 = (x+4)^2
2X^2 + 4X + 4 = X^2 + 8X + 16
X^2 - 4X - 12 = 0
(X + 2) (X - 6) = 0
X = - 2 (rejected) or X = 6

So X = 6

2. let x be the lenth of rectangle
width of the rectangle = (52 - 2x)/2
X * (52 - 2x)/2 = 153
X^2 - 26X + 153 = 0
(x - 17)(x-9) = 0
x = 17 or x = 9

3. 5(3)^2 + a(3) - 12 = 0
3a = -33
a = -11

2006-10-17 20:23:38 補充：
3. 5x^2 - 11 x-12=0 (5x + 4) (x-3) = 0 X = -4/5

1.In a right-angled triangle,the length of the three sides are x,x+2 and x+4,find x.
x²+(x+2)²=(x+4)²
x²+(x²+4x+4)=x²+8x+16
2x²+4x+4=x²+8x+16
x²-4x-12=0
(x-6)(x+2)=0
x=6 or x= -2(rejected)
so x=6

2.If the area and the perimeter of a rectangle are 153m square and 52m respectively,find
the length of the rectangle.
let x be the length of the rectangle,then,the width will be (52/2)-x=26-x
x(26-x)=153
-x²+26x=153
x²-26x+153=0
(x-9)(x-17)=0
x=9 or x=17
so the length of the rectangle is 9 or 17m


3.If x=3 is one of the solution of 5x^2+ax-12=0,find the value of a and the other solution.
then,5(3)²+3a-12=0
45+3a-12=0
3a=-33
a=-11
5x²-11x-12=0
(5x+4)(x-3)=0
x=3 or x=-4/5
so the other root is -4/5