MATHS;1元2次方程

[匿名]

2008-09-23 8:05 am

1.已知y=-x-2x+6的圖像與x軸相交於A(a,0)及B(b,0),其a中<b。不需要繪畫圖像,求a及b的值。

2.己知y=2x^2-11x+12的圖像與x軸相交於P及Q。不需要繪畫圖像,求P及Q的坐標。

3.考慮y=ax^2+bx-8的圖像。若圖像與x軸相交於A(4,0)及B(-2,0)兩點,求a及b的值。

教教我thx>_<要有步驟

回答 (1)

Train

2008-09-23 8:33 am

✔ 最佳答案

1. The equation y=-x-2x+6 is a linear one and will cross x-axis at one point only. I think the equation should be y=-x^2-x+6. Only quadratic equation will cross x-axis twice.
-x^2-x+6=0
x^2+x-6=0
(x+3)(x-2)=0
x+3=0 or x-2=0
therefore, x=-3 or x=2
As a<b, a=-3 and b=2

2.Given y=2x^2-11x+12
2x^2-11x+12=0
(2x-3)(x-4)=0
2x-3=0 or x-4=0
therefore x=1.5 or x=4
The co-ordinates of P and Q are (1.5,0) and (4,0) respectively.

3. Given y=ax^2+bx-8
a(4)^2+b(4)-8=0
16a+4b-8=0
4a+b-2=0
b=2-4a

a(-2)^2+b(-2)-8=0
4a-2b-8=0
2a-b-4=0
2a-(2-4a)-4=0
2a-2+4a-4=0
6a-6=0
a=1

b=2-4a=2-4(1)=-2

👍 0 👎 0