Question

解方程：.

Answer 1

x₁=2,x₂=.

試題分析：解一元二次方程的方法有:1.配方法;2.公式法;3.十字相乘法;方法一(求根公式):去括號,將方程化為2x²-x+6=0,x=, x₁=2,x₂=;方法二(因式分解):x(2x+3)="2(2x+3)," x(2x+3)-2(2x+3)="0,(x-2)(2x+3)=0," x₁=2,x₂=.
試題解析：方法一(求根公式):
去括號,將方程化為2x²-x-6="0,"
b²-4ac=(-1)²-4×2×(-6)=49,
x=,
x₁=2,x₂=.
方法二(因式分解):
x(2x+3)="2(2x+3),"
x(2x+3)-2(2x+3)=0,
(x-2)(2x+3)="0,"
x₁=2,x₂=.