◆用配方法解ax²+bx+c=0.(a≠0)
原方程兩邊同除以a,得:x²+(b/a)x+c/a=0.
配方,得:[x²+(b/a)x+(b/2a)²]-(b/2a)²+c/a=0.
即:(x+b/2a)²=b²/4a²-c/a=(b²-4ac)/4a².
當(dāng)b²-4ac0時(shí),(x+b/2a)²=(b²-4ac)/4a²,x+b/2a=±√(b²-4ac)/2a.
即:x1=-b/2a+√(b²-4ac)/2a=[-b+√(b²-4ac)]/2a;
x2=-b/2a-√(b²-4ac)/2a=[-b-√(b²-4ac)]/2a.