一元二次方程ax²+bx+c=0(a≠0)的两个根为x₁=2a分之 -b+根号b²-4ac,x₂=2a分之-b-根号b²-4ac,所以x₁+x₂=2a分之 -b+根号b²-4ac+2a分之-b-根号b²-4ac=-a分之b;x₁*x₂
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![一元二次方程ax²+bx+c=0(a≠0)的两个根为x₁=2a分之 -b+根号b²-4ac,x₂=2a分之-b-根号b²-4ac,所以x₁+x₂=2a分之 -b+根号b²-4ac+2a分之-b-根号b²-4ac=-a分之b;x₁*x₂](/uploads/image/z/2085930-18-0.jpg?t=%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bax%26%23178%3B%2Bbx%2Bc%3D0%EF%BC%88a%E2%89%A00%EF%BC%89%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%E4%B8%BAx%26%238321%3B%3D2a%E5%88%86%E4%B9%8B+-b%2B%E6%A0%B9%E5%8F%B7b%26%23178%3B-4ac%2Cx%26%238322%3B%3D2a%E5%88%86%E4%B9%8B-b-%E6%A0%B9%E5%8F%B7b%26%23178%3B-4ac%2C%E6%89%80%E4%BB%A5x%26%238321%3B%2Bx%26%238322%3B%3D2a%E5%88%86%E4%B9%8B+-b%2B%E6%A0%B9%E5%8F%B7b%26%23178%3B-4ac%2B2a%E5%88%86%E4%B9%8B-b-%E6%A0%B9%E5%8F%B7b%26%23178%3B-4ac%3D-a%E5%88%86%E4%B9%8Bb%EF%BC%9Bx%26%238321%3B%2Ax%26%238322%3B%26%23)
一元二次方程ax²+bx+c=0(a≠0)的两个根为x₁=2a分之 -b+根号b²-4ac,x₂=2a分之-b-根号b²-4ac,所以x₁+x₂=2a分之 -b+根号b²-4ac+2a分之-b-根号b²-4ac=-a分之b;x₁*x₂
一元二次方程ax²+bx+c=0(a≠0)的两个根为x₁=2a分之 -b+根号b²-4ac,x₂=2a分之-b-根号b²-4ac,所以x₁+x₂=2a分之 -b+根号b²-4ac+2a分之-b-根号b²-4ac=-a分之b;
x₁*x₂₂=2a分之 -b+根号b²-4ac*2a分之-b-根号b²-4ac=-a分之b=4a²分之(-b﹚²-(根号b²-4ac)²=a分之c
(1)已知方程x²-3x-1=0的两个根是x₁,x₂,根据韦达定理,不解方程,求(x₁+1)(x2+1)的值.(2)已知两个数的和等于-5,积等于6,求这两个数.
一元二次方程ax²+bx+c=0(a≠0)的两个根为x₁=2a分之 -b+根号b²-4ac,x₂=2a分之-b-根号b²-4ac,所以x₁+x₂=2a分之 -b+根号b²-4ac+2a分之-b-根号b²-4ac=-a分之b;x₁*x₂
(1)(x₁+1)(x2+1)=x₁+x2+x₁*x₂+1=3+(-1)+1=3,其中x₁+x2=-a 分之b,因a=1,b=-3所以x₁+x2=-a 分之b=3;x₁*x₂=a分之c,把a=1,c=-1代人式中x₁*x₂=a分之c=-1
(2)一元二次方程ax²+bx+c=0(a≠0),两边除以a,方程式两边变为:x²+b/a*x+c/a=0,已知两个数的和等于-5,即x₁+x₂=-b/a=-5,得出b/a=5,积等于6即x₁*x₂=c/a=6,得出c/a=6,把b/a=5,c/a=6代人方程x²+b/a*x+c/a=0,方程变为x²+5x+6=0,求解方程的两个根分别为2和3