已知x1=-b+根号下b²-4ac/2a已知x1=(-b+根号(b^2-4ac))/2a,x2=(-b-根号(b^2-4ac))/2a,则x1+x2=?x1*x2=?
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![已知x1=-b+根号下b²-4ac/2a已知x1=(-b+根号(b^2-4ac))/2a,x2=(-b-根号(b^2-4ac))/2a,则x1+x2=?x1*x2=?](/uploads/image/z/3810679-7-9.jpg?t=%E5%B7%B2%E7%9F%A5x1%3D-b%2B%E6%A0%B9%E5%8F%B7%E4%B8%8Bb%26%23178%3B-4ac%2F2a%E5%B7%B2%E7%9F%A5x1%3D%EF%BC%88-b%2B%E6%A0%B9%E5%8F%B7%EF%BC%88b%5E2-4ac%EF%BC%89%EF%BC%89%2F2a%2Cx2%3D%28-b-%E6%A0%B9%E5%8F%B7%EF%BC%88b%5E2-4ac%29%EF%BC%89%2F2a%2C%E5%88%99x1%2Bx2%3D%EF%BC%9Fx1%2Ax2%3D%EF%BC%9F)
已知x1=-b+根号下b²-4ac/2a已知x1=(-b+根号(b^2-4ac))/2a,x2=(-b-根号(b^2-4ac))/2a,则x1+x2=?x1*x2=?
已知x1=-b+根号下b²-4ac/2a
已知x1=(-b+根号(b^2-4ac))/2a,x2=(-b-根号(b^2-4ac))/2a,则x1+x2=?x1*x2=?
已知x1=-b+根号下b²-4ac/2a已知x1=(-b+根号(b^2-4ac))/2a,x2=(-b-根号(b^2-4ac))/2a,则x1+x2=?x1*x2=?
x1+2x=-b/a
x1x2=c/a