Question

已知x1,x2是方程2x^2+3x-1=0的两个根,用韦达定理求x1-x2的值

Answer 1

由韦达定理有x1+x2=-3/2,x1x2=-1/2,
所以x1-x2
=±√(x1-x2)^2
=±√(x1^2+x2^2-2x1x2)
=±√[(x1^2+x2^2)-2(x1x2)]
=±√[(x1+x2)^2-2x1x2-2x1x2]
=±√[(x1+x2)^2-4x1x2]
=±√[(-3/2)^2-4*(-1/2)]
=±√[(9/4)+2]
=±√(17/4)
=±√17/2.

Answer 2

韦达定理
设x_1，x_2是一元二次方程ax^2+bx+c=0的两个解。
根据求根公式，有

x_1=[-b + \sqrt (b^2-4ac)]/2a,
x_2=[-b -\sqrt (b^2-4ac)]/2a,

所以

x_1+x_2=[-b +(-) \sqrt (b^2-4ac)]/2a+[-b - \sqrt (b^2-4ac)]/2a=-b/a

由于a=2;b=3;c=-1; sqrt是开二次方根号的意思
x1=[-3+sqrt(17)]/4=0.281
x2=[-3-sqrt(17)]/4=-1.781

Answer 3

(x1-x2)^2=(x1+x2)^2-4*x1*x2=17/4
x1-x2=±√17/2