Question

设x1，x2是方程2x∧2+4x-3=0的两个根，利用根与系数的关系求下列各式的值（过程要详细）在线等

解：x1的平方+3x2的平方+4x2-3

Answer 1

解：∵x1，x2是方程2x∧2+4x-3=0的两个根
∴x1+x2=-2, x1·x2=-3/2
(x1)²+(x2)²闭宽尺告
=(x1+x2)²-2x1·x2
=(-2)²-2×(-3/2)
=4+3
=7
∴x1的平方+3x2的平方+4x2-3
=(x1)²+(x2)²轿困亮+2(x2)²+4x2-3
=[(x1)²+(x2)²]+[2(x2)²+4x2-3]
=7+0
=7

Answer 2

把x=x2代入,得:
2x2²+4x2-3=0
2x2²+4x2=3
根据码庆敬韦达差岩定迟慎理,
x1+x2=-2
x1*x2=-3/2

x1²+3x2²+4x2-3
=(x1²+x2²)+(2x2²+4x2-3)
=(x1+x2)²-2x1*x2+(3-3)
=(-2)²-2*(-3/2)+0
=4+3
=7

Answer 3

x1，x2是方程2x∧伏燃游2+4x-3=0的两个根段顷，
则x1+x2=-2，x1x2=-3/2
2x2²+4x2-3=0
x1的平方+3x2的平缺销方+4x2-3
=x1²+x2²+2x2²+4x-3
=x1²+2x1x2+x2²-2x1x2+0
=(x1+x2)²-2x1x2
=(-2)²-2*(-3/2)
=7

Answer 4

由韦达定理x1+x2=负的a分之b，x1乘以x2=a分之c，便可求解。。