Question

已知a2+b2+2a+b+(5/4)=0,求方程ax2+bx+1=0的解

尽快 拜托了

Answer 1

原式=（a²+2a+1）+（b²+b+1/4）=0
=（a+1)²+(b+1/2)²=0
所以a+1=0.a=-1
b+1/2=0.b=-1/2

方程是:-x^2-1/2x+1=0
x^2+1/2x=1
(x+1/4)^2=17/16
x+1/4=(+/-)根号17/4
X1=-1/4+根号17/4
X2=-1/4-根号17/4

Answer 2

a^2+b^2+2a+b+(5/4)=0
a^2+2a+1+b^2+b+1/4=0
(a+1)^2+(b+1/2)^2=0
(a+1)^2=0
a=-1
(b+1/2)^2=0
b=-1/2

ax^2+bx+1=0
-x^2-1/2x+1=0
x^2+1/2x-1=0
x^2+1/2x+1/16-1/16-1=0
(x+1/4)^2-17/16=0
(x+1/4+√17/4)(x+1/4-√17/4)=0
x=-1/4-√17/4或x=-1/4+√17/4

Answer 3

a=-1
b=-1/2

方程自己解