方程2x^2-(k+1)x+k+3=0的两根之差为1,求k值. 一元二次方程ax^2+bx+c=0配方法变形为?要连步骤,谢啦!

Question

方程2x^2-(k+1)x+k+3=0的两根之差为1,求k值. 一元二次方程ax^2+bx+c=0配方法变形为?
要连步骤,谢啦!

Answer 1

x1+x2=(k+1)/2
x1*x2=(k+3)/2
(x1-x2)^2=(x1+x2)^-4x1x2=(k+1)^2/4-2(k+3)=1
(k+1)^2-8(k+3)-4=0
k^2-6k-27=0
k=9或k=-3
a[x^2+b/a*x+(b/2a)^2]+c-b^2/4a=0
(x+b/2a)^2=b^2/4a^2-c/a