Question

一元二次方程ax^2+bx+c=0中,abc都是实数,且(2-a)^2+√下(a^2+b+c)+|c+8|=0,求代数式x^2+x+1的值

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Answer 1

平方和绝对值和根号都是大于或等于0的 所以只能都等于0才能相加为0
2-a＝0得 a＝2 c+8＝0 得c＝-8
a²+b+c＝0 得b＝4
2x²+4x-8＝0 得x²+2x-4＝0
（x+1）²＝5 x＝±√5-1
x²+x+1＝4-2x+x+1＝5-x＝6±√5

Answer 2

∵(2-a)^2+√下(a^2+b+c)+|c+8|=0 ∴ 2-a=0 (a^2+b+c）=0 c+8=0
得a=2 b=4 c=-8
原式ax^2+bx+c=0 即为 x^2+2x-4=0 解得X1=-1+√5 X2=-1-√5
则x^2+x+1=6-√5或6+√5