f(x)=x^2+2(a+b+c)x+3(ab+bc+ca)则f(x)=0有实根吗另有一个f（x）=x^2+2(a+b+1)x+3(a+1)(b+1)-3则f（x）=0有实根吗?

Question

f(x) = x^2+2(a+b+c)x+3(ab+bc+ca) 则f(x)=0有实根吗
另有一个f（x）=x^2+2(a+b+1)x+3(a+1)(b+1)-3 则f（x）=0有实根吗?

Answer 1

x²+2(a+b+c)x+3(ab+bc+ca)=0
判别式△=[2(a+b+c)]²-4×1×[3(ab+bc+ca)]
=4(a²+b²+c²+2ab+2bc+2ca)-12(ab+bc+ca)
=4a²+4b²+4c²-4ab-4bc-4ca
=2(a²-2ab+b²)+2(b²-2bc+c²)+2(c²-2ca+a²)
=2(a-b)²+2(b-c)²+2(c-a)²
平方项恒非负,(a-b)²≥0,2(a-b)²≥0,同理,2(b-c)²≥0 2(c-a)²≥0
2(a-b)²+2(b-c)²+2(c-a)²≥0 △≥0,方程恒有实根.
当a=b=c时,方程有两相等实根,当a、b、c不全相等时,方程有两不等实根.
x²+2(a+b+1)x+3(a+1)(b+1)-3=0
判别式△=[2(a+b+1)]²-4×1×[3(a+1)(b+1)-3]
=4(a²+b²+1+2ab+2a+2b)-12(ab+a+b+1)+12
=4a²+4b²+8ab+8a+8b+4-12ab-12a-12b-12+12
=4a²+4b²-4ab-4a-4b+4
=(2a²-4a+2)+(2b²-4b+2)+(2a²-4ab+2b²)
=2(a-1)²+2(b-1)²+2(a-b)²
平方项恒非负,(a-1)²≥0,2(a-1)²≥0,同理,2(b-1)²≥0 2(a-b)²≥0
2(a-1)²+2(b-1)²+2(a-b)²≥0 △≥0,方程恒有实根.
当a=b=1时,方程有两相等实根,其余情况下,方程有两不等实根.