a-maths難題

prove that, for all real values of a ,
the equation x2+2ax+2a2+a+1=0 has no real roots for x
x2+2ax+2a2+a+1=0
則判別式
B2 – 4AC
= (-2a)2 – 4(1)(2a2+a+1)
= 4a2 – 8a2 – 4a – 4
= -4a2 – 4a – 4
= -4(a2 + a) – 4
= -4(a2 + a + 0.52 – 0.52) – 4
= -4[(a + 0.5)2 – 0.52] – 4
= -4(a + 0.5)2 + 4*0.52 – 4
= -4(a + 0.5)2 – 3
所以當無論 a 為何值，
判別式必小於零，所以這方程式沒有實根。

prove that, for all real values of a ,
the equation x^2+2ax+2a^2+a+1=0 has no real roots for x

Put x^2+2ax+2a^2+a+1=0 into B^2-4AC.
B^2-4AC
=(2a)^2-(4)(1)(2a^2+a+1)
=4a^2-8a^2-4a-4
=-4a^2-4a-4
=-4a^2-4a+8-12
=-4(a^2+a-2)-12
=-4(a+2)(a-1)-12

The largest value of -4(a+2)(a-1)-12 is, -4[(-0.5)+2][(-0.5)-1)-12=(-4)(-1.5)(1.5)-12=9-12=-3
So, the equation x^2+2ax+2a^2+a+1=0 has no real roots for x