幫計2題判別式

(a) x^2+(b-a)x+(a-b)^2=0
△= (b-a)^2 - 4(1)(a-b)^2
=(a-b)^2 - 4(a-b)^2
= (a-b)^2 (1-4)
= -3(a-b)^2
(a-b)^2 > 0, ∵a ≠ b
-3(a-b)^2 < 0
△< 0
∴There are no real roots. （沒有實根）
(b) -2(a+1)x^2+(a-4)x+3a-2=0, where a ≠ -1
△= (a-4)^2 - 4[-2(a+1)](3a-2)
= (a^2 - 8a + 16) + 8(a+1)(3a-2)
= a^2 - 8a + 16 + 8(3a^2 + a - 2)
= a^2 - 8a + 16 + 24a^2 + 8a - 16
= 25a^2
= (5a)^2 > 0, ∵a ≠ 0
∴There are two unequal rational real roots. （兩個不同的有理實根）

(a) x^2 + (b-a)x + (a-b)^2 = 0
判別式 = (b-a)^2 - 4(a-b)^2
= (a - b)^2 - 4(a-b)^2
= -3(a - b)^2
< 0
無實根

(b)-2(a+1)x^2 + (a-4)x + 3a-2 = 0 ,其中a不等於-1
判別式 = (a-4)^2 - 4(3a-2)[-2(a+1)]
= a^2 - 8a + 16 + 8(3a-2)(a+1)
= a^2 - 8a + 16 + 8(3a^2 + a - 2)
= a^2 - 8a + 16 + 24a^2 + 8a - 16
= 25a^2
> 0
有兩個不同實根