F.4 a.maths

METHOD 1: To prove that the roots are rational, we must show that the discriminant is a perfect square.
Discriminant = (4b)^2 - 4a[-(a - 4b)]
= 16b^2 + 4a(a - 4b)
= 16b^2 - 16ab + 4a^2
= (4b - a)^2
Thus the equation has rational roots.
METHOD 2: Solve the equation directly.
ax^2 + 4bx - a + 4b = 0
ax^2 - a + 4bx + 4b = 0
a(x^2 - 1) + 4b(x + 1) = 0
a(x + 1)(x - 1) + 4b(x + 1) = 0
(x + 1)[a(x - 1) + 4b] = 0
x + 1 = 0 or ax - a + 4b = 0
x = -1 or x = (4b - a) / a.
Both roots are rational numbers.

when b^2-4ac>0, the equation has 2 distinct real roots
when b^2-4ac<0,the equation has no real roots.
when b^2-4ac=0,the equation has one repeated root.

so a=a, b=4b c=-(a-4b)

要計就要計到
(4b)^2-4a(-[a-4b])>0 and (4b)^2-4a(-[a-4b])的開方等於整數