A. maths (F.4) 急急急

Given that the quadratic equation x2+kx-(k-3)=0 has real roots a and b . If lal=lbl, find the values of k .
若 │a│=│b│則
a = b 或 a = -b
若 a = b 則有重根則
判別式 B2 – 4AC = 0
k2 – 4(1)(k-3) = 0
k2 – 4k + 12 = 0
但因k2 – 4k + 12 = 0 的判別式大於零所以無解。
若 a = -b
則
a + b = -k
但 a = - b
即
k = 0