Hi guys,
I am confused on how to solve this question:
Prove that 3x2-2mx+m-1 = 0 has real solutions for all real values of m
From my understanding, "Real solutions" suggests that I should use the discriminant, for when b2 - 4ac is either = or > 0.
So, this is where a = 3, b = -2m and c = m-1,
and so: (-2m)2 - 4(3)(m-1) > 0, or (-2m)2 - 4(3)(m-1) = 0
also, -4m2-12m + 12 > 0
From here, I am stuck, as solving this quadratic will not answer the original question.
Is this actually the way to prove this, or have I made a mistake.
Thanks
I am confused on how to solve this question:
Prove that 3x2-2mx+m-1 = 0 has real solutions for all real values of m
From my understanding, "Real solutions" suggests that I should use the discriminant, for when b2 - 4ac is either = or > 0.
So, this is where a = 3, b = -2m and c = m-1,
and so: (-2m)2 - 4(3)(m-1) > 0, or (-2m)2 - 4(3)(m-1) = 0
also, -4m2-12m + 12 > 0
From here, I am stuck, as solving this quadratic will not answer the original question.
Is this actually the way to prove this, or have I made a mistake.
Thanks