# Prove that 3x^2 - 2mx + m -1 = 0 has real solutions for all real values of m

#### LucasE

##### New member
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

#### Subhotosh Khan

##### Super Moderator
Staff member
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,

4m2 - 12m + 12 > 0

m2 - 3m + 3 > 0

(m - 3/2)2 + (3/4) > 0 .......................... 

Now continue.....

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
.

Last edited:

#### LucasE

##### New member
so,
if m > -(m-2)2, then for all values of m will be greater than -(m-2)2?
How should I proceed?

#### Dr.Peterson

##### Elite Member
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
Of course, you mean to allow the discriminant to be 0, so you mean -4m2 - 12m + 12 >= 0. But that can't be true for all m, since the LHS (as a function of m) is a parabola opening downward.

So what's wrong? When you squared -2, you wrote -4 instead of +4. You want 4m2 - 12m + 12 >= 0 for all m, that is, 4x2 - 12x + 12 >= 0 for all x.

There are several ways to show this to be true. One is to use the discriminant again to show that the LHS never crosses the x-axis, so that it is always positive. Another is to find the minimum value (the vertex). Or you could complete the square.

#### lookagain

##### Senior Member
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.
You want 4m^2 - 12m + 12 >= 0

OR

m^2 - 3m + 3u >= 0

It is more direct to complete the square.

$$\displaystyle (m - 3/2)^2 - 9/4 + 12/4 \ \ge \ 0$$

$$\displaystyle (m - 3/2)^2 + 3/4 \ \ge \ 0 \ \ \ \ \$$(edit)

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#### LucasE

##### New member
Thank you guys,
this is really good!

#### Otis

##### Senior Member
… m^2 - 3m + 4 >= 0 …
That line ought to be:

m^2 - 3m + 3 >= 0

#### Otis

##### Senior Member
so, if m > -(m-2)2
It isn't. Some of the replies contain a mistake.

At the end of post #4, Dr. Peterson gives you some suggestions on how to show that for all Real m:

m^2 - 3m + 3 ≥ 0

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#### LucasE

##### New member
You want 4m^2 - 12m + 12 >= 0

OR

m^2 - 3m + 4 >= 0

It is more direct to complete the square.

$$\displaystyle (m - 3/2)^2 - 9/4 + 16/4 \ \ge \ 0$$

$$\displaystyle (m - 3/2)^2 + 7/4 \ \ge \ 0$$
but, should it not be m2-3m + 3
and so (m-3/2)2 +3/4 >= 0?

#### Otis

##### Senior Member
… should it [be] m2 - 3m + 3 … ?
Did you see posts #7, #8 and #9? :-?