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Pranav

Well-known member
Problem:

If the curve $y=ax^2+2bx+c$, ($a,b,c \,\in\,\mathbb{R},\,a,b,c \neq 0$) never meet the x-axis, then a,b,c can't be in

A)Arithmetic Progression

B)Geometric Progression

C)Harmonic Progression

D)All of these

Attempt:

Since, the curve never meets the x-axis, we have the condition that the discriminant of the quadratic is less than zero, hence,

$$b^2<ac$$

The above shows that a,b,c can't be in geometric progression. But the given answers are A and B, how do I show that they are not in arithmetic progression? Any help is appreciated. Thanks!

Klaas van Aarsen

MHB Seeker
Staff member
Problem:

If the curve $y=ax^2+2bx+c$, ($a,b,c \,\in\,\mathbb{R},\,a,b,c \neq 0$) never meet the x-axis, then a,b,c can't be in

A)Arithmetic Progression

B)Geometric Progression

C)Harmonic Progression

D)All of these

Attempt:

Since, the curve never meets the x-axis, we have the condition that the discriminant of the quadratic is less than zero, hence,

$$b^2<ac$$

The above shows that a,b,c can't be in geometric progression. But the given answers are A and B, how do I show that they are not in arithmetic progression? Any help is appreciated. Thanks!
Hey Pranav! If they were in arithmetic progression, you would have that:
$$b = \frac{a+c}{2}$$
Since $ac$ is greater than a square, $ac$ has to be positive, so we have:
$$\frac{a+c}{2}<\sqrt{ac}$$
According to the AM-GM inequality, this is a contradiction if $a$ and $c$ are both non-negative.
In the other case where $a$ and $c$ are both negative, we have:
$$\frac{(-a)+(-c)}{2}<\sqrt{(-a)(-c)}$$
which is again a contradiction according to the AM-GM inequality.
Therefore a,b,c can't be in arithmetic progression.

As for harmonic progression, let's try $a=1,b=\frac 1 2, c=\frac 1 3$.
That gives us:
$$\left(\frac 1 2\right)^2 < 1 \cdot \frac 1 3$$
Yep. A harmonic progression is possible. Oh, and even though my new year hasn't started yet, yours has.
So happy new year! Last edited:

Pranav

Well-known member
Hi ILS! If they were in arithmetic progression, you would have that:
$$b = \frac{a+c}{2}$$
Since $ac$ is greater than a square, $ac$ has to be positive, so we have:
$$\frac{a+c}{2}<\sqrt{ac}$$
According to the AM-GM inequality, this is a contradiction.
Therefore a,b,c can't be in arithmetic progression.
Ah, I was thinking about a contradiction proof. Since this is an exam problem, I wonder if it would have hit me during the exam to check the other cases after looking at $b^2<ac$.

Thanks a lot! Oh, and even though my new year hasn't started yet, yours has.
So happy new year! Yes, its been two hours since midnight. Happy New Year! Klaas van Aarsen

MHB Seeker
Staff member
Hi ILS! Ah, I was thinking about a contradiction proof. Since this is an exam problem, I wonder if it would have hit me during the exam to check the other cases after looking at $b^2<ac$.

Thanks a lot! Btw, I have just edited my post, since we have to distinguish positive and negative cases.
(The AM-GM inequality has the condition that the numbers have to be non-negative.)

Yes, its been two hours since midnight. Happy New Year! Pranav

Well-known member
Btw, I have just edited my post, since we have to distinguish positive and negative cases.
(The AM-GM inequality has the condition that the numbers have to be non-negative.)
Yes, thanks a lot ILS! I really need to be careful with such case-checking problems. 