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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. 