Question

Consider a quadratic equation ax²+2bx+c=0 where a,b,c are positive real numbers. If the equation has no real root, then which of the following is true?

• A. a,b,c cannot be in A.P. or H.P. but can be in G.P.
• B. a,b,c cannot be in G.P. or H.P. but can be in A.P.
• C. a,b,c cannot be in A.P. or G.P. but can be in H.P.
• D. a,b,c cannot be in A.P.,G.P or H.P.

Answer 1

The Correct Option is C

Step 1: Given a quadratic expression: $ax^2 + bx + c$ has no real roots 

Step 2: For no real roots, the discriminant must be less than 0: 
$D = b^2 - 4ac < 0$ 

Step 3: Let us check which sequence among A.P., G.P., and H.P. satisfies this condition when $a$, $b$, $c$ are in that sequence. 

Case 1: A.P.
If $a$, $b$, $c$ are in Arithmetic Progression, then:
$\displaystyle b = \frac{a + c}{2}$ 
Substitute into discriminant:
$b^2 - 4ac = \left(\frac{a + c}{2}\right)^2 - 4ac = \frac{(a + c)^2 - 16ac}{4}$ 
This value is not necessarily negative for all $a$, $c$. 

Case 2: G.P.
If $a$, $b$, $c$ are in Geometric Progression, then:
$b = \sqrt{ac}$ 
Substitute into discriminant:
$b^2 - 4ac = ac - 4ac = -3ac$ 
This is negative if $a$, $c > 0$. 

However, the question specifies the condition $b^2 = ac$, i.e.,
$D = 0$ — this contradicts the requirement that $D < 0$ 

Case 3: H.P.
If $a$, $b$, $c$ are in Harmonic Progression, then:
$\displaystyle \frac{1}{b} = \frac{1}{2} \left( \frac{1}{a} + \frac{1}{c} \right) \Rightarrow b = \frac{2ac}{a + c}$ 
Substitute into discriminant:
$b^2 = \left(\frac{2ac}{a + c}\right)^2$, and we test whether:
$b^2 - 4ac < 0$ 
We simplify:
$\left(\frac{2ac}{a + c}\right)^2 - 4ac = \frac{4a^2c^2}{(a + c)^2} - 4ac$ 
Take common denominator:
$= \frac{4a^2c^2 - 4ac(a + c)^2}{(a + c)^2}$ 
$= \frac{4ac \left(ac - (a + c)^2\right)}{(a + c)^2}$ 
Since $(a + c)^2 > ac$ for $a \ne c$, the numerator is negative, so
$D < 0$ 

Conclusion: The condition $D < 0$ is only satisfied when $a$, $b$, $c$ are in H.P. 

Final Answer: (C): $a$, $b$, $c$ cannot be in A.P. or G.P., but can be in H.P.

Answer 2

Step 1: Given a quadratic expression: $ax^2 + bx + c$ has no real roots 

Step 2: For this to be true, the discriminant must be less than 0: 
$D = b^2 - 4ac < 0$ 

Step 3: We are given that: $4b^2 - 4ac = 0 \Rightarrow b^2 = ac$ 

Step 4: We test different progressions to see which satisfy this: 

Case 1: A.P.
If $a$, $b$, $c$ are in Arithmetic Progression:
$\displaystyle b = \frac{a + c}{2}$ 
Now compute $b^2$:
$\displaystyle b^2 = \left( \frac{a + c}{2} \right)^2 = \frac{(a + c)^2}{4}$ 
Check if $b^2 = ac$:
$\displaystyle \frac{(a + c)^2}{4} = ac$ ⟹ $(a + c)^2 = 4ac$ 
This is not always true for all real $a$, $c$, so A.P. is not valid. 

Case 2: G.P.
If $a$, $b$, $c$ are in Geometric Progression:
$\displaystyle b = \sqrt{ac}$ 
Then clearly: $b^2 = ac$ 
This gives $D = b^2 - 4ac = ac - 4ac = -3ac < 0$ 
But the question gives $b^2 = ac$, so discriminant becomes 0, not less than 0. So G.P. is not valid. 

Case 3: H.P.
If $a$, $b$, $c$ are in Harmonic Progression:
$\displaystyle \frac{1}{b} = \frac{1}{2} \left( \frac{1}{a} + \frac{1}{c} \right) \Rightarrow b = \frac{2ac}{a + c}$ 
Now compute $b^2$:
$\displaystyle b^2 = \left( \frac{2ac}{a + c} \right)^2 = \frac{4a^2c^2}{(a + c)^2}$ 
We want $b^2 = ac$, so we check: $\displaystyle \frac{4a^2c^2}{(a + c)^2} = ac$ 
Multiply both sides by $(a + c)^2$:
$4a^2c^2 = ac(a + c)^2$ 
Divide both sides by $ac$:
$4ac = (a + c)^2$ 
This is only true for specific values of $a$ and $c$. But since $b^2 = ac$ matches the discriminant condition, H.P. is valid. 

Conclusion: The condition $b^2 = ac$ is satisfied only when $a$, $b$, $c$ are in H.P. 

Final Answer: (C): $a$, $b$, $c$ cannot be in A.P. or G.P., but can be in H.P.