If a + b + c = 0, then the quadratic equation 3ax² + 2bx + c = 0 has (A) at least one root in [0, 1]

Q: Why did we integrate the quadratic? Is this a standard trick?

It is not just a trick; it is a powerful technique for proving the existence of roots. When you see a condition like $a+b+c=0$, it often relates to the sum of coefficients of a polynomial. By integrating, we create an auxiliary function $f(x)$ whose derivative is our quadratic. This allows us to use the Mean Value Theorem or Rolle's Theorem to prove the existence of roots without solving the quadratic directly.

Q: Can we just solve for $c = -a - b$ and substitute?

You absolutely can, but that leads to a messy quadratic expression that is difficult to analyze for roots in a specific interval. The beauty of the auxiliary function method is that it bypasses the need for the quadratic formula entirely, providing an elegant, conceptual proof that is much faster and less prone to calculation errors.

Q: Does Rolle's Theorem guarantee only one root?

No, Rolle's Theorem guarantees *at least* one root in the open interval $(0, 1)$. Since a quadratic equation can have at most two roots, this theorem confirms that at least one of them must lie within the specified interval. It is a powerful existence proof, not a complete solution for all roots.

Question

If $a + b + c = 0$, then the quadratic equation $3ax^2 + 2bx + c = 0$ has

Options:

(A) at least one root in $[0, 1]$

(B) one root in $[2, 3]$ and the other in $[-2, -1]$

(C) imaginary roots

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to unravel a beautiful mystery hidden within a simple quadratic equation. In the world of JEE Advanced, problems are rarely just about solving for $x$; they are about seeing the hidden structure of the universe.

We are given the quadratic equation $3ax^2 + 2bx + c = 0$ and the intriguing condition $a + b + c = 0$. At first glance, you might be tempted to use the quadratic formula. But stop! That is the path of brute force. We want the path of elegance.

### The Calculus Bridge

How do we connect our condition $a + b + c = 0$ to the equation $3ax^2 + 2bx + c = 0$? Instead of solving the quadratic directly, let us think in reverse.

What if we integrate the given quadratic expression? Integrating $3ax^2$ gives $ax^3$, integrating $2bx$ gives $bx^2$, and integrating $c$ gives $cx$. This hints at a very powerful calculus tool.

Let us define a brand new auxiliary function based on that integration:
$$f(x) = ax^3 + bx^2 + cx$$

Since this is a polynomial, we know for a fact that it is continuous and differentiable everywhere. This is the foundation we need.

### The Geometric Intuition

Now, let us quickly verify our logic. If we differentiate our new function $f(x)$ with respect to $x$, we obtain:
$$f'(x) = 3ax^2 + 2bx + c$$

Our original quadratic is just the derivative of $f(x)$! Now, let us substitute some simple values into our function to see its behavior.

At $x = 0$, every term has an $x$, so $a(0)^3 + b(0)^2 + c(0)$ collapses to zero. Thus, $f(0) = 0$.

Moving forward, let us check the point $x = 1$:
$$f(1) = a(1)^3 + b(1)^2 + c(1) = a + b + c$$

### The Power of Rolle's Theorem

Now comes the main point. Remember that special condition given in the problem: $a + b + c = 0$. This means $f(1)$ is also exactly zero.

So, our function starts at zero when $x = 0$ and returns to zero when $x = 1$. Visualize this: a smooth curve that starts at zero and ends at zero must turn around somewhere in between.

This is exactly what **Rolle's Theorem** tells us. If a function is continuous, differentiable, and its values at two points are equal, its derivative must be zero at some point between them.

Let us apply this theorem to our function on the interval $[0, 1]$. Since $f(0) = f(1) = 0$, Rolle's Theorem guarantees that there is at least one point, let us call it $x_0$, strictly between zero and one, where the tangent is perfectly horizontal:
$$f'(x_0) = 0$$

### The Final Conclusion

And what was $f'(x)$ again? It was our original quadratic equation! So, setting $f'(x_0) = 0$ means:
$$3ax_0^2 + 2bx_0 + c = 0$$

This proves that our original equation has at least one valid root in the open interval $(0, 1)$. The problem is solved!

You have just used the power of calculus to prove the existence of a root without ever needing to touch the quadratic formula. Keep this **'Auxiliary Function'** method in your toolkit; it is a secret weapon for many JEE problems.

Animated Solution for Mathematics - Differentiation: If $a + b + c = 0$ , then the quadratic equation $3 a x^{2} + 2 b x + c = 0$ has

Select Answer:

Visualized Solution

Analyze the Quadratic Equation

The Intuition - Reverse Engineering

Define the Auxiliary Function

Verify the Derivative

Evaluate $f (0)$

Evaluate $f (1)$

Apply the Given Condition

Introduce Rolle's Theorem

Apply Rolle's Theorem to our Function

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Calculus Bridge

The Geometric Intuition

The Power of Rolle's Theorem

The Final Conclusion

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a+b+c=0, then the quadratic equation 3ax2+2bx+c=0 has