If $$a, b, c \in R$$ and 1 is a root of equation $$ax^2 + bx + c = 0$$, then the curve $$y = 4ax^2 + 3bx + 2c$$, $$a \neq 0$$ intersect $$x$$-axis at

Given:

We are given the quadratic equation $$ax^2 + bx + c = 0$$ which has $$1$$ as one of its roots.

By substituting $$x = 1$$ into the equation, we obtain the fundamental relationship between the coefficients:

$$a(1)^2 + b(1) + c = 0$$

$$a + b + c = 0$$

From this, we can express $$c$$ in terms of $$a$$ and $$b$$:

$$c = -(a + b)$$

To Find:

We need to determine the number of intersection points of the curve $$y = 4ax^2 + 3bx + 2c$$ with the $$x$$ axis.

At the $$x$$ axis, the $$y$$ coordinate is zero. Thus, we need to analyze the nature of the roots of the following quadratic equation:

$$4ax^2 + 3bx + 2c = 0$$

Discriminant Analysis:

Let $$D$$ be the discriminant of this new quadratic equation. The formula for the discriminant is $$B^2 - 4AC$$.

$$D = (3b)^2 - 4(4a)(2c)$$

$$D = 9b^2 - 32ac$$

Now, substitute the value of $$c = -(a + b)$$ that we derived earlier:

$$D = 9b^2 - 32a(-(a + b))$$

$$D = 9b^2 + 32a^2 + 32ab$$

To conclusively determine the sign of $$D$$, we will complete the square by grouping the terms with $$a$$ and $$b$$:

$$D = 32(a^2 + ab) + 9b^2$$

$$D = 32(a^2 + ab + \frac{b^2}{4}) - 32(\frac{b^2}{4}) + 9b^2$$

$$D = 32(a + \frac{b}{2})^2 - 8b^2 + 9b^2$$

$$D = 32(a + \frac{b}{2})^2 + b^2$$

Logical Deduction:

• The square of any real number is always positive or zero. Hence, $$32(a + \frac{b}{2})^2 \ge 0$$ and $$b^2 \ge 0$$.
• The problem explicitly states the condition $$a \neq 0$$.
• For the entire discriminant $$D$$ to be equal to zero, both squared terms must be simultaneously zero. This would require $$b = 0$$ and $$(a + \frac{b}{2}) = 0$$, which consequently means $$a = 0$$.
• Since the problem restricts $$a$$ from being zero, the sum of these squared terms will always be strictly greater than zero.

Therefore, $$D > 0$$ for all real values of $$a$$ and $$b$$ where $$a \neq 0$$.

A strictly positive discriminant proves that the quadratic equation has exactly two distinct real roots. Consequently, the given curve will intersect the $$x$$ axis at exactly two distinct points.

Verification of Options & The Marked Answer

Let us test a counter example to verify our result and check Option A.

Assume $$a = 1$$, $$b = 0$$, and $$c = -1$$.

These values satisfy the given primary condition $$a + b + c = 0$$.

Substituting these into our curve equation gives $$y = 4x^2 - 2$$.

Equating to zero to find the $$x$$ axis intersection points:

$$4x^2 - 2 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$$

These roots are real and distinct but irrational. Thus, we can safely eliminate Option A.

Final Conclusion: The correct choice mathematically is Option C.