Let a, b, c be non zero numbers such that
`int_(0)^(3)sqrt(x^(2)+x+1)(ax^(2)+bx+c)dx=int_(0)^(5)sqrt(1+x^(2)+x)(ax^(2)+bx+c)dx`. Then the quadratic equation `ax^(2)+bx+c=0` has

Options

A. no root in (0, 3)
B. at least one root in (3, 5)
C. a double root in (0, 3)
D. two imaginary roots

Answer

To solve the problem, we start with the given equation:

\[ \int_{0}^{3} \sqrt{x^2 + x + 1} (ax^2 + bx + c) \, dx = \int_{0}^{5} \sqrt{x^2 + x + 1} (ax^2 + bx + c) \, dx \]

Let \( f(x) = \sqrt{x^2 + x + 1} (ax^2 + bx + c) \).

### Step 1: Set up the equation
From the equality of the two integrals, we have:

\[ \int_{0}^{3} f(x) \, dx = \int_{0}^{5} f(x) \, dx \]

### Step 2: Rearranging the equation
This implies:

\[ \int_{3}^{5} f(x) \, dx = 0 \]

### Step 3: Analyze the integral
Since \( f(x) \) is continuous (as it is a product of continuous functions), the integral being zero suggests that the function \( f(x) \) must change sign over the interval \([3, 5]\).

### Step 4: Apply Rolle's Theorem
By Rolle's Theorem, since \( f(x) \) is continuous on \([3, 5]\) and differentiable on \((3, 5)\), there exists at least one \( c \in (3, 5) \) such that:

\[ f(c) = 0 \]

### Step 5: Setting the quadratic to zero
Thus, we have:

\[ \sqrt{c^2 + c + 1} (ac^2 + bc + c) = 0 \]

Since \( \sqrt{c^2 + c + 1} \neq 0 \) for all \( c \), we must have:

\[ ac^2 + bc + c = 0 \]

### Step 6: Factor out \( c \)
This can be factored as:

\[ c(ac + b + 1) = 0 \]

Since \( c \) is non-zero (as given in the problem), we have:

\[ ac + b + 1 = 0 \]

### Step 7: Conclusion about roots
This means that the quadratic equation \( ax^2 + bx + c = 0 \) must have at least one root in the interval \((3, 5)\). Therefore, we conclude that the quadratic equation has roots.

### Final Result
The quadratic equation \( ax^2 + bx + c = 0 \) has at least one root in the interval \((3, 5)\).

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