if `a gt b gt c` and the system of equations `ax + by + cz = 0, bx + cy + az 0 and cx + ay + bz = 0` has a non-trivial solution, then the quadratic equation `ax^(2) + bx + c =0` has

Options

A. at least one positive root
B. roots opposite in sign
C. positive roots
D. imaginary roots

Explanation

To solve the problem step by step, we will analyze the given system of equations and the conditions provided.

### Step 1: Write down the system of equations
We have the following system of equations:
1. \( ax + by + cz = 0 \)
2. \( bx + cy + az = 0 \)
3. \( cx + ay + bz = 0 \)

### Step 2: Formulate the determinant
To find the condition for a non-trivial solution, we need to calculate the determinant of the coefficients of the variables \(x\), \(y\), and \(z\):
\[ \Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula for a \(3 \times 3\) matrix, we expand it:
\[ \Delta = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \)
2. \( \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac \)
3. \( \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2 \)

Substituting these back into the determinant:
\[ \Delta = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]
\[ = acb - a^3 - b^3 + abc + abc - c^3 \]
\[ = -a^3 + b^3 + c^3 - 3abc \]

### Step 4: Set the determinant to zero for non-trivial solutions
For the system to have a non-trivial solution, we set the determinant to zero:
\[ -a^3 + b^3 + c^3 - 3abc = 0 \]
This can be rearranged to:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]

### Step 5: Analyze the implications
From the identity \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)\), we can conclude that either:
1. \( a + b + c = 0 \)
2. \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \)

### Step 6: Determine the case
Given the condition \(a > b > c\), the second case cannot hold (as it would imply \(a = b = c\)). Therefore, we must have:
\[ a + b + c = 0 \]

### Step 7: Substitute into the quadratic equation
Now, we consider the quadratic equation:
\[ ax^2 + bx + c = 0 \]
Substituting \(x = 1\):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c = 0 \]

### Conclusion
Since \(x = 1\) is a root of the quadratic equation, we conclude that the quadratic equation \(ax^2 + bx + c = 0\) has at least one root, which is \(x = 1\).

### Final Answer
Thus, the quadratic equation \(ax^2 + bx + c = 0\) has at least one positive root.

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