If `a gt b gt c` and the system of equtions `ax +by +cz =0`, `bx +cy+az=0`, `cx+ay+bz=0` has a non-trivial solution then both the roots of the quadratic equation `at^(2)+bt+c` are

Options

A. (a) real
B. (b) of opposite sign
C. (c) positive
D. (d) complex

Answer

To solve the given problem step by step, we need to analyze the conditions of the system of equations and the quadratic equation provided.

### Step 1: Understanding Non-Trivial Solutions
The system of equations:
1. \( ax + by + cz = 0 \)
2. \( bx + cy + az = 0 \)
3. \( cx + ay + bz = 0 \)

has a non-trivial solution if the determinant of the coefficients is zero. This means that the determinant formed by the coefficients of \( x, y, z \) must equal zero.

### Step 2: Setting Up the Determinant
The determinant of the coefficients can be expressed as follows:
\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

### Step 3: Calculating the Determinant
Calculating this determinant, we have:
\[ D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating each of these 2x2 determinants:
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, we get:
\[ D = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]

### Step 4: Setting the Determinant to Zero
For a non-trivial solution, we set \( D = 0 \):
\[ a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = 0 \]

### Step 5: Simplifying the Expression
After simplifying, we arrive at the identity:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]

### Step 6: Using the Identity
This identity can be factored as:
\[ (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 0 \]
Thus, either \( a + b + c = 0 \) or \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \).

### Step 7: Analyzing the Quadratic Equation
Now, we need to analyze the quadratic equation:
\[ at^2 + bt + c = 0 \]
The discriminant \( D' \) of this quadratic is given by:
\[ D' = b^2 - 4ac \]

### Step 8: Substituting \( b \)
From \( a + b + c = 0 \), we can express \( b \) as:
\[ b = -a - c \]
Substituting this into the discriminant:
\[ D' = (-a - c)^2 - 4ac = a^2 + 2ac + c^2 - 4ac = a^2 - 2ac + c^2 = (a - c)^2 \]

### Step 9: Analyzing the Roots
Since \( a > b > c \), we know \( a - c > 0 \). Therefore, \( D' > 0 \), which implies that the roots of the quadratic equation are real and distinct.

### Conclusion
Thus, the roots of the quadratic equation \( at^2 + bt + c = 0 \) are real and distinct.