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Step-by-Step Solution
Step 1: Write down the given determinant
We are given the determinant:
$
\begin{vmatrix}
a & b & ax + b \\
b & c & bx + c \\
ax + b & bx + c & 0
\end{vmatrix}
$
Step 2: Apply a suitable row operation
Perform the operation
$
R_{3} \to R_{3} - (x\,R_{1} + R_{2}),
$
which modifies the third row. Let us see how each entry in the third row changes:
The first entry becomes
$
(ax + b) - [x \cdot a + b] = ax + b - ax - b = 0.
$
The second entry becomes
$
(bx + c) - [x \cdot b + c] = bx + c - bx - c = 0.
$
The third entry becomes
$
0 - [x \cdot (ax+b) + (bx+c)].
$
Simplify inside:
$
x\,(ax + b) = a x^2 + b x, \quad
bx + c = bx + c,
$
so the sum inside is
$
(a x^2 + b x) + (bx + c) = a x^2 + 2bx + c.
$
Hence the third entry in the last row is
$
- (a x^2 + 2bx + c).
$
Thus the determinant becomes
$
\begin{vmatrix}
a & b & ax + b \\
b & c & bx + c \\
0 & 0 & - (a x^2 + 2bx + c)
\end{vmatrix}.
$
Step 3: Evaluate the resulting determinant
Since the first two columns in the last row are zero, the determinant simplifies to the product of the diagonal entries in a block-like form:
$
= \bigl(- (a x^2 + 2bx + c)\bigr)
\begin{vmatrix}
a & b \\
b & c
\end{vmatrix}.
$
The determinant
$
\begin{vmatrix}
a & b \\
b & c
\end{vmatrix}
$
is
$
ac - b^2.
$
Therefore, the original determinant becomes
$
= - (a x^2 + 2bx + c)\,(ac - b^2)
= (a x^2 + 2bx + c)\,(b^2 - ac).
$
Step 4: Determine the sign using the given conditions
We are given:
$a > 0$ and the discriminant of $a x^2 + 2bx + c$ is negative.
A negative discriminant of $a x^2 + 2bx + c$ means
$
(2b)^2 - 4ac < 0 \implies 4b^2 - 4ac < 0 \implies b^2 - ac < 0.
$
Because $a > 0$ and the parabola $a x^2 + 2bx + c$ has a negative discriminant, the expression
$
a x^2 + 2bx + c
$
is always positive for real $x$ (it never touches the x-axis and opens upwards). Meanwhile,
$
b^2 - ac < 0.
$
Hence their product
$
(a x^2 + 2bx + c)\,(b^2 - ac)
$
is a positive number times a negative number, which is overall negative. Thus, the value of the determinant is
$
\text{-ve}.
$
