If `sin theta and cos theta ` are the roots of the quadratic equation `ax ^(2) +bx + c=0(a ne 0).` Then find the value of `(b ^(2) -a^(2))/(ac ).`

Answer

To solve the problem, we need to find the value of \(\frac{b^2 - a^2}{ac}\) given that \(\sin \theta\) and \(\cos \theta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\).

### Step-by-Step Solution:

1. **Identify the roots and their properties**:
Since \(\sin \theta\) and \(\cos \theta\) are the roots of the quadratic equation, we can use Vieta's formulas:
- The sum of the roots:
\[ \sin \theta + \cos \theta = -\frac{b}{a} \]
- The product of the roots:
\[ \sin \theta \cdot \cos \theta = \frac{c}{a} \]

2. **Square the sum of the roots**:
We will square the sum of the roots to find \(b^2\):
\[ (\sin \theta + \cos \theta)^2 = \left(-\frac{b}{a}\right)^2 \]
Expanding the left side:
\[ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = \frac{b^2}{a^2} \]
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ 1 + 2\sin \theta \cos \theta = \frac{b^2}{a^2} \]

3. **Express \(b^2\)**:
Rearranging the equation gives us:
\[ b^2 = a^2(1 + 2\sin \theta \cos \theta) \]

4. **Substituting for \(c\)**:
From the product of the roots, we have:
\[ c = a \sin \theta \cos \theta \]

5. **Substituting \(b^2\) and \(c\) into the expression**:
Now we substitute \(b^2\) and \(c\) into the expression \(\frac{b^2 - a^2}{ac}\):
\[ \frac{b^2 - a^2}{ac} = \frac{a^2(1 + 2\sin \theta \cos \theta) - a^2}{a(a \sin \theta \cos \theta)} \]
Simplifying the numerator:
\[ = \frac{a^2(2\sin \theta \cos \theta)}{a^2 \sin \theta \cos \theta} \]
Canceling \(a^2 \sin \theta \cos \theta\) from numerator and denominator:
\[ = 2 \]

### Final Answer:
Thus, the value of \(\frac{b^2 - a^2}{ac}\) is \(2\).