If the quadratic equations `3x^2 + ax + 1 = 0` and `2x^2 + bx + 1 = 0` have a common root then the value of `5ab-2a^2-3b^2` , where `a b in R` , is equal to

Options

A. zero
B. 1
C. `-1`
D. none of these

Answer

To solve the problem, we need to find the value of \(5ab - 2a^2 - 3b^2\) given that the quadratic equations \(3x^2 + ax + 1 = 0\) and \(2x^2 + bx + 1 = 0\) have a common root.

### Step 1: Let the common root be \(r\).
Since \(r\) is a common root, it satisfies both equations:
1. \(3r^2 + ar + 1 = 0\)
2. \(2r^2 + br + 1 = 0\)

### Step 2: Rearranging both equations.
From the first equation, we can express \(ar\):
\[ ar = -3r^2 - 1 \quad \text{(1)} \]
From the second equation, we can express \(br\):
\[ br = -2r^2 - 1 \quad \text{(2)} \]

### Step 3: Express \(a\) and \(b\) in terms of \(r\).
From equation (1):
\[ a = \frac{-3r^2 - 1}{r} \quad \text{(if } r \neq 0\text{)} \]
From equation (2):
\[ b = \frac{-2r^2 - 1}{r} \quad \text{(if } r \neq 0\text{)} \]

### Step 4: Substitute \(a\) and \(b\) into the expression \(5ab - 2a^2 - 3b^2\).
Substituting \(a\) and \(b\):
\[ 5ab = 5 \left(\frac{-3r^2 - 1}{r}\right)\left(\frac{-2r^2 - 1}{r}\right) \]
\[ = 5 \cdot \frac{(3r^2 + 1)(2r^2 + 1)}{r^2} \]
\[ = 5 \cdot \frac{6r^4 + 5r^2 + 1}{r^2} \]
\[ = \frac{30r^4 + 25r^2 + 5}{r^2} \]

Now for \(2a^2\):
\[ 2a^2 = 2\left(\frac{-3r^2 - 1}{r}\right)^2 = 2 \cdot \frac{(3r^2 + 1)^2}{r^2} = 2 \cdot \frac{9r^4 + 6r^2 + 1}{r^2} = \frac{18r^4 + 12r^2 + 2}{r^2} \]

And for \(3b^2\):
\[ 3b^2 = 3\left(\frac{-2r^2 - 1}{r}\right)^2 = 3 \cdot \frac{(2r^2 + 1)^2}{r^2} = 3 \cdot \frac{4r^4 + 4r^2 + 1}{r^2} = \frac{12r^4 + 12r^2 + 3}{r^2} \]

### Step 5: Combine everything.
Now we can substitute back into the expression:
\[ 5ab - 2a^2 - 3b^2 = \frac{30r^4 + 25r^2 + 5}{r^2} - \frac{18r^4 + 12r^2 + 2}{r^2} - \frac{12r^4 + 12r^2 + 3}{r^2} \]
Combining the fractions:
\[ = \frac{(30r^4 + 25r^2 + 5) - (18r^4 + 12r^2 + 2) - (12r^4 + 12r^2 + 3)}{r^2} \]
\[ = \frac{30r^4 + 25r^2 + 5 - 18r^4 - 12r^2 - 2 - 12r^4 - 12r^2 - 3}{r^2} \]
\[ = \frac{0r^4 + 1r^2 + 0}{r^2} = 1 \]

### Final Result:
Thus, the value of \(5ab - 2a^2 - 3b^2\) is \(1\).