Find the value of `q/p` , if `px^2+x-15=0` and `18x^2+3x+q=0` have both the roots common.

Options

A. `(-15)/2`
B. `-9`
C. `-27/4`
D. 14/3`

Answer

To find the value of \( \frac{q}{p} \) given the equations \( px^2 + x - 15 = 0 \) and \( 18x^2 + 3x + q = 0 \) with both equations having common roots, we can follow these steps:

### Step 1: Identify the coefficients
For the first equation \( px^2 + x - 15 = 0 \):
- \( a_1 = p \)
- \( b_1 = 1 \)
- \( c_1 = -15 \)

For the second equation \( 18x^2 + 3x + q = 0 \):
- \( a_2 = 18 \)
- \( b_2 = 3 \)
- \( c_2 = q \)

### Step 2: Use the relationships for roots
The sum of the roots \( \alpha + \beta \) and product of the roots \( \alpha \beta \) for a quadratic equation \( ax^2 + bx + c = 0 \) are given by:
- Sum of roots: \( -\frac{b}{a} \)
- Product of roots: \( \frac{c}{a} \)

### Step 3: Calculate the sum and product for the first equation
For the first equation:
- Sum of roots:
\[ \alpha + \beta = -\frac{b_1}{a_1} = -\frac{1}{p} \]
- Product of roots:
\[ \alpha \beta = \frac{c_1}{a_1} = \frac{-15}{p} \]

### Step 4: Calculate the sum and product for the second equation
For the second equation:
- Sum of roots:
\[ \alpha + \beta = -\frac{b_2}{a_2} = -\frac{3}{18} = -\frac{1}{6} \]
- Product of roots:
\[ \alpha \beta = \frac{c_2}{a_2} = \frac{q}{18} \]

### Step 5: Set the sums equal to each other
Since the roots are common, we have:
\[ -\frac{1}{p} = -\frac{1}{6} \]
This simplifies to:
\[ \frac{1}{p} = \frac{1}{6} \implies p = 6 \]

### Step 6: Set the products equal to each other
Now, equate the products:
\[ \frac{-15}{p} = \frac{q}{18} \]
Substituting \( p = 6 \):
\[ \frac{-15}{6} = \frac{q}{18} \]
This simplifies to:
\[ -\frac{5}{2} = \frac{q}{18} \]

### Step 7: Solve for \( q \)
Cross-multiplying gives:
\[ -5 \cdot 18 = 2q \implies -90 = 2q \implies q = -45 \]

### Step 8: Calculate \( \frac{q}{p} \)
Now we can find \( \frac{q}{p} \):
\[ \frac{q}{p} = \frac{-45}{6} = -\frac{15}{2} \]

Thus, the final answer is:
\[ \frac{q}{p} = -\frac{15}{2} \]