The square of difference of the roots of the equattion `x^(2)+px+45=0` is 144. The roots are

Options

A. `-3,-15`
B. 2,5
C. `-12,12`
D. `18,-18`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Identify the given quadratic equation
The given quadratic equation is:
\[ x^2 + px + 45 = 0 \]

### Step 2: Use Vieta's formulas
According to Vieta's formulas, for the quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

For our equation:
- \( \alpha + \beta = -p \)
- \( \alpha \beta = 45 \)

### Step 3: Use the information about the difference of the roots
We are given that the square of the difference of the roots is 144:
\[ (\alpha - \beta)^2 = 144 \]

### Step 4: Relate the difference of the roots to the sum and product
We can express the square of the difference of the roots using the formula:
\[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \]

Substituting the values we have:
\[ (\alpha - \beta)^2 = (-p)^2 - 4 \cdot 45 \]
\[ 144 = p^2 - 180 \]

### Step 5: Solve for \( p^2 \)
Rearranging the equation gives:
\[ p^2 = 144 + 180 \]
\[ p^2 = 324 \]

### Step 6: Find the value of \( p \)
Taking the square root of both sides:
\[ p = \pm 18 \]

### Step 7: Determine the roots using \( p \)
Now we will find the roots using both values of \( p \).

1. **If \( p = 18 \)**:
- The sum of the roots \( \alpha + \beta = -18 \)
- The product of the roots \( \alpha \beta = 45 \)

We can set up the equations:
\[ \alpha + \beta = -18 \quad (1) \]
\[ \alpha \beta = 45 \quad (2) \]

Let \( \alpha = -x \) and \( \beta = -y \) (where \( x + y = 18 \) and \( xy = 45 \)). The roots can be found by solving:
\[ t^2 - (x+y)t + xy = 0 \]
\[ t^2 - 18t + 45 = 0 \]

The roots of this quadratic can be found using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \]
\[ = \frac{18 \pm \sqrt{324 - 180}}{2} = \frac{18 \pm \sqrt{144}}{2} = \frac{18 \pm 12}{2} \]

This gives us:
\[ t = \frac{30}{2} = 15 \quad \text{or} \quad t = \frac{6}{2} = 3 \]

Thus, the roots are \( \alpha = -15 \) and \( \beta = -3 \).

2. **If \( p = -18 \)**:
- The sum of the roots \( \alpha + \beta = 18 \)
- The product of the roots \( \alpha \beta = 45 \)

Using the same method as above:
\[ t^2 - 18t + 45 = 0 \]
The roots will again be \( 15 \) and \( 3 \), but since we need the roots in the context of the original equation, they will be \( -15 \) and \( -3 \).

### Final Answer
The roots of the equation \( x^2 + px + 45 = 0 \) are:
\[ -15 \text{ and } -3 \]