The quadratic equation whose roots are the arithmetic mean and the harmonic mean of the roots of the equation `x^(2)+7x-1=0` is

Options

A. `14x^(2)+14x-45=0`
B. `45x^(2)-14x+14=0`
C. `14x^(2)+45x-14=0`
D. `45x^(2)+14x-45=0`

Answer

To find the quadratic equation whose roots are the arithmetic mean and the harmonic mean of the roots of the equation \(x^2 + 7x - 1 = 0\), we can follow these steps:

### Step 1: Find the roots of the given quadratic equation.
The roots of the quadratic equation \(ax^2 + bx + c = 0\) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \(x^2 + 7x - 1 = 0\), we have \(a = 1\), \(b = 7\), and \(c = -1\).

Calculating the discriminant:
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 1 \cdot (-1) = 49 + 4 = 53 \]

Now, substituting into the quadratic formula:
\[ x = \frac{-7 \pm \sqrt{53}}{2} \]
Thus, the roots are:
\[ a = \frac{-7 + \sqrt{53}}{2}, \quad b = \frac{-7 - \sqrt{53}}{2} \]

### Step 2: Calculate the arithmetic mean (AM) of the roots.
The arithmetic mean of the roots \(a\) and \(b\) is given by:
\[ AM = \frac{a + b}{2} \]
Using the sum of the roots:
\[ a + b = -\frac{b}{a} = -\frac{7}{1} = -7 \]
Thus,
\[ AM = \frac{-7}{2} \]

### Step 3: Calculate the harmonic mean (HM) of the roots.
The harmonic mean of the roots \(a\) and \(b\) is given by:
\[ HM = \frac{2ab}{a + b} \]
We already have \(a + b = -7\) and now we need to find \(ab\):
\[ ab = \frac{c}{a} = \frac{-1}{1} = -1 \]
Now substituting into the formula for HM:
\[ HM = \frac{2 \cdot (-1)}{-7} = \frac{-2}{-7} = \frac{2}{7} \]

### Step 4: Form the quadratic equation with roots AM and HM.
We have the roots:
\[ AM = \frac{-7}{2}, \quad HM = \frac{2}{7} \]
The quadratic equation with roots \(p\) and \(q\) can be formed as:
\[ x^2 - (p + q)x + pq = 0 \]
Calculating \(p + q\):
\[ p + q = \frac{-7}{2} + \frac{2}{7} \]
Finding a common denominator (14):
\[ p + q = \frac{-49}{14} + \frac{4}{14} = \frac{-45}{14} \]

Now calculating \(pq\):
\[ pq = \left(\frac{-7}{2}\right) \left(\frac{2}{7}\right) = \frac{-14}{14} = -1 \]

### Step 5: Write the quadratic equation.
Substituting \(p + q\) and \(pq\) into the quadratic equation:
\[ x^2 - \left(\frac{-45}{14}\right)x - 1 = 0 \]
Multiplying through by 14 to eliminate the fraction:
\[ 14x^2 + 45x - 14 = 0 \]

### Final Answer:
The quadratic equation whose roots are the arithmetic mean and the harmonic mean of the roots of the equation \(x^2 + 7x - 1 = 0\) is:
\[ \boxed{14x^2 + 45x - 14 = 0} \]