If `alpha and beta` are the zeroes of the quadratic equation `x^2 – 12x + 32 = 0`, then a quadratic equation whose zeroes are `1/(2alpha+beta)and 1/(2beta+alpha) ` is

Options

A. `320x^2 + 36x + 1 = 0`
B. `320x^2 – 36x – 1 = 0`
C. `320x^2 - 36x + 1 = 0`
D. `320x^2 + 36x - 1 = 0`

Answer

To solve the problem step by step, we will first find the roots (zeroes) of the given quadratic equation and then derive the new quadratic equation based on the transformed roots.

### Step 1: Find the roots of the quadratic equation
The given quadratic equation is:
\[ x^2 - 12x + 32 = 0 \]

We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -12, c = 32 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot 32 = 144 - 128 = 16 \]

Now, substituting the values into the quadratic formula:
\[ x = \frac{12 \pm \sqrt{16}}{2 \cdot 1} = \frac{12 \pm 4}{2} \]

Calculating the two roots:
1. \( x_1 = \frac{12 + 4}{2} = \frac{16}{2} = 8 \)
2. \( x_2 = \frac{12 - 4}{2} = \frac{8}{2} = 4 \)

Thus, the roots (zeroes) are:
\[ \alpha = 8, \quad \beta = 4 \]

### Step 2: Calculate the new roots
We need to find the new roots:
1. \( \frac{1}{2\alpha + \beta} \)
2. \( \frac{1}{2\beta + \alpha} \)

Calculating each:
1. For \( \frac{1}{2\alpha + \beta} \):
\[ 2\alpha + \beta = 2 \cdot 8 + 4 = 16 + 4 = 20 \]
\[ \text{So, } \frac{1}{2\alpha + \beta} = \frac{1}{20} \]

2. For \( \frac{1}{2\beta + \alpha} \):
\[ 2\beta + \alpha = 2 \cdot 4 + 8 = 8 + 8 = 16 \]
\[ \text{So, } \frac{1}{2\beta + \alpha} = \frac{1}{16} \]

### Step 3: Form the new quadratic equation
Let the new roots be \( p = \frac{1}{20} \) and \( q = \frac{1}{16} \).

The sum of the roots \( p + q \):
\[ p + q = \frac{1}{20} + \frac{1}{16} \]
Finding a common denominator (LCM of 20 and 16 is 80):
\[ p + q = \frac{4}{80} + \frac{5}{80} = \frac{9}{80} \]

The product of the roots \( p \cdot q \):
\[ p \cdot q = \frac{1}{20} \cdot \frac{1}{16} = \frac{1}{320} \]

Using the sum and product of the roots, we can write the quadratic equation:
\[ x^2 - (p + q)x + (p \cdot q) = 0 \]
Substituting the values:
\[ x^2 - \frac{9}{80}x + \frac{1}{320} = 0 \]

### Step 4: Clear the fractions
To eliminate the fractions, multiply the entire equation by 320 (the least common multiple of the denominators):
\[ 320x^2 - 36x + 1 = 0 \]

### Final Answer
The quadratic equation whose zeroes are \( \frac{1}{2\alpha + \beta} \) and \( \frac{1}{2\beta + \alpha} \) is:
\[ 320x^2 - 36x + 1 = 0 \]