If `alpha and beta` are the roots of the equation `2x^(2)-3x+1=0`, form the equation whose roots are `(alpha)/(2beta+3) and (beta)/(2alpha+3)`.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the roots of the given quadratic equation
The given quadratic equation is:
\[ 2x^2 - 3x + 1 = 0 \]

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -3, c = 1 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \]

Now substituting into the quadratic formula:
\[ x = \frac{3 \pm \sqrt{1}}{2 \cdot 2} = \frac{3 \pm 1}{4} \]

This gives us the roots:
\[ x_1 = \frac{4}{4} = 1 \quad \text{and} \quad x_2 = \frac{2}{4} = \frac{1}{2} \]

Thus, we have:
\[ \alpha = \frac{1}{2}, \quad \beta = 1 \]

### Step 2: Find the new roots
We need to form a new equation whose roots are:
\[ \frac{\alpha}{2\beta + 3} \quad \text{and} \quad \frac{\beta}{2\alpha + 3} \]

Calculating the first root:
\[ \frac{\alpha}{2\beta + 3} = \frac{\frac{1}{2}}{2 \cdot 1 + 3} = \frac{\frac{1}{2}}{2 + 3} = \frac{\frac{1}{2}}{5} = \frac{1}{10} \]

Calculating the second root:
\[ \frac{\beta}{2\alpha + 3} = \frac{1}{2 \cdot \frac{1}{2} + 3} = \frac{1}{1 + 3} = \frac{1}{4} \]

Thus, the new roots are:
\[ \alpha' = \frac{1}{10}, \quad \beta' = \frac{1}{4} \]

### Step 3: Form the new quadratic equation
Using the formula for a quadratic equation based on its roots:
\[ x^2 - (s)x + p = 0 \]
where \( s = \alpha' + \beta' \) and \( p = \alpha' \cdot \beta' \).

Calculating \( s \):
\[ s = \frac{1}{10} + \frac{1}{4} \]
Finding a common denominator (which is 20):
\[ s = \frac{2}{20} + \frac{5}{20} = \frac{7}{20} \]

Calculating \( p \):
\[ p = \frac{1}{10} \cdot \frac{1}{4} = \frac{1}{40} \]

### Step 4: Write the quadratic equation
Substituting \( s \) and \( p \) into the quadratic equation:
\[ x^2 - \left(\frac{7}{20}\right)x + \frac{1}{40} = 0 \]

To eliminate the fractions, multiply through by 40 (the least common multiple of the denominators):
\[ 40x^2 - 14x + 1 = 0 \]

### Final Answer
The required quadratic equation is:
\[ 40x^2 - 14x + 1 = 0 \]

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