If `alpha` and `beta` are the roots of the equation `2x^(2)+4x-5=0`, then the equation whose roots are `(1)/(2alpha-3) and (1)/(2beta-3)` is

Options

A. `x^(2)+10x-11=0`
B. `11x^(2)+10x+1=0`
C. `x^(2)+10x+11=0`
D. `11x^(2)-10x+1=0`

Answer

To solve the problem, we need to find the equation whose roots are given by \( \frac{1}{2\alpha - 3} \) and \( \frac{1}{2\beta - 3} \), where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( 2x^2 + 4x - 5 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \) of the equation \( 2x^2 + 4x - 5 = 0 \).

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = 4 \), and \( c = -5 \).

Calculating the discriminant:
\[ b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot (-5) = 16 + 40 = 56 \]

Now substituting into the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{56}}{2 \cdot 2} = \frac{-4 \pm 2\sqrt{14}}{4} = \frac{-2 \pm \sqrt{14}}{2} \]

Thus, the roots are:
\[ \alpha = \frac{-2 + \sqrt{14}}{2}, \quad \beta = \frac{-2 - \sqrt{14}}{2} \]

### Step 2: Express the new roots in terms of \( \alpha \) and \( \beta \).

The new roots are:
\[ y_1 = \frac{1}{2\alpha - 3}, \quad y_2 = \frac{1}{2\beta - 3} \]

### Step 3: Find \( 2\alpha - 3 \) and \( 2\beta - 3 \).

Calculating \( 2\alpha \) and \( 2\beta \):
\[ 2\alpha = -2 + \sqrt{14}, \quad 2\beta = -2 - \sqrt{14} \]
Thus,
\[ 2\alpha - 3 = -5 + \sqrt{14}, \quad 2\beta - 3 = -5 - \sqrt{14} \]

### Step 4: Find the new roots \( y_1 \) and \( y_2 \).

Calculating \( y_1 \) and \( y_2 \):
\[ y_1 = \frac{1}{-5 + \sqrt{14}}, \quad y_2 = \frac{1}{-5 - \sqrt{14}} \]

### Step 5: Find the sum and product of the new roots.

Using the formulas for the sum and product of roots:
\[ \text{Sum} = y_1 + y_2 = \frac{1}{-5 + \sqrt{14}} + \frac{1}{-5 - \sqrt{14}} = \frac{(-5 - \sqrt{14}) + (-5 + \sqrt{14})}{(-5 + \sqrt{14})(-5 - \sqrt{14})} \]
\[ = \frac{-10}{(-5)^2 - (\sqrt{14})^2} = \frac{-10}{25 - 14} = \frac{-10}{11} \]

For the product:
\[ \text{Product} = y_1 \cdot y_2 = \left(\frac{1}{-5 + \sqrt{14}}\right) \left(\frac{1}{-5 - \sqrt{14}}\right) = \frac{1}{(-5 + \sqrt{14})(-5 - \sqrt{14})} = \frac{1}{25 - 14} = \frac{1}{11} \]

### Step 6: Form the new quadratic equation.

Using the sum and product of the roots, we can form the quadratic equation:
\[ x^2 - \left(\text{Sum}\right)x + \text{Product} = 0 \]
Substituting the values:
\[ x^2 + \frac{10}{11}x + \frac{1}{11} = 0 \]

Multiplying through by 11 to eliminate the fractions:
\[ 11x^2 + 10x + 1 = 0 \]

### Final Answer:
The required equation is:
\[ \boxed{11x^2 + 10x + 1 = 0} \]