If 8,2 are the roots of `x^(2)+ax+beta=0` and 3,3 are the roots of `x^(2)+alphax+b=0`, then the roots of `x^(2)+ax+b=0` are

Options

A. `8,-1`
B. `-9,2`
C. `-8,-2`
D. 9,1

Answer

To solve the problem, we need to find the roots of the quadratic equation \(x^2 + ax + b = 0\) given that the roots of two other quadratic equations are provided.

### Step 1: Identify the roots and their sums/products
The roots of the first equation \(x^2 + ax + \beta = 0\) are given as 8 and 2.

Using the properties of roots:
- Sum of roots \( = 8 + 2 = 10\)
- Product of roots \( = 8 \times 2 = 16\)

From the standard form of a quadratic equation \(x^2 + px + q = 0\), we know:
- The sum of roots is given by \(-a\) (where \(a\) is the coefficient of \(x\)).
- The product of roots is given by \(\beta\) (the constant term).

Thus, we have:
1. \( -a = 10 \) ⇒ \( a = -10 \)
2. \( \beta = 16 \)

### Step 2: Analyze the second equation
The roots of the second equation \(x^2 + \alpha x + b = 0\) are given as 3 and 3 (a repeated root).

Using the properties of roots:
- Sum of roots \( = 3 + 3 = 6\)
- Product of roots \( = 3 \times 3 = 9\)

From the standard form:
1. \( -\alpha = 6 \) ⇒ \( \alpha = -6 \)
2. \( b = 9 \)

### Step 3: Substitute \(a\) and \(b\) into the final equation
Now we substitute \(a\) and \(b\) into the equation \(x^2 + ax + b = 0\):
\[ x^2 - 10x + 9 = 0 \]

### Step 4: Factor the quadratic equation
To find the roots, we can factor the equation:
\[ x^2 - 10x + 9 = 0 \]

We look for two numbers that multiply to \(9\) and add to \(-10\). These numbers are \(-1\) and \(-9\):
\[ x^2 - 9x - x + 9 = 0 \]
\[ x(x - 9) - 1(x - 9) = 0 \]
\[ (x - 9)(x - 1) = 0 \]

### Step 5: Solve for \(x\)
Setting each factor to zero gives us the roots:
1. \( x - 9 = 0 \) ⇒ \( x = 9 \)
2. \( x - 1 = 0 \) ⇒ \( x = 1 \)

Thus, the roots of the equation \(x^2 + ax + b = 0\) are \(x = 1\) and \(x = 9\).

### Final Answer
The roots of \(x^2 + ax + b = 0\) are \(1\) and \(9\).