Find the roots of the equation
`2y ^(2) - 9y + 9 = 0.`

Options

A. `3,2`
B. `3, (3)/(2)`
C. `2, (2)/(3)`
D. `-3, (5)/(2)`

Answer

To find the roots of the equation \( 2y^2 - 9y + 9 = 0 \), we can use the factorization method. Let's go through the steps:

### Step 1: Identify coefficients
The given quadratic equation is in the standard form \( ay^2 + by + c = 0 \), where:
- \( a = 2 \)
- \( b = -9 \)
- \( c = 9 \)

### Step 2: Calculate the product \( ac \)
We need to calculate \( ac \):
\[ ac = 2 \times 9 = 18 \]

### Step 3: Find two numbers that multiply to \( ac \) and add to \( b \)
We need to find two numbers that multiply to \( 18 \) (the value of \( ac \)) and add up to \( -9 \) (the value of \( b \)). The numbers are \( -6 \) and \( -3 \) since:
\[ -6 \times -3 = 18 \quad \text{and} \quad -6 + (-3) = -9 \]

### Step 4: Rewrite the equation
Now we can rewrite the middle term \( -9y \) using \( -6y \) and \( -3y \):
\[ 2y^2 - 6y - 3y + 9 = 0 \]

### Step 5: Factor by grouping
Next, we group the terms:
\[ (2y^2 - 6y) + (-3y + 9) = 0 \]
Now, factor out the common factors in each group:
\[ 2y(y - 3) - 3(y - 3) = 0 \]

### Step 6: Factor out the common binomial
Now we can factor out the common binomial \( (y - 3) \):
\[ (2y - 3)(y - 3) = 0 \]

### Step 7: Set each factor to zero
Now, we set each factor equal to zero:
1. \( 2y - 3 = 0 \)
2. \( y - 3 = 0 \)

### Step 8: Solve for \( y \)
Solving the first equation:
\[ 2y - 3 = 0 \implies 2y = 3 \implies y = \frac{3}{2} \]
Solving the second equation:
\[ y - 3 = 0 \implies y = 3 \]

### Step 9: Write the roots
The roots of the equation \( 2y^2 - 9y + 9 = 0 \) are:
\[ y = 3 \quad \text{and} \quad y = \frac{3}{2} \]

### Final Answer
Thus, the roots are \( y = 3 \) and \( y = \frac{3}{2} \).

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