Consider the following for the next two items that follow :
`2x^(2) + 3x - alpha = 0` has roots - 2 and `beta` while the equation `x^(2) - 3mx + 2m^(2) = 0` has both roots positive, where `alpha gt 0 and beta gt 0` .
What is the value of `alpha^(2)` ?

Options

A. `(1)/(2)`
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to analyze the first quadratic equation and find the value of \(\alpha^2\).

### Step 1: Use the given roots in the first equation
The first equation is given as:
\[ 2x^2 + 3x - \alpha = 0 \]
We know that one of the roots is \(-2\) and the other root is \(\beta\). Since \(-2\) is a root, we can substitute \(x = -2\) into the equation to find \(\alpha\).

### Step 2: Substitute \(-2\) into the equation
Substituting \(x = -2\):
\[ 2(-2)^2 + 3(-2) - \alpha = 0 \]
Calculating this gives:
\[ 2(4) - 6 - \alpha = 0 \]
\[ 8 - 6 - \alpha = 0 \]
\[ 2 - \alpha = 0 \]

### Step 3: Solve for \(\alpha\)
From the equation \(2 - \alpha = 0\), we can solve for \(\alpha\):
\[ \alpha = 2 \]

### Step 4: Calculate \(\alpha^2\)
Now, we need to find \(\alpha^2\):
\[ \alpha^2 = 2^2 = 4 \]

### Final Answer
Thus, the value of \(\alpha^2\) is:
\[ \boxed{4} \]

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