The equation `(a + 2)x^2 + (a-3)x = 2a - 1, a != -2` has roots rational for

Options

A. all rational values of a except a = -2
B. all real values of a except a = - 2
C. rational values of `a gt 1//2`
D. none of these

Answer

To determine the values of \( a \) for which the equation

\[ (a + 2)x^2 + (a - 3)x = 2a - 1 \]

has rational roots, we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the given equation into standard quadratic form:

\[ (a + 2)x^2 + (a - 3)x - (2a - 1) = 0 \]

This simplifies to:

\[ (a + 2)x^2 + (a - 3)x + (1 - 2a) = 0 \]

### Step 2: Identifying Coefficients
From the standard form \( Ax^2 + Bx + C = 0 \), we identify:

- \( A = a + 2 \)
- \( B = a - 3 \)
- \( C = 1 - 2a \)

### Step 3: Condition for Rational Roots
For the quadratic equation to have rational roots, the discriminant \( D \) must be a perfect square. The discriminant is given by:

\[ D = B^2 - 4AC \]

Substituting the values of \( A \), \( B \), and \( C \):

\[ D = (a - 3)^2 - 4(a + 2)(1 - 2a) \]

### Step 4: Expanding the Discriminant
Now we expand \( D \):

\[ D = (a^2 - 6a + 9) - 4[(a + 2)(1 - 2a)] \]

Calculating \( (a + 2)(1 - 2a) \):

\[ (a + 2)(1 - 2a) = a - 2a^2 + 2 - 4a = -2a^2 - 3a + 2 \]

Now substituting this back into the discriminant:

\[ D = a^2 - 6a + 9 - 4(-2a^2 - 3a + 2) \]

Expanding further:

\[ D = a^2 - 6a + 9 + 8a^2 + 12a - 8 \]

Combining like terms:

\[ D = 9a^2 + 6a + 1 \]

### Step 5: Setting the Discriminant Greater than or Equal to Zero
For \( D \) to be a perfect square, we set:

\[ 9a^2 + 6a + 1 \geq 0 \]

### Step 6: Analyzing the Quadratic
The expression \( 9a^2 + 6a + 1 \) is always non-negative because its discriminant is:

\[ D' = 6^2 - 4 \cdot 9 \cdot 1 = 36 - 36 = 0 \]

Since the discriminant is zero, the quadratic has a double root at:

\[ a = -\frac{6}{2 \cdot 9} = -\frac{1}{3} \]

### Step 7: Conclusion
Thus, the quadratic \( 9a^2 + 6a + 1 \) is always greater than or equal to zero for all \( a \) except \( a = -2 \) (as given in the problem). Therefore, the values of \( a \) for which the original equation has rational roots are:

\[ \text{All rational values of } a \text{ except } a = -2. \]