Find a if `ax^(2)-4x+9=0` has integral roots.

Answer

To find the values of \( a \) such that the quadratic equation \( ax^2 - 4x + 9 = 0 \) has integral roots, we need to ensure that the discriminant of the equation is a perfect square. The discriminant \( D \) for a quadratic equation of the form \( Ax^2 + Bx + C = 0 \) is given by:

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

In our case, \( A = a \), \( B = -4 \), and \( C = 9 \). Thus, the discriminant becomes:

\[ D = (-4)^2 - 4 \cdot a \cdot 9 \]
\[ D = 16 - 36a \]

For the roots to be integral, \( D \) must be a perfect square. Therefore, we set:

\[ 16 - 36a = k^2 \]

where \( k \) is some integer. Rearranging gives:

\[ 36a = 16 - k^2 \]
\[ a = \frac{16 - k^2}{36} \]

Next, we need \( 16 - k^2 \) to be a non-negative integer, which means:

\[ 16 - k^2 \geq 0 \implies k^2 \leq 16 \]

Thus, \( k \) can take the values \( -4, -3, -2, -1, 0, 1, 2, 3, 4 \). We will evaluate \( a \) for each of these values of \( k \):

1. For \( k = 0 \):
\[ a = \frac{16 - 0^2}{36} = \frac{16}{36} = \frac{4}{9} \]

2. For \( k = 1 \):
\[ a = \frac{16 - 1^2}{36} = \frac{15}{36} = \frac{5}{12} \]

3. For \( k = 2 \):
\[ a = \frac{16 - 2^2}{36} = \frac{12}{36} = \frac{1}{3} \]

4. For \( k = 3 \):
\[ a = \frac{16 - 3^2}{36} = \frac{7}{36} \]

5. For \( k = 4 \):
\[ a = \frac{16 - 4^2}{36} = \frac{0}{36} = 0 \]

6. For \( k = -1 \):
\[ a = \frac{16 - (-1)^2}{36} = \frac{15}{36} = \frac{5}{12} \]

7. For \( k = -2 \):
\[ a = \frac{16 - (-2)^2}{36} = \frac{12}{36} = \frac{1}{3} \]

8. For \( k = -3 \):
\[ a = \frac{16 - (-3)^2}{36} = \frac{7}{36} \]

9. For \( k = -4 \):
\[ a = \frac{16 - (-4)^2}{36} = \frac{0}{36} = 0 \]

Now, we summarize the values of \( a \) obtained:

- \( a = \frac{4}{9} \)
- \( a = \frac{5}{12} \)
- \( a = \frac{1}{3} \)
- \( a = \frac{7}{36} \)
- \( a = 0 \)

However, \( a = 0 \) is not valid as it would lead to an equation that does not have a quadratic term.

Next, we need to check which of these values of \( a \) yield integral roots:

1. **For \( a = \frac{1}{3} \)**:
\[ D = 16 - 36 \cdot \frac{1}{3} = 16 - 12 = 4 \quad (\text{perfect square}) \]

2. **For \( a = \frac{4}{9} \)**:
\[ D = 16 - 36 \cdot \frac{4}{9} = 16 - 16 = 0 \quad (\text{perfect square}) \]

3. **For \( a = \frac{5}{12} \)**:
\[ D = 16 - 36 \cdot \frac{5}{12} = 16 - 15 = 1 \quad (\text{perfect square}) \]

4. **For \( a = \frac{7}{36} \)**:
\[ D = 16 - 36 \cdot \frac{7}{36} = 16 - 7 = 9 \quad (\text{perfect square}) \]

Thus, the values of \( a \) that yield integral roots are:

- \( a = \frac{1}{3} \)
- \( a = \frac{4}{9} \)
- \( a = \frac{5}{12} \)
- \( a = \frac{7}{36} \)

**Final Answer:**
The values of \( a \) for which the equation \( ax^2 - 4x + 9 = 0 \) has integral roots are \( a = \frac{1}{3}, \frac{4}{9}, \frac{5}{12}, \frac{7}{36} \).

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