If `3x + 4y= 16 and xy= 4`, find the value of `9x^(2) + 16y^(2)`

Answer

To solve the problem, we need to find the value of \(9x^2 + 16y^2\) given the equations \(3x + 4y = 16\) and \(xy = 4\).

### Step-by-Step Solution:

1. **Start with the equations:**
We have two equations:
\[ (1) \quad 3x + 4y = 16 \]
\[ (2) \quad xy = 4 \]

2. **Express \(y\) in terms of \(x\):**
From equation (2), we can express \(y\) as:
\[ y = \frac{4}{x} \]

3. **Substitute \(y\) in equation (1):**
Substitute \(y\) in equation (1):
\[ 3x + 4\left(\frac{4}{x}\right) = 16 \]
Simplifying this gives:
\[ 3x + \frac{16}{x} = 16 \]

4. **Multiply through by \(x\) to eliminate the fraction:**
Multiply the entire equation by \(x\):
\[ 3x^2 + 16 = 16x \]

5. **Rearrange the equation:**
Rearranging gives:
\[ 3x^2 - 16x + 16 = 0 \]

6. **Use the quadratic formula to solve for \(x\):**
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 3\), \(b = -16\), and \(c = 16\). Plugging in these values:
\[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 16}}{2 \cdot 3} \]
Simplifying the discriminant:
\[ x = \frac{16 \pm \sqrt{256 - 192}}{6} = \frac{16 \pm \sqrt{64}}{6} = \frac{16 \pm 8}{6} \]
This gives us two possible values for \(x\):
\[ x = \frac{24}{6} = 4 \quad \text{or} \quad x = \frac{8}{6} = \frac{4}{3} \]

7. **Find corresponding values of \(y\):**
For \(x = 4\):
\[ y = \frac{4}{4} = 1 \]
For \(x = \frac{4}{3}\):
\[ y = \frac{4}{\frac{4}{3}} = 3 \]

8. **Calculate \(9x^2 + 16y^2\) for both pairs \((x, y)\):**
For \((x, y) = (4, 1)\):
\[ 9x^2 + 16y^2 = 9(4^2) + 16(1^2) = 9 \cdot 16 + 16 \cdot 1 = 144 + 16 = 160 \]

For \((x, y) = \left(\frac{4}{3}, 3\right)\):
\[ 9x^2 + 16y^2 = 9\left(\frac{4}{3}\right)^2 + 16(3^2) = 9 \cdot \frac{16}{9} + 16 \cdot 9 = 16 + 144 = 160 \]

9. **Conclusion:**
In both cases, we find that:
\[ 9x^2 + 16y^2 = 160 \]

### Final Answer:
\[ \boxed{160} \]