If x2 + kx + $\frac{16}{9}$ is a perfect square, find the value of k

Analyze the question: We need k so that x² + kx + 16/9 is a perfect square, meaning it can be written as (x + a)².

Concept: (x + a)² = x² + 2ax + a². Comparing the terms:

• the constant term gives a² = 16/9
• the coefficient of x gives k = 2a

Solution:

a² = 16/9, so a = ±4/3

k = 2a = ±8/3

The positive value, k = 8/3, is option A. Check: (x + 4/3)² = x² + (8/3)x + 16/9 ✓

(The value k = -8/3 also works, but it is not among the options.)

Alternative method (discriminant): A quadratic is a perfect square when b² − 4ac = 0: k² − 4(1)(16/9) = 0 k² = 64/9 k = ±8/3

Tip: For x² + kx + c to be a perfect square, use k = 2√c, or "half the coefficient of x, squared, equals the constant." Here (k/2)² = 16/9, so k/2 = 4/3 and k = 8/3.