Find the value of k such that $t^2$ + kt + $\frac{25}{4}$ is a perfect square

Analyze the question: We need k so that t² + kt + 25/4 is a perfect square, meaning it can be written as (t + a)².

Concept: (t + a)² = t² + 2at + a². Comparing this with t² + kt + 25/4:

• a² = 25/4, so a = ±5/2
• k = 2a

Solution:

a² = 25/4, so a = ±5/2

k = 2a = 2(5/2) = 5, or k = 2(−5/2) = −5

Another way is the discriminant method. A quadratic is a perfect square when b² − 4ac = 0:

k² − 4(1)(25/4) = 0 k² − 25 = 0 k² = 25 k = ±5

The positive value, k = 5, is among the options. Check: t² + 5t + 25/4 = (t + 5/2)². ✓

(The other value, −5, is not listed.)

Answer: A. 5

Tip: For x² + bx + c to be a perfect square, c must equal (b/2)². Equivalently, b = 2√c. Here, 2 × √(25/4) = 2 × 5/2 = 5. Remember that k can also be negative, so check which values appear in the options.