If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k
Answer Details
We can approach this problem using a well-known algebraic identity:
(a + b)^2 = a^2 + 2ab + b^2
If we let a = x and b = \(\frac{4}{3}\), we can write:
x^2 + kx + \(\frac{16}{9}\) = (x + \(\frac{4}{3}\))^2
Expanding the right side of the equation, we get:
x^2 + 2x(\(\frac{4}{3}\)) + \(\frac{16}{9}\) = x^2 + kx + \(\frac{16}{9}\)
Simplifying, we get:
2x(\(\frac{4}{3}\)) = kx
k = 2(\(\frac{4}{3}\)) = \(\frac{8}{3}\)
Therefore, the value of k is \(\frac{8}{3}\).