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

Assessment: WAEC SSCE - General Mathematics - 2010 (Objective) Subject: General Mathematics

Question 1 Report

\(\frac{8}{3}\)

\(\frac{7}{3}\)

\(\frac{5}{3}\)

\(\frac{2}{3}\)

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 \(\frac{8}{3}\).

Read lesson note on Quadratic Equations (WAEC)

Quadratic Equations

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

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