If \(x\) and \(y\) are real numbers such that \(4x^2 + 4y^2 - 4xy - 6y + 3 = 0\), then the value of \((4x + 5y)\) is
Correct Answer: 7
Solution and Explanation
Given: \(4x^2 + 4y^2 - 4xy - 6y + 3 = 0\)
Group terms: \[ 4x^2 - 4xy + 4y^2 - 6y + 3 = 0 \] Try completing the square or rewriting in a suitable form. Let us try substituting: Let \(4x + 5y = z\), and try to express original equation in terms of \(z\). But first, complete the square.
Group as: \[ 4x^2 - 4xy + 4y^2 - 6y + 3 \] Note that: \[ 4x^2 - 4xy + 4y^2 = 4(x - y)^2 \] So, \[ 4(x - y)^2 - 6y + 3 = 0 \] Now expand: \[ 4(x - y)^2 = 6y - 3 \] Let’s try a substitution: Let \(x = y + a\), then \(x - y = a\), so: \[ 4a^2 = 6y - 3 \Rightarrow a^2 = \frac{6y - 3}{4} \] Now, \[ 4x + 5y = 4(y + a) + 5y = 9y + 4a \] We want the value of \(4x + 5y = z = 9y + 4a\) Substitute \(a = \sqrt{\frac{6y - 3}{4}}\) Try values to get rational result. Try \(y = 1\): Then LHS: \[ 4x^2 + 4y^2 - 4xy - 6y + 3 = 4x^2 + 4 - 4x - 6 + 3 = 4x^2 - 4x + 1 \] Set equal to 0: \[ 4x^2 - 4x + 1 = 0 \Rightarrow (2x - 1)^2 = 0 \Rightarrow x = \frac{1}{2} \] Then \(4x + 5y = 4 \cdot \frac{1}{2} + 5 \cdot 1 = 2 + 5 = \boxed{7}\)
∴ The value of \(4x + 5y\) is \(\boxed{7}\)
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