If $x$ and $y$ are real numbers such that $4x^2 + 4y^2 - 4xy - 6y + 3 = 0$, then the value of $(4x + 5y)$ is

We need to find the value of $(4x + 5y)$ when $4x^2 + 4y^2 - 4xy - 6y + 3 = 0$.

The key insight here is to rearrange and group terms to create perfect squares. When we have a sum of squares equal to zero, we can find exact values for our variables.

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Starting with: $4x^2 + 4y^2 - 4xy - 6y + 3 = 0$

We'll split the $4y^2$ term as $y^2 + 3y^2$:

$4x^2 + y^2 + 3y^2 - 4xy - 6y + 3 = 0$

Why this split? We want to create two separate perfect square expressions. The $y^2$ will help us form a perfect square with the $x$ terms, while $3y^2$ will work with the remaining $y$ terms.

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$(4x^2 - 4xy + y^2) + (3y^2 - 6y + 3) = 0$

Now we have two groups that we can convert into perfect squares.

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For $4x^2 - 4xy + y^2$:

This matches the pattern $(a - b)^2 = a^2 - 2ab + b^2$

$a^2 = 4x^2$, so $a = 2x$

$b^2 = y^2$, so $b = y$

$-2ab = -2(2x)(y) = -4xy$

Therefore: $4x^2 - 4xy + y^2 = (2x - y)^2$

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For $3y^2 - 6y + 3$:

Factor out 3: $3(y^2 - 2y + 1)$

The expression $y^2 - 2y + 1$ is a perfect square: $(y - 1)^2$

Therefore: $3y^2 - 6y + 3 = 3(y - 1)^2$

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Our equation becomes: $(2x - y)^2 + 3(y - 1)^2 = 0$

Key insight: We have a sum of two non-negative terms (squares are always ≥ 0) that equals zero.

$(2x - y)^2 \geq 0$ for all real $x, y$

$3(y - 1)^2 \geq 0$ for all real $y$

The only way their sum can be zero is if both terms are individually zero.

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From $(2x - y)^2 = 0$: $2x - y = 0$

$\rightarrow y = 2x$

From $3(y - 1)^2 = 0$: $y - 1 = 0$

$\rightarrow y = 1$

Combining these: $y = 1$ and $y = 2x$

Therefore: $1 = 2x$

$\rightarrow x = \frac{1}{2}$

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$4x + 5y = 4 \times \frac{1}{2} + 5 \times 1 = 2 + 5 = 7$

Therefore, $(4x + 5y) = 7$

Key Takeaway: When we see a quadratic equation with mixed terms, we try grouping to form perfect squares. The property that "sum of squares equals zero only when each square is zero" is a powerful tool for finding exact solutions.