Question : If $(2 x+3 y+4)(2 x+3 y-5)$ is equivalent to $\left(a x^2+b y^2+2 h x y+2 g x+2 f y+c\right)$, then what is the value of $\frac{g + f - c}{abh}$?
Option 1: $\frac{37}{216}$
Option 2: $\frac{35}{432}$
Option 3: $\frac{19}{108}$
Option 4: $\frac{19}{216}$
Correct Answer: $\frac{35}{432}$
Solution :
According to the question
$(2x + 3y + 4)(2x + 3y - 5) = (ax^2 + by^2 + 2hxy + 2gx + 2fy + c)$
⇒ $4x^2 + 6xy - 10x + 6xy + 9y^2 - 15y + 8x + 12y - 20 = (ax^2 + by^2 + 2hxy + 2gx + 2fy + c)$
⇒ $4x^2 + 12xy - 2x + 9y^2 - 3y - 20 = (ax^2 + by^2 + 2hxy + 2gx + 2fy + c)$
On comparing,
⇒ $a = 4, h = 6, g = -1, b = 9, f =\frac{-3}{2}$ and $c = -20$
Now,
$\frac{g + f - c}{abh}$
$=\frac{-1 - \frac{3}{2} + 20}{4 × 9 × 6}$
$=\frac{\frac{-2 - 3 + 40}{2}}{4 × 6 × 9}$
$=\frac{\frac{35}{2}}{216}$
$=\frac{35}{432}$
Hence, the correct answer is $\frac{35}{432}$.
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