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Unique (x,y) with (4x²+6x+4)(4y²-12y+25)=28 | IOQM

JEE Maths question with a full step-by-step solution.

Question

Determine the unique pair of real numbers

(x, y)

that satisfy

(4x^2 + 6x + 4)(4y^2 - 12y + 25) = 28

(x, y) = \left(-\dfrac{3}{4},\ \dfrac{5}{2}\right)

(x, y) = \left(-\dfrac{5}{4},\ \dfrac{3}{2}\right)

(x, y) = \left(-\dfrac{-3}{4},\ \dfrac{-3}{2}\right)

(x, y) = \left(-\dfrac{3}{4},\ \dfrac{3}{2}\right)

.correct

Solution

Step 1: Complete the square in each factor.

4x^2 + 6x + 4 = \left(2x + \frac{3}{2}\right)^2 + \frac{7}{4}, \qquad 4y^2 - 12y + 25 = (2y - 3)^2 + 16.

Step 2: Bound each factor. Since squares are non-negative,

\left(2x + \frac{3}{2}\right)^2 + \frac{7}{4} \ge \frac{7}{4}, \qquad (2y - 3)^2 + 16 \ge 16.

Step 3: Therefore the product satisfies

(4x^2 + 6x + 4)(4y^2 - 12y + 25) \ge \frac{7}{4} \times 16 = 28.

The given product equals

28

, so equality must hold in both factors. Step 4: Equality requires

2x + \dfrac{3}{2} = 0

and

2y - 3 = 0

, that is

x = -\dfrac{3}{4}

and

y = \dfrac{3}{2}

. Answer:

(x, y) = \left(-\dfrac{3}{4},\ \dfrac{3}{2}\right)

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