Algebra (Olympiad)mediumFree

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)(x, y) that satisfy (4x2+6x+4)(4y2−12y+25)=28(4x^2 + 6x + 4)(4y^2 - 12y + 25) = 28.
A(x,y)=(−34, 52)(x, y) = \left(-\dfrac{3}{4},\ \dfrac{5}{2}\right).
B(x,y)=(−54, 32)(x, y) = \left(-\dfrac{5}{4},\ \dfrac{3}{2}\right).
C(x,y)=(−−34, −32)(x, y) = \left(-\dfrac{-3}{4},\ \dfrac{-3}{2}\right).
D(x,y)=(−34, 32)(x, y) = \left(-\dfrac{3}{4},\ \dfrac{3}{2}\right).correct
Solution
Step 1: Complete the square in each factor.
4x2+6x+4=(2x+32)2+74,4y2−12y+25=(2y−3)2+16.4x^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,
(2x+32)2+74≥74,(2y−3)2+16≥16.\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
(4x2+6x+4)(4y2−12y+25)≥74×16=28.(4x^2 + 6x + 4)(4y^2 - 12y + 25) \ge \frac{7}{4} \times 16 = 28.
 The given product equals 2828, so equality must hold in both factors. Step 4: Equality requires 2x+32=02x + \dfrac{3}{2} = 0 and 2y−3=02y - 3 = 0, that is x=−34x = -\dfrac{3}{4} and y=32y = \dfrac{3}{2}. Answer: (x,y)=(−34, 32)(x, y) = \left(-\dfrac{3}{4},\ \dfrac{3}{2}\right).
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Algebra (Olympiad) · medium
The all pairs of real numbers (x,y)(x, y) satisfying (2x+1)2+y2+(y−2x)2=13(2x + 1)^2 + y^2 + (y - 2x)^2 = \dfrac{1}{3}.
Algebra (Olympiad) · medium
The number of all real pairs (a,b)(a, b) satisfying a2+b2=25a^2 + b^2 = 25 and 3(a+b)−ab=153(a + b) - ab = 15.
Algebra (Olympiad) · hard
Let {an}\{a_n\} be a sequence of positive integers such that a1=a2=2a_1=a_2=2. For n≥1n\ge1, an+2an−an+12=an+1an.a_{n+2}a_n-a_{n+1}^2=a_{n+1}a_n. Determine the value of a1a2a1032 a7\dfrac{a_1a_2a_{10}}{32\,a_7}.
Algebra (Olympiad) · hard
Consider the polynomial (x2+1)(x2+4)(x2−2x+2)(x2+2x+2).(x^2+1)(x^2+4)(x^2-2x+2)(x^2+2x+2). Let f(x)=(x4+2x2+2x)2+(P(x))2f(x)=\big(x^4+2x^2+2x\big)^2+\big(P(x)\big)^2, where P(x)P(x) is a cubic polynomial. Find the value of P(1) P(2)P(1)\,P(2)

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath