Maximum Solution

Algebra Level 3

If x x and y y are real numbers that satisfy the equation 16 x 2 + 37 x y + 25 y 2 + 24 x + 48 y + 9 = 0 16x^2 + 37xy + 25 y^2 + 24 x + 48 y + 9 = 0 what is the maximum value of x x ?


The answer is 6.

Calvin Lin by Calvin Lin

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1 solution

Ercole Suppa
May 7, 2014

We have 25 y 2 + ( 37 x + 48 ) y + 16 x 2 + 24 x + 9 = 0 25y^2+(37x+48)y+16x^2+24x+9=0 hence Δ = ( 37 x + 48 ) 2 − 100 ( 16 x 2 + 24 x + 9 ) ≥ 0 ⇔ \Delta=(37x+48)^2-100(16x^2+24x+9)\geq 0 \quad \ \Leftrightarrow ( 18 − 3 x ) ( 78 + 77 x ) ≥ 0 ⇔ − 78 77 ≤ x ≤ 6 (18-3x)(78+77x) \geq 0 \quad \ \Leftrightarrow\ \quad -\frac{78}{77}\leq x \leq 6 Therefore the maximum value of x x is 6 \boxed{6}

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