Where's y 2 y^2

Algebra Level 4

If real numbers x x and y y satisfy the equation

x 2 + 3 x − y = 12 x^2+3x-y=12 ,

then find the smallest possible value of x + y x+y .


The answer is -16.

Tom Zhou by Tom Zhou , 21, Canada

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2 solutions

Boris Barron
Aug 24, 2014

x 2 + 3 x − y = 12 − − − − > y = x 2 + 3 x − 12 f ( x ) = x + y = x + ( x 2 + 3 x − 12 ) = x 2 + 4 x − 12 d f ( x ) d x = 2 x + 4 = 0 S o i t i s m i n i m u m w h e n x = − 2 S o x + y = − 16 x^{ 2 }+3x-y=12\quad ---->\quad y\quad =\quad x^{ 2 }+3x-12\\ f(x)\quad =\quad x\quad +\quad y\\ \qquad \quad =\quad x\quad +\quad (x^{ 2 }+3x-12)\\ \quad \quad \quad \quad =\quad x^{ 2 }+4x-12\quad \\ \frac { df(x) }{ dx } \quad =\quad 2x\quad +\quad 4\quad =\quad 0\\ So\quad it\quad is\quad minimum\quad when\quad x\quad =\quad -2\\ So\quad x\quad +\quad y\quad =\quad -16

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@Zhou ZeHao Thanks for editing the question. Those who previously answered 16 (answer to original question) have been marked correct.

Calvin Lin Staff - 6 years, 9 months ago

Rewrite it as x 2 + 4 x + 4 − ( x + y ) = 16 x^{2} + 4x + 4 - (x+y) = 16

x + y = ( x + 2 ) 2 − 16 x+y = (x+2)^{2} - 16

Which gives the min of x + y = − 16 x+y = \boxed{-16} at x + 2 = 0 x+2 = 0

x = − 2 , y = − 14 x = -2, y = -14 ~~~

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