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Algebra Level 4

{ ( x + 2 ) ( y + 2 ) = 3 ( x 2 + y 2 + x y ) ( x + y ) 3 = 4 ( x 3 + y 3 ) \begin{aligned} \begin{cases} (x+2)(y+2) = 3(x^{2}+y^{2}+\sqrt{xy}) \\ (\sqrt{x}+\sqrt{y})^{3} = 4(x^{3}+ y^{3}) \end{cases} \end{aligned}

Given that positive reals x , y x,y satisfy both equations. Determine the value of x + y \sqrt{x}+\sqrt{y} .


The answer is 2.

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Leah Jurgens by Leah Jurgens , 25, Vietnam

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1 solution

Leah Jurgens
Mar 30, 2016

{ ( x + 2 ) ( y + 2 ) = 3 ( x 2 + y 2 + x y ) ( i ) ( x + y ) 3 = 4 ( x 3 + y 3 ) ( i i ) \begin{aligned} \begin{cases} (x+2)(y+2) = 3(x^{2}+y^{2}+\sqrt{xy}) (i) \\ (\sqrt{x}+\sqrt{y})^{3} = 4(x^{3}+ y^{3}) (ii) \end{cases} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I again try to solve a difficult problem but on a lower scale, here's my simple but incorrect solution, feel free to correct me below: (x+2)(y+2) = 3(x²+y²+√xy) (√x+√y)³ = 4(x³+y³)

Let's say the min value of x and y is 2.

  1. (2+2)(2+2)=3(2²+2²+√2×2)= 16=(4+4+2) = 16+10=26

2.(1.4+1.4)³ = 4(16+16) = 2.7=4(32)

Add: 26+131=157 √157 ≈ 13

Garrett O’Brien - 5 years, 2 months ago

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