COULD U SOLVE THIS ONE??

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16^(x^2+y)+16^(y^2+x)=1 FIND ALL REAL SOLUTIONS FOR X AND Y.


The answer is -0.5.

Snehil Sinha by SNEHIL SINHA , 21, India

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1 solution

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Feb 26, 2014

By Inequality Cauchy: 1 = 1 6 x 2 + y + 1 6 y 2 + x > = 2 × 1 6 x 2 + y × 1 6 y 2 + x = 2 × 4 x 2 + y + y 2 + x 1= 16 ^ {x ^ {2}+y} + 16 ^ {y ^ {2}+x} >= 2 \times \sqrt{16 ^ {x ^ {2}+y} \times 16 ^ {y ^ {2}+x}} = 2 \times 4 ^ {x ^ {2}+y+y ^ {2}+x} = > 4 x 2 + y + y 2 + x < = 1 2 => 4 ^ {x ^ {2}+y+y ^ {2}+x} <= \frac{1}{2}
= > x 2 + y + y 2 + x < = − 1 2 = > ( x + 1 2 ) 2 + ( y + 1 2 ) 2 < = 0 = > x = y = − 1 2 => x ^ {2}+y+y ^ {2}+x <= \frac{-1}{2} => (x+\frac{1}{2}) ^ {2} + (y+\frac{1}{2}) ^ {2} <=0 => x=y=\frac{-1}{2}

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