Root operation

Algebra Level 3


The answer is 45.

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1 solution

x = 2017 + 2017 + 2017 + . . . + 2017 x 2 = 2017 + x x 2 x 2017 = 0 x=\sqrt { 2017+\sqrt { 2017+\sqrt { 2017+...+\sqrt { 2017 } } } } \\ { x }^{ 2 }=2017+x\\ { x }^{ 2 }-x-2017=0

Based on b ± b 2 4 a c 2 a \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x 1 , x 2 = ( 1 ) ± ( 1 ) 2 4 1 ( 2017 ) 2 1 x 1 , x 2 = 1 ± 1 + 8068 2 x 1 , x 2 = 1 ± 8067 2 x 1 , x 2 1 ± 89.82 2 x 1 1 + 89.82 2 45.43 x 2 1 89.82 2 44.43 { x }_{ 1 },{ x }_{ 2 }=\frac { -(-1)\pm \sqrt { { (-1) }^{ 2 }-4\cdot 1\cdot (-2017) } }{ 2\cdot 1 } \\ { x }_{ 1 },{ x }_{ 2 }=\frac { 1\pm \sqrt { 1+8068 } }{ 2 } \\ { x }_{ 1 },{ x }_{ 2 }=\frac { 1\pm \sqrt { 8067 } }{ 2 } \\ { x }_{ 1 },{ x }_{ 2 }\approx \frac { 1\pm 89.82 }{ 2 } \\ { x }_{ 1 }\approx \frac { 1+89.82 }{ 2 } \Rightarrow 45.43\\ { x }_{ 2 }\approx \frac { 1-89.82 }{ 2 } \Rightarrow -44.43

Since the result of the square root is always positive, so the value of x x that satisfies is 45.43 45.43 . So, x = 45 \left\lfloor x \right\rfloor =45

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