System of equations

Algebra Level 4

{ y = x + 1 + x 1 4 y 4 + 2 0 = x 2 + 2 x ( y 1 ) + y 2 6 y + 1 \begin{cases} {y = \sqrt { x+1 } +\sqrt [ 4 ]{ x-1 } -\sqrt { { y }^{ 4 }+2 } } \\ {0=x^2+2x(y-1)+y^2-6y+1} \end{cases}

Let ( x 1 , y 1 ) , ( x 2 , y 2 ) , , ( x n , y n ) (x_1,y_1), (x_2,y_2), \ldots , (x_n,y_n) be the solution set of the above system of equations for real x , y x,y

Evaluate n + i = 1 n ( x i + y i ) \displaystyle n+ \sum _{ i=1 }^{ n }{ (x_i+y_i) }


The answer is 6.0.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Nguyen Thanh Long
Feb 18, 2015

According to second equation, having: x = 2 y ( y 1 ) x=2\sqrt{y}-(y-1) Replacing to first equation: y + 2 y + 2 + ( y + 2 y ) 1 4 = y 4 + 2 + y \sqrt{-y+2\sqrt{y}+2}+(-y+2\sqrt{y})^{\frac{1}{4}}=\sqrt{y^4+2}+y 0 y 1 \Rightarrow 0 \le y \le 1 It is easy to get two roots of (x, y): (1, 0), (2, 1) n + Σ i = 1 n ( x i + y i ) = 6 n+\Sigma^{n}_{i=1} (x_{i} + y_{i})=\boxed{6}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...