Question 68

If $$x^2 + y^2 + 2x + 1 = 0$$, then the value of $$x^{31} + y{35}$$ is

Solution

$$x^2 + y^2 + 2x + 1 = 0$$

$$x^2 + 2x + 1 + y^2 = 0$$ {$$\because (x+1)^2 = x^2 + 2x + 1$$}

$$(x+1)^2 + y^2 = 0$$

$$\therefore (x+1)^2 = 0 , y^2 = 0 $$

x+1 = 0 y = 0

x = -1 y = 0

$$x^{31} + y{35}$$ = $$-1^{31} + 0^{35}$$ = -1

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