Question

X^4=y^4 и (y+1)^2=16. Чему равно минимальное возможное значение (x+1)^2 ?

Answer 1

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Answer 2

To find the minimum possible value of (x+1)^2, we first need to solve the given system of equations:

1. x^4 = y^4
2. (y+1)^2 = 16

Let's solve the second equation for y:

(y+1)^2 = 16

Taking the square root of both sides:

y + 1 = ±√16

y + 1 = ±4

Now we have two cases to consider:

Case 1: y + 1 = 4 If y + 1 = 4, then y = 3.

Case 2: y + 1 = -4 If y + 1 = -4, then y = -5.

Now, let's find the corresponding values of x for each case:

Case 1: y = 3 From the first equation, x^4 = y^4 = 3^4 = 81. Taking the fourth root of both sides:

x = ±√81 x = ±9

Case 2: y = -5 From the first equation, x^4 = y^4 = (-5)^4 = 625. Taking the fourth root of both sides:

x = ±√625 x = ±25

Now, we need to find the minimum value of (x+1)^2 for both cases:

Case 1: (x+1)^2 = (9+1)^2 = 10^2 = 100 Case 2: (x+1)^2 = (25+1)^2 = 26^2 = 676

The minimum value among these two cases is 100.

Therefore, the minimum possible value of (x+1)^2 is 100.