Question

(х+1)^2=16 (X-2)^2=3 (y-2)^2=25 (m+2)^2=0

Answer 1

1)2х+2=16
2х=16-2
2х=14
х=7
2)2х-4=3
2х=3+4
2х=7
х=3,5
3)2у-4=25
2у=25+4
2у=29
у=14,5
4)2m+4=0
2m=4
m=2

Answer 2

To solve each of these equations, we'll isolate the variable on one side of the equation and find its value. Let's go through each equation one by one:

1. (x + 1)^2 = 16 First, take the square root of both sides to eliminate the square: √((x + 1)^2) = √(16) x + 1 = ±√(16)

Now, let's solve for x: x + 1 = ±4

Case 1: x = 4 - 1 x = 3

Case 2: x = -4 - 1 x = -5

So the solutions to the equation are x = 3 and x = -5.

2. (X - 2)^2 = 3 Following a similar process, we'll take the square root of both sides: √((X - 2)^2) = √(3) X - 2 = ±√(3)

Now, let's solve for X: X = 2 ± √(3)

So the solutions to the equation are X = 2 + √(3) and X = 2 - √(3).

3. (y - 2)^2 = 25 Again, take the square root of both sides: √((y - 2)^2) = √(25) y - 2 = ±√(25)

Now, let's solve for y: y = 2 ± 5

Case 1: y = 2 + 5 y = 7

Case 2: y = 2 - 5 y = -3

So the solutions to the equation are y = 7 and y = -3.

4. (m + 2)^2 = 0 Here, we have a special case where the square of a real number is zero. The only real number whose square is zero is 0 itself. Therefore, the equation (m + 2)^2 = 0 has only one solution:

m + 2 = 0 m = -2

So the solution to the equation is m = -2.

In summary, the solutions to the given equations are:

1. x = 3 and x = -5
2. X = 2 + √(3) and X = 2 - √(3)
3. y = 7 and y = -3
4. m = -2