1) 25x^2 = 1 First, divide both sides by 25: x^2 = 1/25 Taking the square root of both sides: x = ±√(1/25) x = ±1/5
So the solutions are x = 1/5 and x = -1/5
2) 2x^2 - 4 = 0 Add 4 to both sides: 2x^2 = 4 Divide both sides by 2: x^2 = 2 Taking the square root of both sides: x = ±√2
So the solutions are x = √2 and x = -√2
3) 2x^2 + 6 = 0 Subtract 6 from both sides: 2x^2 = -6 Divide both sides by 2: x^2 = -3 Since the square of a real number is always non-negative, there are no real solutions to this equation.
To solve the equations:
1) 25x^2 = 1
First, divide both sides by 25:
x^2 = 1/25
Taking the square root of both sides:
x = ±√(1/25)
x = ±1/5
So the solutions are x = 1/5 and x = -1/5
2) 2x^2 - 4 = 0
Add 4 to both sides:
2x^2 = 4
Divide both sides by 2:
x^2 = 2
Taking the square root of both sides:
x = ±√2
So the solutions are x = √2 and x = -√2
3) 2x^2 + 6 = 0
Subtract 6 from both sides:
2x^2 = -6
Divide both sides by 2:
x^2 = -3
Since the square of a real number is always non-negative, there are no real solutions to this equation.