To solve this equation, we can start by simplifying each square root separately:
Simplify √(3x+9-4√(3x+5)): Let y = √(3x+5) Therefore, y^2 = 3x + 5 Now, the given equation becomes: √(y^2 + 4) = 1 y^2 + 4 = 1 y^2 = -3, which is not possible since the square root of a number cannot be negative. Therefore, there are no solutions for this part of the equation.
Next, we simplify √(3x+14-6√(3x+5)): Let z = √(3x+5) Therefore, z^2 = 3x + 5 Now, the given equation becomes: √(2z+9) = 1 2z + 9 = 1 2z = -8 z = -4 Since the square root of a number cannot be negative, there are no solutions for this part of the equation either.
Therefore, the original equation has no solutions.
To solve this equation, we can start by simplifying each square root separately:
Simplify √(3x+9-4√(3x+5)):
Let y = √(3x+5)
Therefore, y^2 = 3x + 5
Now, the given equation becomes:
√(y^2 + 4) = 1
y^2 + 4 = 1
y^2 = -3, which is not possible since the square root of a number cannot be negative. Therefore, there are no solutions for this part of the equation.
Next, we simplify √(3x+14-6√(3x+5)):
Let z = √(3x+5)
Therefore, z^2 = 3x + 5
Now, the given equation becomes:
√(2z+9) = 1
2z + 9 = 1
2z = -8
z = -4
Since the square root of a number cannot be negative, there are no solutions for this part of the equation either.
Therefore, the original equation has no solutions.