This is a quadratic equation in terms of x and involves a square root, so we can solve it by making a substitution. Let y = √(x + 5), then y^2 = x + 5:
This equation can be solved using numerical methods, such as graphing or using a calculator, to find approximate values for y. Once you have found y, substitute back into the equation y = √(x + 5) to solve for x.
To solve this equation, we first combine like terms:
2 + x + 2√(x + 5) = 0
Next, let's isolate the square root term by moving the other terms to the other side of the equation:
x + 2√(x + 5) = -2
Now, square both sides to eliminate the square root:
(x + 2√(x + 5))^2 = (-2)^2
x^2 + 4x√(x + 5) + 4(x + 5) = 4
x^2 + 4x√(x + 5) + 4x + 20 = 4
x^2 + 4x√(x + 5) + 4x + 16 = 0
This is a quadratic equation in terms of x and involves a square root, so we can solve it by making a substitution. Let y = √(x + 5), then y^2 = x + 5:
Substitute y^2 for x + 5 in the equation:
y^2 + 4y√y^2 + 4(y^2 - 5) + 16 = 0
y^2 + 4y^3 + 4y^2 - 20 + 16 = 0
4y^3 + 5y^2 + 16 = 0
This equation can be solved using numerical methods, such as graphing or using a calculator, to find approximate values for y. Once you have found y, substitute back into the equation y = √(x + 5) to solve for x.