The equation from the image is:

$x + 1 - 2\sqrt{x + 4} = 0$

To solve this, we can follow these steps:

Step 1: Isolate the square root term

Move the other terms to the right-hand side of the equation: $2\sqrt{x + 4} = x + 1$

Step 2: Square both sides

Squaring both sides eliminates the square root: $(2\sqrt{x + 4})^2 = (x + 1)^2$ $4(x + 4) = (x + 1)^2$

Step 3: Expand both sides

Expand both the left and right-hand sides: $4(x + 4) = 4x + 16$ On the right-hand side: $(x + 1)^2 = x^2 + 2x + 1$

So, the equation becomes: $4x + 16 = x^2 + 2x + 1$

Step 4: Move all terms to one side

Rearrange the equation to set it equal to 0: $0 = x^2 + 2x + 1 - 4x - 16$ Simplify: $x^2 - 2x - 15 = 0$

Step 5: Solve the quadratic equation

This is a standard quadratic equation. We can factor it: $(x - 5)(x + 3) = 0$

So, the solutions are: $x = 5 \quad \text{or} \quad x = -3$

Step 6: Check for extraneous solutions

Since we squared both sides in the process, we need to check if both solutions are valid by substituting them back into the original equation:

• For $x = 5$: $5 + 1 - 2\sqrt{5 + 4} = 6 - 2\sqrt{9} = 6 - 6 = 0$ This is valid.

• For $x = -3$: $-3 + 1 - 2\sqrt{-3 + 4} = -2 - 2\sqrt{1} = -2 - 2 = -4$ This does not satisfy the original equation.

Thus, the only valid solution is: $x = 5$

Would you like more details or have any questions?

Related Questions:

1. How do you solve an equation with multiple square root terms?
2. How does squaring both sides of an equation affect possible solutions?
3. Can a quadratic equation have more than two solutions?
4. What are the conditions for a solution to be extraneous?
5. How can you solve equations with cube roots?

Tip:

Always check for extraneous solutions when solving equations that involve squaring both sides!