Let's go step by step to solve each of these equations by taking the square roots, as instructed.

Problem 19:

$(x + 5)^2 - 6 = 43$

Step 1: Isolate the squared term.

First, add 6 to both sides of the equation: $(x + 5)^2 = 43 + 6$ $(x + 5)^2 = 49$

Step 2: Take the square root of both sides.

$x + 5 = \pm \sqrt{49}$ $x + 5 = \pm 7$

Step 3: Solve for $x$.

There are two cases to consider:

1. $x + 5 = 7$ $x = 7 - 5$ $x = 2$

2. $x + 5 = -7$ $x = -7 - 5$ $x = -12$

Solution for Problem 19: $x = 2 \quad \text{or} \quad x = -12$

Problem 20:

$(x - 1)^2 - 19 = 81$

Step 1: Isolate the squared term.

First, add 19 to both sides of the equation: $(x - 1)^2 = 81 + 19$ $(x - 1)^2 = 100$

Step 2: Take the square root of both sides.

$x - 1 = \pm \sqrt{100}$ $x - 1 = \pm 10$

Step 3: Solve for $x$.

There are two cases to consider:

1. $x - 1 = 10$ $x = 10 + 1$ $x = 11$

2. $x - 1 = -10$ $x = -10 + 1$ $x = -9$

Solution for Problem 20: $x = 11 \quad \text{or} \quad x = -9$

Would you like more details on any of the steps?

Additional Questions:

1. How do we handle square root equations when the radicand is negative?
2. What if the equation included a different exponent, say $x^3$? How would solving change?
3. How can you check if the solutions obtained are correct?
4. Why do we consider both positive and negative square roots?
5. Can you apply the same method for inequalities involving squares?

Tip:

Always remember to check your solutions by substituting them back into the original equation!