The problem asks us to find the solution set for the given quadratic equations by factoring. The equations are: a) $x^2 + 9x + 14 = 0$ b) $x^2 - 4x = 12$ c) $x^2 = 49$ d) $x^2 - 81 = 0$

1. Problem Description

The problem asks us to find the solution set for the given quadratic equations by factoring. The equations are:

a) $x^2 + 9x + 14 = 0$

b) $x^2 - 4x = 12$

c) $x^2 = 49$

d) $x^2 - 81 = 0$

2. Solution Steps

a) $x^2 + 9x + 14 = 0$

We need to find two numbers that multiply to 14 and add up to

9. These numbers are 2 and

7. So, we can factor the quadratic equation as:

$(x + 2)(x + 7) = 0$

Setting each factor to zero, we get:

$x + 2 = 0 \Rightarrow x = -2$

$x + 7 = 0 \Rightarrow x = -7$

Thus, the solution set is {-2, -7}.

b) $x^2 - 4x = 12$

First, rewrite the equation as:

$x^2 - 4x - 12 = 0$

We need to find two numbers that multiply to -12 and add up to -

4. These numbers are -6 and

2. So, we can factor the quadratic equation as:

$(x - 6)(x + 2) = 0$

Setting each factor to zero, we get:

$x - 6 = 0 \Rightarrow x = 6$

$x + 2 = 0 \Rightarrow x = -2$

Thus, the solution set is {6, -2}.

c) $x^2 = 49$

Rewrite the equation as:

$x^2 - 49 = 0$

This is a difference of squares, so we can factor it as:

$(x - 7)(x + 7) = 0$

Setting each factor to zero, we get:

$x - 7 = 0 \Rightarrow x = 7$

$x + 7 = 0 \Rightarrow x = -7$

Thus, the solution set is {7, -7}.

d) $x^2 - 81 = 0$

This is a difference of squares, so we can factor it as:

$(x - 9)(x + 9) = 0$

Setting each factor to zero, we get:

$x - 9 = 0 \Rightarrow x = 9$

$x + 9 = 0 \Rightarrow x = -9$

Thus, the solution set is {9, -9}.

3. Final Answer

a) {-2, -7}

b) {6, -2}

c) {7, -7}

d) {9, -9}