Chapter 3 - Parallel and Perpendicular Lines - 3-8 Slopes of Parallel and Perpendicular Lines - Practice and Problem-Solving Exercises - Page 204: 51

$x = 7$

If $a \parallel b$, and the lines are cut by a transversal, then alternate exterior angles are congruent. The measures of the alternate exterior angles are given. We can now set them equal to one another: $x^2 - 14 = 5x$ If we subtract $5x$ from both sides, we can set the equation equal to zero, and then we will have a quadratic equation, which is given by the formula $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real numbers: $x^2 - 5x - 14 = 0$ We can now factor this equation. We need to find the factors of $ac$ or $-6$ that, when added together, will equal $b$ or $-1$. We need one negative factor and one positive factor because a positive times a negative will equal a negative; when they are added together, either a negative or a positive number can result, depending on the absolute value of the negative and positive number. In this case, we need for the negative number to have the greater absolute value. We found the following possibilities: $(a)(c) = (-14)(1)$ $b = -13$ $(a)(c) = (-7)(2)$ $b = -5$ The second option works for us. Now we can factor the quadratic equation: $(x - 7)(x + 2) = 0$ According to the zero product property, if the product of two factors is $0$, then each factor is equal to $0$ or both factors equal $0$. Let us now set each factor equal to zero: First factor: $x - 7 = 0$ Add $7$ to each side to solve for $x$: $x = 7$ Second factor: $x + 2 = 0$ Subtract $2$ from each side to solve for $x$: $x = -2$ We cannot use the second solution because if we have the measure of an angle equal to $5x$, and $x = -2$, then that angle would measure $-10^{\circ}$, and we cannot have negative angle measures.