## Precalculus: Mathematics for Calculus, 7th Edition

$x^{2}-4x-5=0$ Part A: By Factoring 1) Factor the equation: $(x-5)(x+1) = 0$ 2) Using the Zero-Product Property we get: $x-5 = 0$ or $x+1 = 0$ 3) Solve the two simpler equations: $x = 5$ or $x = -1$ Part B: By Completing the Squares 1) To make it a perfect square we need to first move the (-5) to the other side by adding 5 to each side. $x^{2}-4x = 5$ 2) Then we make it square by adding 4 to each side. $x^{2}-4x+4 = 9$ 3) We can now factor that giving us: $(x-2)^{2} = 9$ 4) Get the square root of each side: $x-2 = \frac{+}{-}\sqrt 9$ $x-2 = 3$ or $x-2 = -3$ 5) Solve the simple equations: $x = 5$ or $x = -1$ Part C: Using the Quadratic Formula $x=\frac{-b\frac{+}{-}\sqrt (b^{2}-4ac)}{2a}$ 1) Plug the equation into the formula: $x=\frac{-(-4)\frac{+}{-}\sqrt ((-4)^{2}-4(1)(-5))}{2(1)}$ 2) Solve for x: $x=\frac{4\frac{+}{-}\sqrt (16+20)}{2}$ $x=\frac{4\frac{+}{-}\sqrt (36)}{2}$ $x=\frac{4\frac{+}{-}6}{2}$ $x=\frac{10}{2}$ or $x=\frac{-2}{2}$ $x=5$ or $x=-1$