x² - 60 = 4

x² = 64

x = ±√(64)

x = ±8

x = 8, x = -8

What method of solving is used?

2x² + 3x - 18 = 52

Can we solve by completing the square?

No!

Because a = 2

We can only complete the square when a = 1

ex: x² + 3x - 18 = 52

x² + 12x - 50 = 0

x² + 12x = 50

If we want to solve by completing the square, what do we need to add to both sides?

(b/2)² = (12/2)² = 6² = 36

and the rest...

x² + 12x + 36 = 50 + 36

(x + 6)² = 85

√((x + 6)²) =±√(85)

x + 6 = ±√(85)

x = ±√(85) - 6

x = √(85) - 6, x = -√(85) - 6

You are solving this problem by completing the square

x² + 12x = 50

x² + 12x + 36 = 50 + 36

Now what?

(x + 6)² = 85

and then...

√((x + 6)²) =±√(85)

x + 6 = ±√(85)

x = ±√(85) - 6

x = √(85) - 6, x = -√(85) - 6

x² + 6x = 12

If we want to solve using the quadratic formula, what is 'c'?

x² + 6x = 12

x² + 6x - 12 = 0

x² + 6x + (-12) = 0

c =-12

We drop a bowling ball from 10,000 feet. Recall we can find the distance something has fallen with d = 16t²

Create a function h(t) that expresses the bowling balls height as a function of time

h(t) = h₀ - 16t²

h(t) = -16t² + h₀

h(t) = -16t² + 10,000

the height of a ball dropped of a 16' high building as a function of time is given by the quadratic equation

h(t) = -16t² + 16

How long does it take for the ball to hit the ground?

When the ball hits the ground the height, h(t), equals 0

aka

0 = h(t) = -16t² + 16

0 = -16t² + 16

16t² = 16

t² = 1

t = ±1

We have solutions t = 1 and t = -1. We know the ball didn't hit the ground a second before we dropped it, so we discard t = -1, and accept t = 1 as our answer.