9-6a Solving Quadratic Equations Using Square ... - Math Slide Show

Lesson 9-6a
Objective - To solve quadratic equations using the
square root method.
Solve.
x 2 = 9
2
x =
9
x = 3
Solve each equation below.
1) 2)
x 2 = 36 n 2 =169
2
x = 36
x = 6
2
n = 169
n = 13
x = ±6 n =±13
x =±3
Two Solutions
Solve each equation below.
3) y 2 = 7 4) m 2 − 6 = 7
+6 + 6
2
y = 7
m 2 =13
y = 7
y =± 7
2
m = 13
m = 13
m =± 13
Solve each equation below.
5) 3y 2 − 20 = 28 6) 2x 2 −16 = 4
+20 + 20 +16 +16
3y 2 = 48
2x 2 = 20
3 3
x 2 = 10
y
2 = 16
2
y = 16
y = 4
y = ±4
2
x = 10
x = 10
x =± 10
Solve each equation below.
7) 2y 2 + 30 =16 8) 1 = 5x 2 − 2
−30 − 30 +2 + 2
2y 2 =−14 3 = 5x 2
2 2 5 5
y 2 =−77
3 2
5 = x
No Solution
x
=
2 3
x
=
5
3 ⎛ 5⎞
5
⎜ ⎟
⎝ 5⎠
x =±
15
5
Solve each equation below.
9) y+ 1 = 16 10)
( ) 2
( y+ 1) 2
= 16
y+ 1 = 4
y + 1= ±
4
−1 −1
y = −1 ± 4
−1+ 4 −1− 4
3 or − 5
( x − 3) 2
= 25
( x − 3) 2
= 25
x − 3 = 5
x − 3 = ±5
+3 + 3
x = 3±
5
3 + 5 3− 5
8 or − 2
Algebra Slide Show: Teaching Made Easy As Pi, by James Wenk © 2010

Lesson 9-6a (cont.)
Height of a Falling Object
Acceleration Due to Gravity ≈16 ft / sec 2
h =−16t 2 + s
h = height of object (in feet)
t = time (in seconds)
s = initial drop height (in feet)
If a penny is dropped from the top of the
Empire State Building which is 1453 feet
tall, how long will it take to hit the ground?
h = −16t 2 + s
0 = −16t 2 +1453
−1453
−1453
−1453 = −16t 2
−16 −16
90.8125 = t 2
t = 90.8125 ≈ 9.5 sec .
s = 1453 ft.
h = 0 ft.
Word Problems Involving Quadratics
If the square of two consecutive integers are added,
the result is 61. Find the integers.
Let x = 1st integer =−6 x = 5
or
x + 1 = 2nd integer =−5 x+ 1=
6
2 2
x + (x+ 1) = 61
2 2
x + x + 2x+ 1=
61
−61 −61
2
2x + 2x − 60 = 0
2 2
2
x + x− 30=
0
2
x + x− 30=
0
(x + 6)(x − 5) = 0
x =−6
or x = 5
The hypotenuse of a right triangle is 20 inches. One
leg of the right triangle is 4 inches longer than the
other leg. Find the lengths of both legs.
20 Let x = smaller leg = 12 inches
x
x + 4 = larger leg = 16 inches
x + 4
a + b = c
2 2 2
2 2 2
x + (x+ 4) = 20
2 2
x + x + 8x + 16 = 400
− 400 − 400
2
2x + 8x − 384 = 0
2 2
2
4 1
x + 4x − 192 = 0
•
192
2•
96
3•
64
+ 16 − 12 = 4•
48
6•
32
8•
24
12 • 16
(x )(x ) 0
x =−16
or 12
Find the dimensions of a rectangle whose perimeter
is 38 inches and whose area is 84 square inches.
19 - x
P = 38 in.
x
Let x = width = 7 inches
19 - x = length = 12 inches
19 in.
l • w = A
2
0= x − 19x+
84
(19 − x) • x = 84
0 = (x −7 )(x −12)
2
19x − x = 84
2 2
+ x + x
x = 7 or 12
2
19x = x + 84
−19x
−19x
2
0= x − 19x+
84
1•
84
2•
42
3•
28
4•
21
6•
14
7•
12
A picture frame measures 30”x 20”. The picture
2
inside the frame has a area of 416 in . Find the
width of the frame.
if if
x Let x = frame width x = 2 x = 23
416 in 2 30 - 2x = picture length = 26 −16
x 20 - 2x = picture width = 16
20 x x
30
l• w=
A
(30− 2x)(20− 2x) = 416
2
600 − 100x + 4x = 416
− 416 − 416
2
4x − 100x + 184 = 0
4 4
2
x − 25x+ 46=
0
(x −2)(x − 23) = 0
x = 2 or x = 23
frame width = 2 in.
Algebra Slide Show: Teaching Made Easy As Pi, by James Wenk © 2010