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Mr. Simonds MTH 65
Page 1 of 2
- a. (^8 448) 4 11
6 6 8 4 11 6 8 2 11 6 4 11 3
− ± −^ ±^ ⋅
=−^ ±
=−^ +
=−^ ±
Scratch work – divide 2 out of each term.
4
1 11 6 3
b. (^10 2710) 9 3 2 2 10 9 3 2 10 3 3 2
± ±^ ⋅
c. (^12 100 12 ) 4 4
(^12 10 1) and 12 10 11 4 2 4 2
2. a. ( 5 2 )^288
w w w w
w
The solutions are 2 22 5
c. ( 6 10 )^272
6 10 72 6 10 6 2 6 10 6 2 10 6 2 6 5 3 2 3
t t t t t t
=−^ ±
=−^ ±
The solutions are 5 3 2 3
b. ( )
2 2
x x
The equation has no real number solutions.
5 ± 6
3 2 6 3
Mr. Simonds MTH 65
Page 2 of 2
Problem 3 Let x^ represent the length (ft) of each side of the original pen.^ Then the lengths of the sides of the new pen are shown in the diagram below.
Problem 4 Let h represent the height of the TV screen (in) and w represent the width of the screen (in).
From the aspect ratio we have 16 16 9 9
w (^) w h h
= ⇒ =. From the Pythagorean Theorem we have:
A =774.4 ft^2 x
x + 0.6 x =1.6 x
2 2
x x x x x x
Since the length cannot be negative, each side of the pen was originally 22 ft.
h^ 40 in
h
2 2 2
2 2
2 2
2
2
h h
h h
h h
h
h
h
h
+ ⎛^ ⎞ =
= ⎛^ ⎞
= ± ⎛^ ⎞
Since the height cannot be negative, the height must be about 19.
inches. This makes the width about 16 ( 19.6 in ) 34.8 in
