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4.8 Polynomial Word Problems
a) Define the variable,
b) Write the equation, and
c) Solve the problem.
1)
The sum of a number and its square is 42.
Find the number.
x2
a) Let x = a number
b) x + x2 = 42
x2 + x – 42 = 0
(x + 7)(x – 6) =0
x+7=0 x–6=0
x = –7
x=6
Check!
(–7)2 + 7 = 49 + 7 = 42
(6)2 + 6 = 36 + 6 = 42
c) –7 or 6
2) The length of a rectangle is 11 cm more than its
width. The area of the rectangle is 2040 cm2. Find the
dimensions of the rectangle.
a) Let L = length & W = width
L =( W + 11)
Area of rectangle = L • W
L • W = 2040
Dimensions can’t be (–)
b) (W + 11) • W = 2040
Plug in to find L:
2
W + 11W – 2040 = 0
L = 40 + 11
(W + 51)(W – 40) = 0
= 51
W + 51 = 0 W – 40 = 0
40 cm x 51 cm
W = –51
W = 40
3) The sum of two numbers is 28. The sum of
the squares of the two numbers is 490. Find
the numbers.
a) Let x = one number & y = other number
x + y = 28  y =(28 – x )
x2 + y2 = 490
b) x2 + (28 – x)2 = 490
x2 + 784 – 56x + x2 = 490
2x2 – 56x + 294 = 0
x = 7 or x = 21
x2 – 28x + 147 = 0
Check!
The numbers are
(x – 7)(x – 21) =0
c) 7 & 21
x – 7 = 0 x – 21 = 0
4) The sum of the squares of two consecutive
integers is 313. Find the integers.
a) Let x = 1st integer & x + 1 = 2nd integer
b) x2 + (x + 1)2 = 313
x2 + x2 + 2x + 1 = 313
2x2 + 2x – 312 = 0
When x = –13
2
x + x – 156 = 0
x + 1 = –12
(x + 13)(x – 12) =0
When x = 12
x + 13 = 0 x – 12 = 0
x + 1 = 13
x = –13
x = 12 The numbers are
c) –13 & –12 or 12 & 13
Homework
#8 Polynomial Word Problems WS
Remember Consecutive Odd Integers:
x, x + 2, x + 4, …