Chapter 11 QuaDraTiC appliCaTionS 399- P = 30 + 2x Q = 40 − 5x R = (30 + 2x)(40 − 5x)
1120 = (30 + 2x)(40 − 5x)
1120 = 1200 − 70x − 10x^2
10 x^2 + 70x − 80 = 0
1
10
10 70 80 1
10( xx^2 + – )( = 0 )x^2 + 7x − 8 = 0
(x − 1)(x + 8) = 0
x
x
– =
= 10
1x
xx
( )+ =
= –=8isnotasolution80
8 −The price should be 30 + 2(1) = $32. There would be 40 − 5(1) = 35 oil
changes performed each day.
- P = 1.50 − 0.05x Q = 250 + 10x
R = (1.50 − 0.05x)(250 + 10x)
378 = (1.50 − 0.05x)(250 + 10x)
378 = 375 + 2.5x − 0.5x^2
0.5x^2 − 2.5x + 3 = 0
2(0.5x^2 − 2.5x + 3) = 2(0)
x^2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
x
x
– =
= 20
2x
x– =
= 30
3
If x = 2, the price should be 1.50 − 0.05(2) = $1.40. If x = 3, the price
should be 1.50 − 0.05(3) = $1.35.
- P = 40 + 2x Q = 80 − 1x R = (40 + 2x)(80 − x)
3648 = (40 + 2x)(80 − x)
3648 = 3200 + 120x − 2x^2
2 x^2 − 120x + 448 = 0
1
2
2 120 448 1
2( xx^2 – + )( = 0 )x^2 − 60x + 224 = 0