Download Solving Equations with Substitution Examples

1.5 – Applications of Quadratic Equations
Finding Numbers
Find two consecutive positive even numbers whose product is 168.
1𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟: 𝑥
2𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟: 𝑥 + 2
𝑥 = −14, 12
𝑥 𝑥 + 2 = 168
𝑥 2 + 2𝑥 = 168
𝑥 2 + 2𝑥 − 168 = 0
𝑥=
− 2 ±
2
2
− 4 ∙ 1 ∙ −168
2∙1
−2 ± 4 + 672
−2 ± 676
𝑥=
=
2
2
−2 ± 26
−28 24
𝑥=
𝑥=
,
2
2 2
1𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟: 12
2𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟: 14
1.5 – Applications of Quadratic Equations
Area and Volume
A rectangular field is to be enclosed by fence and then divided into two
by another fence parallel to one of the sides. The area of the field must be
2400 square feet. If there are 240 feet of fencing available, find the
values of x and y.
𝑥 − 30 = 0
x
x
2400 = 2𝑥𝑦
𝑥 = 30 𝑓𝑡.
y
y
y
240 − 4𝑥
2400 = 2𝑥
3
x
x
240 − 4𝑥
𝑦=
7200
=
2𝑥
240
−
4𝑥
𝑦
3
𝐴 = 2𝑥 ∙
2400 = 2𝑥𝑦
7200 = 480𝑥 − 8𝑥 2
4𝑥 +3𝑦 = 240
3𝑦 = 240 − 4𝑥
240 − 4𝑥
𝑦=
3
8𝑥 2 − 480𝑥 + 7200 = 0
240 − 4 ∙ 30
𝑦=
3
8 𝑥 2 − 60𝑥 + 900 = 0
𝑦 = 40 𝑓𝑡.
8 𝑥 − 30 𝑥 − 30 = 0
1.5 – Applications of Quadratic Equations
Vertical Motion
A rocket is launched from a hill 80 feet above a lake. The rocket will fall
into lake after exploding at its maximum height. The rocket’s height at
any time (t in seconds) above the surface of the lake is given by: ℎ =
− 16𝑡 2 + 64𝑡 + 80. (a) What is the height of the rocket after 1.5
second? (b) How long will it take for the rocket to hit 128 feet? (c)
After how many seconds after it is launched will the rocket hit the lake?
𝑡 = 1 𝑢𝑝 𝑎𝑛𝑑 3 (𝑑𝑜𝑤𝑛) 𝑠𝑒𝑐.
(a) ℎ = −16𝑡 2 + 64𝑡 + 80
ℎ = −16 1.5
2
+ 64 1.5 + 80
ℎ = 140 𝑓𝑡.
(b) 128 = −16𝑡 2 + 64𝑡 + 80
16𝑡 2 − 64𝑡 + 48 = 0
16 𝑡 2 − 4𝑡 + 3 = 0
16 𝑡 − 3 𝑡 − 1 = 0
(c) 0 = −16𝑡 2 + 64𝑡 + 80
0 = −16 𝑡 2 − 4𝑡 − 5
−16 𝑡 + 1 𝑡 − 5 = 0
𝑡 = −1 𝑎𝑛𝑑 5 𝑠𝑒𝑐.
𝑡 = 5 𝑠𝑒𝑐.
1.6 – More Equations and Applications
Solving Equations by Grouping
Examples:
6𝑥 2 + 3𝑥 + 20𝑥 + 10 = 0
5𝑣 3 − 2𝑣 2 + 25𝑣 − 10 = 0
3𝑥 2𝑥 + 1 +10 2𝑥 + 1 = 0
𝑣 2 5𝑣 − 2 +5 5𝑣 − 2 = 0
2𝑥 + 1 3𝑥 + 10 = 0
2𝑥 + 1 = 0
1
𝑥=−
2
3𝑥 + 10 = 0
10
𝑥=−
3
5𝑣 − 2 𝑣 2 + 5 = 0
5𝑣 − 2 = 0
2
𝑣=
5
𝑣2 + 5 = 0
𝑣 2 = −5
𝑣 2 = ± −5
𝑣 = ±𝑖 5
1.6 – More Equations and Applications
Solving Rational Equations
Examples:
5𝑥
11𝑥 + 1
12
= 2
−
𝑥 + 2 𝑥 + 7𝑥 + 10 𝑥 + 5
5𝑥
11𝑥 + 1
12
=
−
𝑥+2
𝑥+2 𝑥+5
𝑥+5
5𝑥 2 + 26𝑥 + 23 = 0
𝑥=
LCD: 𝑥 + 2 𝑥 + 5
𝑥+2 𝑥+5
5𝑥
11𝑥 + 1
12
=
−
𝑥+2
𝑥+2 𝑥+5
𝑥+5
5𝑥 𝑥 + 5 = 11𝑥 + 1 − 12 𝑥 + 2
5𝑥 2 + 25𝑥 = 11𝑥 + 1 − 12𝑥 − 24
5𝑥 2 + 25𝑥 = −𝑥 − 23
−26 ±
26 2 − 4 5 23
2 5
−26 ± 216
𝑥=
10
−26 ± 36 ∙ 6
𝑥=
10
−13 ± 3 6
𝑥=
5
1.6 – More Equations and Applications
Solving Radical Equations
Examples:
𝑥+3−1=7
𝑥+3=8
𝑥+3
2
= 8
𝑥 + 3 = 64
𝑥 = 61
2
5x  1  x  2  3
16 x 2  16 x  4  4 x
5x  1  x  1
16 x 2  20 x  4  0
 5 x  1  
2

x 1
2
5x  1  x  x  x  1
5x  1  x  2 x  1
4x  2  2 x
4 x  2
2
 
 2 x
2
44 x 2  5 x  1  0
4x  14x  1  0
x  1 0 4x  1  0
1
x
x 1
4
1.6 – More Equations and Applications
Solving Equations with Fractional Exponents
Examples:
𝑝
𝑝
3
3
2
2
2
3
= 11
𝑝 = 11
𝑝=
𝑝=
𝑝
= 11
3
3
2
3
112
121
2
3
𝑝
4
4
3
3
3
4
= 11
= 11
𝑝 = 11
3
3
4
4
𝑝 = ± 113
4
𝑝 = ± 1331
4
1.6 – More Equations and Applications
Solving Equations with Substitution
Examples:
𝑥 4 − 7𝑥 2 + 12 = 0
𝑙𝑒𝑡 𝑢 = 𝑥 2
𝑡ℎ𝑒𝑛 𝑢2 = 𝑥 2
2
𝑥 2 = 3, 4
= 𝑥4
Substitute
𝑢2 − 7𝑢 + 12 = 0
𝑢−3 𝑢−4 =0
𝑢−3=0 𝑢−4=0
𝑢 = 3, 4
𝑢 = 𝑥2
𝑥2 = 3
𝑥2 = 4
𝑥=± 3
𝑥=± 4
𝑥 = ±2
1.6 – More Equations and Applications
Solving Equations with Substitution
Examples:
𝑥
2
3
− 2𝑥
1
3
𝑙𝑒𝑡 𝑢 = 𝑥
𝑡ℎ𝑒𝑛
𝑢2
= 𝑥
1
𝑢=𝑥
− 15 = 0
1
𝑥
3
2
3
=𝑥
2
Substitute
𝑢2 − 2𝑢 − 15 = 0
𝑢+3 𝑢−5 =0
𝑢+3=0 𝑢−5=0
𝑢 = −3, 5
𝑥
3
𝑥
1
1
3
3
3
1
3
1
3
= −3, 5
= −3
= −3
𝑥 = −27
𝑥
3
𝑥
1
1
3
3
3
=5
= 5
𝑥 = 125
3
1.6 – More Equations and Applications
Solving Equations with Substitution
Examples:
𝑢 = 𝑥 −3 = 1, 8
𝑥 −6 − 9𝑥 −3 + 8 = 0
𝑙𝑒𝑡 𝑢 = 𝑥 −3
2
𝑡ℎ𝑒𝑛 𝑢 = 𝑥
−3 2
=𝑥
−6
Substitute
𝑢2 − 9𝑢 + 8 = 0
𝑢−1 𝑢−8 =0
𝑢−1=0 𝑢−8=0
𝑢 = 1, 8
𝑥 −3 = 8
𝑥 −3 = 1
−1
−3
3
𝑥
= 1
𝑥=
1
1
1
−1
3
−1
−3
3
𝑥
𝑥=
3
𝑥=1
= 8
1
8
1
1
𝑥=
2
3
−1
3