1

𝑥 (𝑥+2 )¿168

Finding Numbers

𝑥2+2𝑥=168

Find two consecutive positive even numbers whose product is 168.

2𝑛𝑑𝑛𝑢𝑚𝑏𝑒𝑟 :𝑥+2

𝑥2+2𝑥−168=0

𝑥=− (2 )±√ (2 )2−4 ∙1 ∙ (−168 )

2 ∙1

𝑥=−2±√4+6722

¿ −2±√6762

𝑥=−2±26

2𝑥=

−282,242

𝑥=−14 ,12

12𝑛𝑑𝑛𝑢𝑚𝑏𝑒𝑟 : 14

1.5 – Applications of Quadratic Equations

1.5 – Applications of Quadratic EquationsArea and Volume

𝐴=¿2 𝑥 ∙𝑦xx

x x

yyy

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.

4 𝑥+3 𝑦¿2402400=¿2 𝑥𝑦

3 𝑦=240−4 𝑥𝑦=

240−4 𝑥  3

2400=2 𝑥𝑦

2400=2 𝑥( 240−4 𝑥  3 )

7200=2 𝑥 (240−4 𝑥 )7200=480𝑥−8 𝑥2

8 𝑥2−480 𝑥+7200=08 (𝑥2−6 0 𝑥+900 )=0

8 (𝑥−30 ) (𝑥−30 )=0

𝑥−30=0𝑥=30𝑓𝑡 .

𝑦=240−4 𝑥  

3

𝑦=240−4 ∙30  

3

𝑦=40𝑓𝑡 .

Vertical Motion

(a)

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: . (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?

h=−16 (1.5 )2+64 (1.5 )+80

h=140 𝑓𝑡 .(b)

16 𝑡 2−64 𝑡+48=016 (𝑡 2−4 𝑡+3 )=016 (𝑡−3 ) (𝑡−1 )=0

𝑡=1 (𝑢𝑝 )𝑎𝑛𝑑 3(𝑑𝑜𝑤𝑛) 𝑠𝑒𝑐 .

(c)

0=−16 (𝑡 2−4 𝑡−5 )−16 (𝑡+1 ) (𝑡−5 )=0𝑡=−1𝑎𝑛𝑑5 𝑠𝑒𝑐 .𝑡=5 𝑠𝑒𝑐 .

1.5 – Applications of Quadratic Equations

1.6 – More Equations and ApplicationsSolving Equations by GroupingExamples:6 𝑥2+3 𝑥+20𝑥+10=0

3 𝑥 (2 𝑥+1 )+10 (2 𝑥+1 )=0

(2 𝑥+1 ) (3 𝑥+10 )=02 𝑥+1=0 3 𝑥+10=0

𝑥=−12

𝑥=−103

5𝑣3−2𝑣2+25𝑣−10=0𝑣2 (5𝑣−2 )+5 (5𝑣−2 )=0

(5𝑣−2 ) (𝑣2+5 )=05𝑣−2=0 𝑣2+5=0

𝑣=25

𝑣=± 𝑖√5

𝑣2=−5

√𝑣2=±√−5

1.6 – More Equations and ApplicationsSolving Rational EquationsExamples:5 𝑥

𝑥+2=

11𝑥+1

𝑥2+7𝑥+10−

12𝑥+5

5 𝑥 (𝑥+5 )=11𝑥+1−12 (𝑥+2 )

5 𝑥𝑥+2

=11𝑥+1

(𝑥+2 ) (𝑥+5 )−

12𝑥+5LCD:

(𝑥+2 ) (𝑥+5 )( 5 𝑥𝑥+2

=11𝑥+1

(𝑥+2 ) (𝑥+5 )−

12𝑥+5 )

5 𝑥2+25𝑥=11𝑥+1−12𝑥−245 𝑥2+25𝑥=−𝑥−23

5 𝑥2+26𝑥+23=0

𝑥=−26±√ (26 )2−4 (5 ) (23 )

2 (5 )

𝑥=−26±√21610

𝑥=−26±√36 ∙610

𝑥=−13±3√65

1.6 – More Equations and ApplicationsSolving Radical EquationsExamples:√𝑥+3−1=7

√𝑥+3=8(√𝑥+3 )2=(8 )2

𝑥+3=64𝑥=61

3215 xx

115 xx

22115 xx

115 xxxx

1215 xxx

xx 224

01544 2 xx

01414 xx

01x

1x

014 x

22 224 xx

xxx 441616 2

042016 2 xx

4

1x

1.6 – More Equations and ApplicationsSolving Equations with Fractional ExponentsExamples:

𝑝32=11

(𝑝 32)

23=(11 )

23

𝑝=1123

𝑝= 3√112

𝑝= 3√121

𝑝43 =11

(𝑝 43 )

34=(11 )

34

𝑝=1134

𝑝=± 4√113

𝑝=± 4√1331

1.6 – More Equations and ApplicationsSolving Equations with SubstitutionExamples: 𝑥4−7 𝑥2+12=0

𝑙𝑒𝑡 𝑢=𝑥2

Substituteh𝑡 𝑒𝑛𝑢2= (𝑥2 )2=𝑥4

𝑢2−7𝑢+12=0

(𝑢−3 ) (𝑢−4 )=0𝑢−3=0𝑢−4=0

𝑢=3 ,4

𝑢=𝑥2

𝑥2=3 , 4𝑥2=3 𝑥2=4𝑥=±√3 𝑥=±√4

𝑥=±2

1.6 – More Equations and ApplicationsSolving Equations with SubstitutionExamples:

𝑥23−2𝑥

13−15=0

𝑙𝑒𝑡 𝑢=𝑥13

Substituteh𝑡 𝑒𝑛𝑢2=(𝑥13 )2=𝑥

23

𝑢2−2𝑢−15=0

(𝑢+3 ) (𝑢−5 )=0

𝑢+3=0𝑢−5=0𝑢=−3 ,5

𝑢=𝑥13

𝑥13=−3 ,5

𝑥13=−3 𝑥

13=5

(𝑥13 )3=(−3 )3

𝑥=−27

(𝑥13 )3=(5 )3

𝑥=125

1.6 – More Equations and ApplicationsSolving Equations with SubstitutionExamples: 𝑥− 6−9 𝑥− 3+8=0

𝑙𝑒𝑡 𝑢=𝑥−3

Substituteh𝑡 𝑒𝑛𝑢2= (𝑥− 3 )2=𝑥−6

𝑢2−9𝑢+8=0

(𝑢−1 ) (𝑢−8 )=0

𝑢−1=0𝑢−8=0𝑢=1 ,8

𝑢=𝑥− 3¿1 ,8𝑥− 3=1

(𝑥−3 )−13 =(1 )

−13

𝑥=1

113

𝑥=12

𝑥=1

(𝑥−3 )−13 =(8 )

−13

𝑥=1

813

𝑥− 3=8