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Mathematics LibreTexts

3.1.1: Exercises 3.1

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Terms and Concepts

Exercise 3.1.1.1

Is it possible to solve a cubic statement?

Answer

Yes, but it can be quite difficult, especially if it has many parameters.

Exercise 3.1.1.2

What are the possible types of solutions when solving a quadratic statement?

Answer

2 distinct real roots; 1 repeated real root; a complex conjugate pair of roots.

Exercise 3.1.1.3

What is the maximum number of different solutions that a seventh degree statement could have?

Answer

7

Exercise 3.1.1.4

T/F: A cubic statement can have only complex solutions. Explain.

Answer

F; it must have at least one real solution since complex solutions come in pairs

Problems

In exercises 3.1.1.5 - 3.1.1.10, determine the type of statement in terms of the given variable.

Exercise 3.1.1.5

x3y+2x2yz6xz2=yz210 in terms of x

Answer

cubic

Exercise 3.1.1.6

x3y+2x2yz6xz2=yz210 in terms of y

Answer

linear

Exercise 3.1.1.7

x3y+2x2yz6xz2=yz210 in terms of z

Answer

quadratic

Exercise 3.1.1.8

xt+cos(θ)=x4t36t in terms of θ

Answer

trigonometric

Exercise 3.1.1.9

xt+cos(θ)=x4t36t in terms of x

Answer

quartic, or a statement of degree 4

Exercise 3.1.1.10

xt+cos(θ)=x4t36t in terms of t

Answer

cubic

In exercises 3.1.1.11 - 3.1.1.19, determine if it is possible to solve the statement for the given variable. If it is possible, solve but do not simplify your answer(s). If it is not possible, explain why.

Exercise 3.1.1.11

xy2xy=5y3x for x

Answer

It is possible to solve; x=5yy2y+3

Exercise 3.1.1.12

xy2xy=5y3x for y

Answer

x+5±11x2+10x+252x

Exercise 3.1.1.13

3t25mq=8qt+2m3 for q

Answer

It is possible to solve; q=3t22m38t+5m

Exercise 3.1.1.14

2a2bc3+3abc2+4a2c23b=4c for a

Answer

It is possible to solve; a=(3bc2)±(3bc2)24(2bc3+4c2)(3b4c)2(2bc3+4c2)

Exercise 3.1.1.15

2a2bc3+3abc2+4a2c23b=4c for b

Answer

It is possible to solve; b=4c4a2c22a2c3+3ac23

Exercise 3.1.1.16

2a2bc3+3abc2+4a2c23b=4c for c

Answer

It is possible to solve; but it would require using the cubic root formula

Exercise 3.1.1.17

log2(xy)=x+ez for x

Answer

Not possible to solve for x; it is inside of a logarithm and has a linear term

Exercise 3.1.1.18

log2(xy)=x+ez for y

Answer

It is possible to solve; y=2x+ezlog2(x) or y=2x+ezx

Exercise 3.1.1.19

log2(xy)=x+ez for z

Answer

It is possible to solve; z=ln[log2(xy)x]

In exercises 3.1.1.20 - 3.1.1.28, solve for x. Be sure to list all possible values of x.

Exercise 3.1.1.20

x216=0

Answer

x=4,4

Exercise 3.1.1.21

x2+16=0

Answer

x=4i,4i

Exercise 3.1.1.22

x24x7=2

Answer

x=2+13,213

Exercise 3.1.1.23

x22x+7=2

Answer

x=1+2i,12i

Exercise 3.1.1.24

5x2+2x=1

Answer

x=1+2i5,12i5

Exercise 3.1.1.25

x3=8

Answer

x=2

Exercise 3.1.1.26

x3+x2=4x+4

Answer

x=2,1,2

Exercise 3.1.1.27

2(x3)27=4x+9

Answer

x=23,2+3

Exercise 3.1.1.28

(x+2)3=2x2+8x+7

Answer

x=1,3+52,352

In exercises 3.1.1.29 - 3.1.1.33, classify the type(s) of solution(s) from the given exercise.

Exercise 3.1.1.29

Exercise 3.1.1.20 

Answer

Two real solutions

Exercise 3.1.1.30

Exercise 3.1.1.21 

Answer

A complex conjugate pair

Exercise 3.1.1.31

Exercise 3.1.1.22 

Answer

Two real solutions

Exercise 3.1.1.32

Exercise 3.1.1.25 

Answer

One repeated solution

Exercise 3.1.1.33

Exercise 3.1.1.26 

Answer

Three real solutions


3.1.1: Exercises 3.1 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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