BASIC ALGEBRA PRACTICE PROBLEMS FOR ALGEBRA 1

Solve the following equations for the unknown x:

Problem 1 :

5 = 7x - 16

Solution:

5 = 7x - 16

7x = 5 + 16

7x = 21

x = 21/7

x = 3

Problem 2 :

2x - 3 = 5 - x

Solution:

2x - 3 = 5 - x

2x + x = 5 + 3

3x = 8

x = 8/3

Multiply the indicated polynomials and simplify.

Problem 3 :

(x + 1)(x² - x + 1)

Solution:

= (x + 1)(x² - x + 1)

= x³ - x² + x + x² - x + 1

= x³ + 1

Problem 4 :

(x - 2)(x + 2)

Solution:

= (x - 2)(x + 2)

= x² + 2x - 2x - 4

= x² - 4

Problem 5 :

(x - 2)(x - 2)

Solution:

= (x - 2)(x - 2)

= x² - 2x - 2x + 4

= x² - 4x + 4

Find the domain of each of the following functions.

Problem 6 :

Solution:

√(1 + x) = 0

x = -1

x ≠ -1

1 + x ≥ 0

x ≥ -1

x ≠ -1 and x ≥ -1

x > -1

Domain = (-1, ∞)

Problem  7 :

Solution:

1 + x² ≠ 0

x ∈ R

Domain = (-∞, ∞)

Problem 8 :

Given that f(x) = x² - 3x + 4, find and simplify f(3), f(a), f(-t) and f(x² + 1).

Solution:

f(x) = x² - 3x + 4

f(3) = 3² - 3(3) + 4

= 9 - 9 + 4

f(3) = 4

f(a) = a² - 3(a) + 4

= a² - 3a + 4

By using quadratic formula,

f(-t) = (-t)² - 3(-t) + 4

= t² + 3t + 4

By using quadratic formula,

f(x² + 1) = (x² + 1)² - 3(x² + 1) + 4

= x⁴ + 2x² + 1 - 3x² - 3 + 4

= x⁴ - x² + 2

Factor the following quadratics.

Problem 9 :

x² - x - 20

Solution:

x² - x - 20 = 0

x² + 4x - 5x - 20 = 0

x(x + 4) - 5(x + 4) = 0

(x - 5)(x + 4) = 0

x - 5 = 0 or x + 4 = 0

x = 5 or x = -4

Problem 10 :

-2x² + 7x + 15

Solution:

-2x² + 7x + 15 = 0

-(2x² - 7x - 15) = 0

-(2x² + 3x - 10x - 15) = 0

-(x(2x + 3) - 5(2x + 3)) = 0

-(x - 5)(2x + 3) = 0

x - 5 = 0 or 2x + 3 = 0

x = 5 or x = -3/2

Problem 11 :

x2 - 2

Solution:

x² - 2 = 0

x² = 2

x = ±√2

Solve the following quadratic equations in three ways.

1) factor

2) quadratic

3) completing the square.

Problem 12 :

-x² - 3x - 2  = 0 

Solution:

Factor:

-x² - 3x - 2  = 0 

-(x² + 3x + 2) = 0

-(x² + 2x + x + 2) = 0

-(x(x + 2) + 1(x + 2)) = 0

-(x + 1) (x + 2) = 0

x + 1 = 0 or x + 2 = 0

x = -1 or x = -2

Quadratic formula  :

-x² - 3x - 2  = 0 

Completing the square method:

-x² - 3x - 2  = 0 

Divide by -1.

x² + 3x + 2 = 0

x² + 3x = -2

x² + 2(x)(3/2) = -2

x² + 2(x)(3/2) + (3/2)² = -2 + (3/2)²

(x + 3/2)² = -2 + 9/4

(x + 3/2)² = 1/4

Taking square root on both sides,

√(x + 3/2)² = √(1/4)

x + 3/2 = ±1/2

x = 1/2 - 3/2 or x = -1/2 - 3/2

x = (1 - 3)/2 or x = (-1 - 3)/2

x = -2/2 or x = -4/2

x = -1 or x = -2

Problem 13 :

2x² + 2x - 4 = 0

Solution:

2x² + 2x - 4 = 0

Divide by 2,

x² + x - 2 = 0

x² + 2x - x - 2 = 0

x(x + 2) - 1(x + 2) = 0

(x - 1)(x + 2) = 0

x = 1 or x = -2

Quadratic formula :

2x² + 2x - 4 = 0

x² + x - 2 = 0

Completing the square method:

2x² + 2x - 4 = 0

x² + x - 2 = 0

x² + x = 2

x2 + 2(x)(1/2) = 2

x² + 2(x)(1/2) + (1/2)² = 2 + (1/2)²

(x + 1/2)² = 2 + 1/4

(x + 1/2)² = 9/4

Take square root on both sides.

√(x + 1/2)² = √9/4

(x + 1/2) = ±3/2

x + 1/2 = 3/2 or x + 1/2 = -3/2

x = 3/2 - 1/2 or x = -3/2 - 1/2

x = (3 - 1)/2 or x = (-3 - 1)/2

x = 2/2 or x = -4/2

x = 1 or x = -2

Solve the following smorgasbord of equations and inequalities.

Problem 14 :

√x = √(2x - 1)

Solution:

√x = √(2x - 1)

Take square on both sides,

x = 2x - 1

x - 2x = -1

x = 1

Checking the original equation:

LHS = √1

RHS = √(2(1) - 1) = √1

Problem 15 :

√x² - 3 = √2x

Solution:

√x² - 3 = √2x

Take square root on both sides,

x² - 3 = 2x

x² - 2x - 3 = 0

(x - 3)(x + 1) = 0

x = 3 or x = -1

Checking the original equation:

LHS = √(3)² - 3

= √(9 - 3)

= √6

= 2√3 = RHS

Problem 16 :

|x - 5| = 4

Solution:

|x - 5| = 4

x - 5 = 4 or x - 5 = -4

x = 9 or x = 1

Checking the original equation:

|9 - 5| = | 4| = 4

|1 - 5| = |-4| = 4

Problem 17 :

2x + 4 ≥ 3

Solution:

2x + 4 ≥ 3

2x ≥ -1

x ≥ -1/2

If x ≥ -1/2 that will imply that 2x + 4 ≥ 3.

So, the interval [-1/2, ∞).

Problem 18 :

-2x + 4 ≥ 3

Solution:

-2x + 4 ≥ 3

-2x ≥ -1

x ≤ 1/2

If x ≤ 1/2 this will imply that -2x + 4 ≥ 3.

So, the interval (-∞, 1/2].

Problem 19 :

Solution:

Checking the original equation: