Intermediate Algebra Practice Questions — ACT — Page 12

An infinite geometric series has a first term of $12$ and a sum of $8$ . What is the common ratio of this series?

-\frac{1}{2}

\frac{1}{2}

\frac{1}{3}

\frac{2}{3}

\frac{5}{2}

Show answer & explanation

Answer: $-\frac{1}{2}$

Answer

The correct common ratio is

-\frac{1}{2}

.

The correct answer is

-\frac{1}{2}

. The formula for the sum of an infinite geometric series is

S = \frac{a_1}{1 - r}

. Substituting the first term

a_1 = 12

and the sum

S = 8

gives

8 = \frac{12}{1 - r}

. Multiplying both sides by

1 - r

yields

8(1 - r) = 12

, which simplifies to

8 - 8r = 12

. Subtracting

8

from both sides gives

-8r = 4

. Dividing by

-8

gives

r = -\frac{1}{2}

. Since

|-\frac{1}{2}| < 1

, the series converges.

Step-by-Step Solution

1

Write the formula for the sum of an infinite geometric series.

S = \frac{a_1}{1 - r}

This formula relates the sum (

S

), the first term (

a_1

), and the common ratio (

r

) of an infinite geometric series where

|r| < 1

.

2

Substitute the given values into the formula.

8 = \frac{12}{1 - r}

Replacing

S

with

8

and

a_1

with

12

leaves the common ratio

r

as the only unknown variable.

3

Solve for the common ratio

r

.

8(1 - r) = 12 \Rightarrow 8 - 8r = 12 \Rightarrow -8r = 4 \Rightarrow r = -\frac{4}{8} = -\frac{1}{2}

Multiplying by the denominator and isolating

r

yields the value of the common ratio.

Key Concept

Sum of an infinite geometric series

Alternative Method

Test the given choices by substituting each value of

r

back into the sum formula

S = \frac{12}{1 - r}

to see which one yields a sum of

8

. For example, testing

-\frac{1}{2}

gives

\frac{12}{1 - (-0.5)} = \frac{12}{1.5} = 8

, which matches the given sum.

Estimated Time:1m 0s

Question 222Question →

A geometric sequence consists of positive terms. The first term of the sequence is $3$ , and the sum of the first $3$ terms is $39$ . What is the value of the $4$ th term of this sequence?

33

81

120

192

243

Show answer & explanation

Answer: 81

Answer

81

The sum of the first three terms of a geometric sequence is given by

S_3 = a_1(1 + r + r^2)

. Substituting the first term

a_1 = 3

and the sum

S_3 = 39

gives

3(1 + r + r^2) = 39

, which simplifies to

r^2 + r - 12 = 0

. Factoring this quadratic equation yields

(r - 3)(r + 4) = 0

. Since the sequence consists of positive terms, the common ratio must be positive, which gives

r = 3

. The fourth term of the sequence is then calculated using the formula

a_4 = a_1 r^3 = 3 \cdot 3^3 = 81

.

Step-by-Step Solution

1

Write the expression for the sum of the first 3 terms of a geometric sequence and set it equal to the given sum.

S_3 = a_1(1 + r + r^2) = 3(1 + r + r^2) = 39

This sets up the equation needed to solve for the common ratio of the sequence.

2

Divide both sides of the equation by 3 and solve the resulting quadratic equation for the common ratio

r

.

1 + r + r^2 = 13 \implies r^2 + r - 12 = 0 \implies (r - 3)(r + 4) = 0 \implies r = 3

(since terms must be positive)

This determines the common ratio of the geometric sequence.

3

Use the formula for the

n

-th term of a geometric sequence,

a_n = a_1 r^{n-1}

, to find the value of the 4th term.

a_4 = 3 \cdot 3^3 = 3 \cdot 27 = 81

This computes the final value requested by the question.

Key Concept

Geometric sequence formulas for the sum of a finite number of terms and the n-th term

Estimated Time:1m 30s

Question 223Question →

In the standard $(x, y)$ coordinate plane, the circle defined by $x^2 + y^2 = 26$ and the line defined by $y = x - 4$ intersect at two points. What is the distance between these two points of intersection?

2\sqrt{10}

4\sqrt{2}

6

6\sqrt{2}

12

Show answer & explanation

Answer: $6\sqrt{2}$

Answer

6\sqrt{2}

Solving the system of equations by substituting

y = x - 4

into

x^2 + y^2 = 26

yields

2x^2 - 8x - 10 = 0

, which simplifies to

x^2 - 4x - 5 = 0

. Factoring gives

(x-5)(x+1) = 0

, leading to the points

(5, 1)

and

(-1, -5)

. The distance between these points is

\sqrt{6^2 + 6^2} = 6\sqrt{2}

.

Step-by-Step Solution

1

Substitute the linear equation into the circle equation.

x^2 + (x - 4)^2 = 26

This reduces the system to a single equation in terms of

x

.

2

Expand the squared binomial and simplify the quadratic equation.

2x^2 - 8x - 10 = 0

, which simplifies to

x^2 - 4x - 5 = 0

.

Expanding

(x - 4)^2

yields

x^2 - 8x + 16

. Setting the equation to zero allows us to solve for

x

.

3

Factor the quadratic equation to find the

x

-coordinates.

x = 5

or

x = -1

.

The factored form is

(x - 5)(x + 1) = 0

.

4

Substitute the

x

-values back into the linear equation to find the corresponding

y

-coordinates.

The intersection points are

(5, 1)

and

(-1, -5)

.

For

x = 5

,

y = 5 - 4 = 1

. For

x = -1

,

y = -1 - 4 = -5

.

5

Apply the distance formula to find the straight-line distance between the two points.

6\sqrt{2}

The distance is

\sqrt{(5 - (-1))^2 + (1 - (-5))^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}

.

Key Concept

Solving systems of linear and non-linear (circular) equations by substitution and finding the distance between intersection points.

Question 224Question →

A geometric sequence has a first term of $2^3$ and a common ratio of $2^2$ . What is the value of the 5th term of this sequence?

24

512

2,048

8,192

16,777,216

Show answer & explanation

Answer: $2,048$

Answer

The 5th term of the geometric sequence is

2,048

.

To find the 5th term of a geometric sequence, we use the formula

a_n = a_1 \cdot r^{n-1}

. Substituting the given first term

a_1 = 2^3 = 8

and the common ratio

r = 2^2 = 4

for

n = 5

gives

a_5 = 2^3 \cdot (2^2)^4

. Applying the power of a power rule,

(2^2)^4 = 2^{2 \times 4} = 2^8

. Then, multiplying the bases by adding the exponents gives

2^3 \cdot 2^8 = 2^{3+8} = 2^{11}

, which evaluates to

2,048

.

Step-by-Step Solution

1

Identify the given values and the formula for the

n

-th term of a geometric sequence.

The first term is

a_1 = 2^3 = 8

, the common ratio is

r = 2^2 = 4

, and we need to find the term for

n = 5

using the formula

a_n = a_1 \cdot r^{n-1}

.

Knowing the correct formula is necessary to compute the specific term of a geometric sequence.

2

Substitute the values into the formula to express the 5th term in terms of base 2.

a_5 = 2^3 \cdot (2^2)^{5-1} = 2^3 \cdot (2^2)^4

This substitutes the specific term number and sequence parameters into the general term formula.

3

Simplify the exponential expression and calculate the final numerical value.

a_5 = 2^3 \cdot 2^8 = 2^{3+8} = 2^{11} = 2,048

Applying exponent rules (multiplying powers of a power and adding exponents when multiplying like bases) allows us to evaluate the expression to a single number.

Key Concept

Finding a specific term in a geometric sequence using exponential properties

Estimated Time:1m 0s

Question 225Question →

A parabola is defined by the equation $y = x^2 - 4x + 3$ , and a line is defined by the equation $y = -x + 7$ . The parabola and the line intersect at two points in the standard $(x, y)$ coordinate plane. What is the distance between these two points of intersection?

5

\sqrt{34}

5\sqrt{2}

3\sqrt{2}

50

Show answer & explanation

Answer: $5\sqrt{2}$

Answer

5\sqrt{2}

The correct answer is the value representing the straight-line distance between the two intersection points. Setting the equations equal to each other gives

x^2 - 3x - 4 = 0

, which factors to

(x - 4)(x + 1) = 0

. This yields

x = 4

and

x = -1

. Evaluating these in

y = -x + 7

yields the points

(4, 3)

and

(-1, 8)

. The distance between them is

\sqrt{(4 - (-1))^2 + (3 - 8)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}

.

Step-by-Step Solution

1

Set the equations of the parabola and line equal to each other to find the

x

-coordinates of the intersection points.

x^2 - 4x + 3 = -x + 7

Intersection points must satisfy both equations simultaneously.

2

Rearrange the equation into standard quadratic form and factor it to solve for

x

.

x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0 \implies x = 4

or

x = -1

Factoring the quadratic equation gives the roots which correspond to the

x

-coordinates of the intersection points.

3

Substitute the

x

-values back into the linear equation to determine the corresponding

y

-coordinates.

For

x = 4

,

y = -(4) + 7 = 3

, yielding point

(4, 3)

. For

x = -1

,

y = -(-1) + 7 = 8

, yielding point

(-1, 8)

.

Substituting into the simpler linear equation provides the

y

-coordinates of the intersection points.

4

Apply the distance formula to calculate the distance between

(4, 3)

and

(-1, 8)

.

d = \sqrt{(4 - (-1))^2 + (3 - 8)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}

The distance formula calculates the straight-line distance between two coordinates in the coordinate plane.

Key Concept

Systems of Linear and Non-Linear Equations

Question 226Question →

A recipe calls for $\frac{1}{2}$ cup of sugar on the first day of a fermentation process. Each day after that, the ratio of the sugar added on that day to the sugar added on the previous day is $1:3$ . What is the total amount of sugar, in cups, added during the first 3 days?

\frac{13}{18}

\frac{3}{26}

\frac{2}{9}

\frac{5}{2}

\frac{21}{32}

Show answer & explanation

Answer: $\frac{13}{18}$

Answer

The total amount of sugar added is

\frac{13}{18}

cups.

The correct answer is

\frac{13}{18}

. The sugar added on Day 1 is

\frac{1}{2}

cup. Since the ratio between consecutive days is

1:3

, the common ratio is

r = \frac{1}{3}

. This gives a Day 2 amount of

\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}

cup, and a Day 3 amount of

\frac{1}{6} \times \frac{1}{3} = \frac{1}{18}

cup. Summing these three amounts using a common denominator of 18 yields

\frac{9}{18} + \frac{3}{18} + \frac{1}{18} = \frac{13}{18}

cups.

Step-by-Step Solution

1

Identify the type of sequence and the first term.

The first term is

a_1 = \frac{1}{2}

. The sequence is geometric because the ratio between the amounts added on consecutive days is constant.

The problem states that the ratio of sugar added on consecutive days is

1:3

, which establishes a common ratio for a geometric sequence.

2

Determine the common ratio and find the terms for the second and third days.

The common ratio is

r = \frac{1}{3}

. The second term is

a_2 = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}

, and the third term is

a_3 = \frac{1}{6} \times \frac{1}{3} = \frac{1}{18}

.

Each subsequent term of a geometric sequence is found by multiplying the previous term by the common ratio.

3

Calculate the sum of the first three terms.

The total sum is

S_3 = \frac{1}{2} + \frac{1}{6} + \frac{1}{18} = \frac{9}{18} + \frac{3}{18} + \frac{1}{18} = \frac{13}{18}

.

Finding the total amount requires summing the individual amounts added over the three days using a common denominator of 18.

Key Concept

Sum of a finite geometric series

Alternative Method

Instead of calculating and adding the individual terms, you can use the sum of a finite geometric series formula:

S_n = \frac{a_1(1-r^n)}{1-r}

. Substituting

a_1 = \frac{1}{2}

,

r = \frac{1}{3}

, and

n=3

yields:

S_3 = \frac{\frac{1}{2}\left(1 - \left(\frac{1}{3}\right)^3\right)}{1 - \frac{1}{3}} = \frac{\frac{1}{2}\left(1 - \frac{1}{27}\right)}{\frac{2}{3}} = \frac{\frac{1}{2}\left(\frac{26}{27}\right)}{\frac{2}{3}} = \frac{13}{27} \times \frac{3}{2} = \frac{13}{18}

.

Estimated Time:1m 30s

Question 227Question →

The first term of an arithmetic sequence is $\frac{1}{2}$ , and the third term of the sequence is $\frac{5}{6}$ . What is the sum of the first 6 terms of this sequence?

\frac{9}{4}

3

5

\frac{11}{2}

\frac{21}{2}

Show answer & explanation

Answer: $\frac{11}{2}$

Answer

The sum of the first 6 terms is

\frac{11}{2}

.

To find the sum of the first 6 terms, we first determine the common difference

d

from the given terms:

a_3 = a_1 + 2d \Rightarrow \frac{5}{6} = \frac{1}{2} + 2d \Rightarrow 2d = \frac{1}{3} \Rightarrow d = \frac{1}{6}

. Then, we apply the arithmetic series sum formula:

S_6 = \frac{6}{2}[2(\frac{1}{2}) + 5(\frac{1}{6})] = 3[1 + \frac{5}{6}] = 3(\frac{11}{6}) = \frac{11}{2}

.

Step-by-Step Solution

1

Find the common difference

d

using the formula for the

n

-th term of an arithmetic sequence

a_n = a_1 + (n - 1)d

with the given values

a_1 = \frac{1}{2}

and

a_3 = \frac{5}{6}

.

2d = \frac{5}{6} - \frac{1}{2} \Rightarrow 2d = \frac{1}{3} \Rightarrow d = \frac{1}{6}

We need to find the common difference to determine the subsequent terms and calculate the sum of the sequence.

2

Use the sum formula

S_n = \frac{n}{2}[2a_1 + (n - 1)d]

with

n = 6

,

a_1 = \frac{1}{2}

, and

d = \frac{1}{6}

to find the sum of the first 6 terms.

S_6 = \frac{6}{2}[2(\frac{1}{2}) + (6 - 1)(\frac{1}{6})] = 3[1 + \frac{5}{6}] = 3(\frac{11}{6}) = \frac{11}{2}

This formula sums the first 6 terms of the arithmetic sequence directly.

Key Concept

Sum of the first

n

terms of an arithmetic sequence using fractional terms

Alternative Method

Alternatively, you can list the first 6 terms of the sequence and add them directly:

\frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \frac{6}{6}, \frac{7}{6}, \frac{8}{6}

. Adding these gives

\frac{33}{6} = \frac{11}{2}

.

Estimated Time:1m 30s

Question 228Question →

One of the solutions to the quadratic equation $2x^2 - kx + 18 = 0$ is exactly four times the other solution. If $k$ is a positive constant, what is the value of $k$ ?

Show answer & explanation

Answer: 15

Answer

15

By representing the roots as

r

and

4r

, we can use Vieta's formulas to find that the product of the roots is

4r^2 = 9

, which gives

r = 1.5

. The sum of the roots is

5r = k/2

, which gives

k = 10(1.5) = 15

.

Step-by-Step Solution

1

Represent the roots of the quadratic equation.

Let the two solutions be

r

and

4r

, where

r

is a real number.

We are given that one solution is exactly four times the other.

2

Apply Vieta's formulas for the product of the roots.

r \cdot 4r = 4r^2 = \frac{18}{2} = 9

For a quadratic equation

ax^2 + bx + c = 0

, the product of the roots is equal to

\frac{c}{a}

.

3

Solve for the root

r

.

r^2 = \frac{9}{4} \Rightarrow r = 1.5

(since

k

is positive,

r

must be positive)

Solving the equation

4r^2 = 9

gives

r = \pm 1.5

. Since the sum of the roots is positive, we select the positive root.

4

Apply Vieta's formulas for the sum of the roots to find

k

.

r + 4r = 5r = -\frac{-k}{2} = \frac{k}{2} \Rightarrow k = 10r = 10(1.5) = 15

For a quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is equal to

-\frac{b}{a}

.

Key Concept

Using Vieta's formulas to relate the roots of a quadratic equation to its coefficients.

Question 229Question →

On the first day of a research project, a student analyzes $\frac{1}{4}$ of a dataset. On each subsequent day, the student analyzes a fraction of the dataset that is exactly $\frac{2}{3}$ of the fraction analyzed on the previous day. What fraction of the dataset does the student analyze on the 4th day of the project?

\frac{1}{216}

\frac{4}{81}

\frac{2}{27}

\frac{1}{2}

\frac{9}{31}

Show answer & explanation

Answer: $\frac{2}{27}$

Answer

The fraction of the dataset analyzed on the 4th day is

\frac{2}{27}

.

The problem describes a geometric sequence where the first term is the fraction of the dataset analyzed on the first day, which is

\frac{1}{4}

, and the common ratio is the multiplier for each subsequent day, which is

\frac{2}{3}

. The general formula for the

n

-th term of a geometric sequence is

a_n = a_1 \cdot r^{n-1}

. For the 4th day, we evaluate

a_4 = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^{4-1} = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^3 = \frac{1}{4} \cdot \frac{8}{27} = \frac{2}{27}

.

Step-by-Step Solution

1

Identify the type of sequence and its parameters from the word problem.

This is a geometric sequence with the first term

a_1 = \frac{1}{4}

and a common ratio

r = \frac{2}{3}

.

Each day's fraction is a constant multiple of the previous day's fraction, which characterizes a geometric sequence.

2

Set up the formula for the

n

-th term of a geometric sequence to find the term for the 4th day.

a_n = a_1 \cdot r^{n-1} \implies a_4 = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^{4-1}

To find the fraction analyzed on the 4th day, we evaluate the 4th term of the sequence (

n = 4

).

3

Simplify the exponent and calculate the final fraction.

a_4 = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^3 = \frac{1}{4} \cdot \frac{8}{27} = \frac{2}{27}

Performing exponentiation before multiplication satisfies the order of operations and yields the correct fraction.

Key Concept

Modeling real-world scenarios using the general term formula of geometric sequences.

Question 230Question →

A rectangular painting is $2$ feet wide and $6$ feet long. A wooden frame of uniform width $x$ feet is placed around the painting. If the total area of the painting and the frame is $21$ square feet, what is the width of the frame, in feet?

1.5

1.0

0.5

4.5

8.0

Show answer & explanation

Answer: 0.5

Answer

The correct width of the frame is 0.5 feet.

The correct answer is 0.5 feet. To find the width of the frame, we define the total dimensions of the framed painting as

(2 + 2x)

and

(6 + 2x)

. Setting their product equal to the total area of 21 square feet gives the quadratic equation

4x^2 + 16x - 9 = 0

. Solving this using the quadratic formula yields

x = 0.5

and

x = -4.5

. Since a physical width must be positive, the width of the frame is 0.5 feet.

Step-by-Step Solution

1

Set up the equation for the total area. The painting's dimensions are

2

feet by

6

feet. Adding a frame of uniform width

x

on all sides increases both the width and the length by

2x

.

The total dimensions are

(2 + 2x)

and

(6 + 2x)

, and the total area is given by the equation:

(2 + 2x)(6 + 2x) = 21

.

To find the width of the frame, we must relate the final total area to the dimensions of the painting plus the frame.

2

Expand the equation and write it in standard quadratic form

ax^2 + bx + c = 0

.

12 + 4x + 12x + 4x^2 = 21 \implies 4x^2 + 16x + 12 = 21 \implies 4x^2 + 16x - 9 = 0

.

Standard form is required to apply the quadratic formula or to factor the quadratic expression.

3

Solve the quadratic equation

4x^2 + 16x - 9 = 0

using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

.

x = \frac{-16 \pm \sqrt{16^2 - 4(4)(-9)}}{2(4)} = \frac{-16 \pm \sqrt{256 + 144}}{8} = \frac{-16 \pm \sqrt{400}}{8} = \frac{-16 \pm 20}{8}

.

This formula provides the solutions to any quadratic equation.

4

Calculate the two possible values for

x

and select the physically meaningful one.

x = \frac{-16 + 20}{8} = 0.5

or

x = \frac{-16 - 20}{8} = -4.5

. Since width must be positive, the only valid solution is

x = 0.5

.

A physical measurement like width cannot be negative.

Key Concept

Quadratic Equations and the Quadratic Formula

Question 231Question →

For the quadratic equation $2x^2 + bx + 16 = 0$ , where $b$ is a positive constant, the sum of the squares of the two complex solutions is equal to $-7$ . What is the value of $b$ ?

Show answer & explanation

Answer: 6

Answer

The value of the positive constant

b

is

6

.

By Vieta's formulas, the sum of the roots of the quadratic equation

2x^2 + bx + 16 = 0

is

-\frac{b}{2}

and the product is

8

. Using the identity

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

, we substitute

-7

for the sum of the squares, yielding

-7 = \left(-\frac{b}{2}\right)^2 - 2(8)

. This simplifies to

-7 = \frac{b^2}{4} - 16

, which leads to

\frac{b^2}{4} = 9

and

b^2 = 36

. Since

b

is a positive constant,

b = 6

.

Step-by-Step Solution

1

Find the sum and product of the roots in terms of

b

using Vieta's formulas.

x_1 + x_2 = -\frac{b}{2}

and

x_1 x_2 = 8

Vieta's formulas state that for a quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a}

and the product of the roots is

\frac{c}{a}

.

2

Relate the sum of the squares of the roots to their sum and product.

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

This algebraic identity allows us to express the sum of the squares of the roots using the sum and product of the roots.

3

Substitute the known values into the identity and solve for

b

.

-7 = \left(-\frac{b}{2}\right)^2 - 2(8) \implies -7 = \frac{b^2}{4} - 16 \implies \frac{b^2}{4} = 9 \implies b^2 = 36 \implies b = 6

Substituting the given sum of squares (

-7

), sum (

-\frac{b}{2}

), and product (

8

) produces a single-variable equation that can be solved for the positive constant

b

.

Key Concept

Using Vieta's formulas and algebraic identities to relate the roots of a quadratic equation to its coefficients.

Question 232Question →

In the standard $(x, y)$ coordinate plane, the graph of the parabola $y = x^2 - 5x + 2$ intersects the line $2x - y = 4$ at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the value of $x_1 x_2 + y_1 y_2$ ?

50

22

-10

6

102

Show answer & explanation

Answer: -10

Answer

The value of

x_1 x_2 + y_1 y_2

is

-10

.

The correct answer is

-10

. To find the intersection points, substitute the linear equation

y = 2x - 4

into the quadratic equation

y = x^2 - 5x + 2

, which simplifies to

x^2 - 7x + 6 = 0

. Factoring gives

(x-1)(x-6) = 0

, yielding

x

-coordinates

x_1 = 1

and

x_2 = 6

. Substituting these back into the linear equation gives the corresponding

y

-coordinates

y_1 = -2

and

y_2 = 8

. Evaluating the requested expression gives

(1)(6) + (-2)(8) = 6 - 16 = -10

.

Step-by-Step Solution

1

Isolate

y

in the linear equation.

y = 2x - 4

This allows for substitution into the quadratic equation.

2

Substitute

y = 2x - 4

into the quadratic equation

y = x^2 - 5x + 2

.

2x - 4 = x^2 - 5x + 2

Setting the two equations equal to each other helps find the x-coordinates of the intersection points.

3

Rearrange the equation into standard quadratic form.

x^2 - 7x + 6 = 0

This sets up the equation for factoring.

4

Factor the quadratic equation.

(x-1)(x-6) = 0

Factoring reveals the solutions for the x-coordinates.

5

Solve for the two x-coordinates.

x_1 = 1

and

x_2 = 6

These are the x-coordinates of the two intersection points.

6

Find the corresponding y-coordinates by substituting the x-values back into the linear equation.

For

x_1 = 1

:

y_1 = 2(1) - 4 = -2

. For

x_2 = 6

:

y_2 = 2(6) - 4 = 8

.

This gives the full coordinate pairs:

(1, -2)

and

(6, 8)

.

7

Calculate the value of the expression

x_1 x_2 + y_1 y_2

.

(1)(6) + (-2)(8) = 6 - 16 = -10

This provides the final required value.

Key Concept

Solving systems of linear and quadratic equations by substitution and evaluating expressions of their coordinate solutions.

Estimated Time:1m 30s

Question 233Question →

A geometric sequence has a second term of $-\frac{3}{2}$ and a fifth term of $12$ . What is the eighth term of this sequence?

-96

-48

\frac{51}{2}

96

192

Show answer & explanation

Answer: $-96$

Answer

-96

The correct term is

-96

. First, find the common ratio

r

by taking the ratio of the fifth term to the second term:

\frac{a_5}{a_2} = \frac{a_1 r^4}{a_1 r} = r^3

. Substituting the given values,

r^3 = \frac{12}{-3/2} = -8

, which gives

r = -2

. To find the eighth term, multiply the fifth term by the common ratio cubed:

a_8 = a_5 r^3 = 12 \times (-2)^3 = 12 \times (-8) = -96

.

Step-by-Step Solution

1

Set up the ratio between the fifth term and the second term using the geometric sequence formula

a_n = a_1 r^{n-1}

.

\frac{a_5}{a_2} = \frac{a_1 r^4}{a_1 r} = r^3

This allows us to isolate the common ratio

r

without needing to calculate the first term

a_1

first.

2

Substitute the given values into the ratio and solve for

r

.

r^3 = \frac{12}{-3/2} = 12 \times \left(-\frac{2}{3}\right) = -8 \Rightarrow r = -2

Finding the common ratio is necessary to compute any subsequent terms in the sequence.

3

Use the common ratio to find the eighth term

a_8

by multiplying the fifth term

a_5

by

r^3

.

a_8 = a_5 r^{8-5} = 12 \times (-2)^3 = 12 \times (-8) = -96

Since

a_8 = a_5 r^3

, multiplying

12

by

-8

directly gives the eighth term.

Key Concept

Finding terms in a geometric sequence using the common ratio.

Question 234Question →

A line is defined by the equation $y = 2x + k$ , where $k$ is a constant. A parabola is defined by the equation $y = x^2 - 4x + 14$ . If the line and the parabola intersect at exactly one point in the standard $(x, y)$ coordinate plane, what is the value of $k$ ?

Show answer & explanation

Answer: 5

Answer

The value of the constant

k

must be 5.

To find the intersection of the line and the parabola, set their equations equal to each other:

x^2 - 4x + 14 = 2x + k

. Rearranging this into standard quadratic form gives

x^2 - 6x + (14 - k) = 0

. For the system to have exactly one solution, the discriminant of this quadratic equation must be zero. The discriminant is

b^2 - 4ac = (-6)^2 - 4(1)(14 - k) = 36 - 56 + 4k = 4k - 20

. Setting

4k - 20 = 0

yields

k = 5

.

Step-by-Step Solution

1

Equate the linear and quadratic expressions to find their intersection.

x^2 - 4x + 14 = 2x + k

Setting the two equations equal to each other allows us to find the

x

-coordinates of any intersection points.

2

Rearrange the terms to write the equation in standard quadratic form,

ax^2 + bx + c = 0

.

x^2 - 6x + (14 - k) = 0

Subtracting

2x

and

k

from both sides simplifies the equation into a quadratic form where

a = 1

,

b = -6

, and

c = 14 - k

.

3

Apply the condition for exactly one intersection point by setting the discriminant to zero.

(-6)^2 - 4(1)(14 - k) = 0

A quadratic equation has exactly one real root if and only if its discriminant,

b^2 - 4ac

, is equal to zero.

4

Solve the linear equation for

k

.

k = 5

Expanding the equation yields

36 - 56 + 4k = 0

, which simplifies to

4k - 20 = 0

, leading to

k = 5

.

Key Concept

Determining the condition for a linear equation to be tangent to a quadratic equation by setting the discriminant of their intersection equation to zero.

Question 235Question →

An arithmetic sequence has a first term of $2$ and a common difference of $d$ . A geometric sequence has a first term of $4$ and a common ratio of $r$ . The third term of the arithmetic sequence is equal to the third term of the geometric sequence. If $d = r$ and $d \neq 1$ , what is the value of $d$ ?

2

\frac{1}{2}

-1

-\frac{1}{2}

\frac{1}{3}

Show answer & explanation

Answer: $-\frac{1}{2}$

Answer

-\frac{1}{2}

The correct answer is

-\frac{1}{2}

. The third term of the arithmetic sequence is

a_3 = a_1 + 2d = 2 + 2d

. The third term of the geometric sequence is

g_3 = g_1 r^2 = 4r^2

. Given that

d = r

, we set the two terms equal:

2 + 2d = 4d^2

. Rearranging and dividing by

2

yields

2d^2 - d - 1 = 0

, which factors as

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

. Since the problem specifies that

d \neq 1

, the only valid solution is

d = -\frac{1}{2}

.

Step-by-Step Solution

1

Write the formulas for the third term of both the arithmetic and geometric sequences.

For the arithmetic sequence:

a_3 = a_1 + (3 - 1)d = 2 + 2d

. For the geometric sequence:

g_3 = g_1 \cdot r^{3 - 1} = 4r^2

.

This establishes the algebraic expressions for the third terms using the given first terms.

2

Substitute

d

for

r

and set the two expressions equal to each other.

Since

d = r

, we substitute

d

into the geometric term to get

g_3 = 4d^2

. Setting them equal gives

2 + 2d = 4d^2

.

The problem states that the third terms are equal and that the common difference equals the common ratio.

3

Solve the quadratic equation

4d^2 - 2d - 2 = 0

for

d

.

Divide the equation by

2

to get

2d^2 - d - 1 = 0

. Factoring this gives

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

. The roots are

d = -\frac{1}{2}

and

d = 1

. Since the problem specifies

d \neq 1

, we have

d = -\frac{1}{2}

.

Solving the quadratic equation gives the possible values of the common difference, and the constraint rules out

d = 1

.

Key Concept

Relating arithmetic and geometric sequence terms and solving the resulting quadratic equation.

Alternative Method

Instead of factoring, the quadratic formula can be used to solve

2d^2 - d - 1 = 0

:

d = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4}

. This yields

d = 1

and

d = -\frac{1}{2}

. Since the problem specifies

d \neq 1

, the only valid solution is

d = -\frac{1}{2}

.

Estimated Time:1m 30s

Question 236Question →

If $k$ is a non-zero constant, for what value of $k$ does the quadratic equation $(x - 3)^2 = kx$ have exactly one real solution?

6

-3

-12

-6

12

Show answer & explanation

Answer: $-12$

Answer

-12

Expanding

(x - 3)^2

yields

x^2 - 6x + 9 = kx

. Subtracting

kx

from both sides and grouping like terms gives

x^2 - (6 + k)x + 9 = 0

. For a quadratic equation to have exactly one real solution, its discriminant must be equal to zero. Thus, we set

(-(6+k))^2 - 4(1)(9) = 0

, which simplifies to

(6+k)^2 - 36 = 0

. Taking the square root of both sides gives

6 + k = 6

or

6 + k = -6

. Solving these equations gives

k = 0

or

k = -12

. Since the problem specifies that

k

is a non-zero constant, the correct value is

-12

.

Step-by-Step Solution

1

Expand the squared binomial on the left side of the equation.

x^2 - 6x + 9 = kx

Expanding the binomial

(x - 3)^2

allows us to rewrite the equation in a form where we can group terms.

2

Move

kx

to the left side and group the

x

terms to write the equation in standard quadratic form

ax^2 + bx + c = 0

.

x^2 - (6 + k)x + 9 = 0

, where

a = 1

,

b = -(6 + k)

, and

c = 9

.

The coefficients

a

,

b

, and

c

must be identified from the standard quadratic form to calculate the discriminant.

3

Set the discriminant

b^2 - 4ac

equal to

0

and solve for

k

.

(-(6+k))^2 - 4(1)(9) = 0 \implies (6+k)^2 - 36 = 0 \implies (6+k)^2 = 36 \implies 6+k = \pm 6

. This yields

k = 0

or

k = -12

. Since

k

is non-zero,

k = -12

.

A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

Key Concept

Using the discriminant (

b^2 - 4ac = 0

) to determine when a quadratic equation has exactly one real solution.

Alternative Method

Alternatively, one can recognize that the equation

(x-3)^2 = kx

can be written as

x^2 - (6+k)x + 9 = 0

. For a quadratic equation with a leading coefficient of

1

and a constant term of

9

to have exactly one real solution, it must be a perfect square trinomial. A perfect square trinomial of the form

x^2 + bx + 9

must have

b = \pm 6

. Setting the middle coefficient equal to these values gives

-(6+k) = 6 \implies k = -12

or

-(6+k) = -6 \implies k = 0

. Since

k

is non-zero,

k = -12

.

Estimated Time:1m 30s

Question 237Question →

In the standard $(x, y)$ coordinate plane, a circular running track is modeled by the equation $(x - 2)^2 + (y - 3)^2 = 25$ . A straight pathway, modeled by the line $y = 2x - 1$ , cuts through the track. What is the distance, in coordinate units, between the two points where the pathway intersects the track?

5

2\sqrt{5}

10

20

50

Show answer & explanation

Answer: $10$

Answer

The distance between the two intersection points is

10

.

The circle

(x - 2)^2 + (y - 3)^2 = 25

has a center of

(2, 3)

and a radius of

r = 5

. Since the line

y = 2x - 1

passes through

(2, 3)

, the line contains a diameter of the circle. The distance between the two intersection points is therefore the diameter of the circle, which is

2r = 10

.

Step-by-Step Solution

1

Identify the center and radius of the circle from its equation

(x - 2)^2 + (y - 3)^2 = 25

.

The center is

(2, 3)

and the radius is

r = \sqrt{25} = 5

.

To understand the geometry of the circle and find its radius.

2

Determine if the line

y = 2x - 1

passes through the center of the circle

(2, 3)

by substituting the coordinates into the equation.

3 = 2(2) - 1 \implies 3 = 3

, which is true.

If the line passes through the center, the distance between the intersection points is simply the diameter of the circle.

3

Calculate the diameter of the circle.

\text{Diameter} = 2r = 2(5) = 10

.

The distance between two opposite points on a circle passing through the center is equal to the diameter.

Key Concept

Systems of Linear and Non-Linear Equations

Question 238Question →

The quadratic equation $0.4x^2 - 2x + c = 0$ has two real solutions. If the difference between these two solutions is exactly $3$ , what is the value of $c$ ?

Show answer & explanation

Answer: 1.6

Answer

1.6

By using the relation between the roots of a quadratic equation

ax^2 + bx + c = 0

, we find that the difference of the roots is given by

|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}

. Substituting

a = 0.4

,

b = -2

, and the difference of

3

yields

3 = \frac{\sqrt{4 - 1.6c}}{0.4}

. Multiplying by

0.4

and squaring both sides gives

1.44 = 4 - 1.6c

. Solving this linear equation gives

c = 1.6

. We can verify this result by substituting

c = 1.6

back into the original equation:

0.4x^2 - 2x + 1.6 = 0

simplifies to

x^2 - 5x + 4 = 0

, which factors as

(x-1)(x-4) = 0

. The roots are

1

and

4

, and their difference is

4 - 1 = 3

.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation.

a = 0.4

,

b = -2

, and the constant term is

c

.

To apply formulas relating the roots to the coefficients of the equation.

2

Apply the formula for the difference of the roots.

|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}

The problem states the difference between the two solutions is

3

.

3

Substitute the known values into the formula and solve.

3 = \frac{\sqrt{(-2)^2 - 4(0.4)c}}{0.4} \Rightarrow 1.2 = \sqrt{4 - 1.6c}

To isolate the square root expression containing the unknown variable.

4

Square both sides and solve the linear equation for

c

.

1.44 = 4 - 1.6c \Rightarrow 1.6c = 2.56 \Rightarrow c = 1.6

To eliminate the square root and find the value of

c

.

Key Concept

Quadratic Equations and the Quadratic Formula

Estimated Time:1m 30s

Question 239Question →

For the quadratic equation $1.5x^2 - 5x + c = 0$ , where $c$ is a real constant, the equation has two non-real complex solutions. Which of the following inequalities represents all possible values of $c$ ?

c > \frac{25}{6}

c < \frac{25}{6}

c > -\frac{25}{6}

c > \frac{50}{3}

c < -\frac{25}{6}

Show answer & explanation

Answer: $c > \frac{25}{6}$

Answer

c > \frac{25}{6}

For a quadratic equation to have two non-real complex solutions, its discriminant must be negative. Substituting

a = 1.5

,

b = -5

, and the constant

c

into the discriminant formula

b^2 - 4ac < 0

gives

25 - 6c < 0

. Solving this inequality results in the requirement that the constant must be strictly greater than twenty-five sixths.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation

1.5x^2 - 5x + c = 0

in the standard form

ax^2 + bx + c = 0

.

a = 1.5

,

b = -5

, and

c = c

.

These coefficients are required to compute the discriminant.

2

Set up the condition for the quadratic equation to have two non-real complex solutions using the discriminant

\Delta = b^2 - 4ac

.

The discriminant must be strictly negative:

(-5)^2 - 4(1.5)(c) < 0

.

A quadratic equation has non-real complex solutions if and only if its discriminant is negative.

3

Simplify the inequality and solve for

c

.

25 - 6c < 0 \implies 25 < 6c \implies c > \frac{25}{6}

.

Isolating

c

yields the range of values that satisfy the condition.

Key Concept

A quadratic equation

ax^2 + bx + c = 0

has two non-real complex solutions if and only if its discriminant,

\Delta = b^2 - 4ac

, is strictly less than zero.

Estimated Time:1m 30s

Question 240Question →

A circle in the standard $(x, y)$ coordinate plane is described by the equation $x^2 + y^2 = 25$ . A line is described by the equation $3x - 4y = c$ , where $c$ is a positive constant. If the system of these two equations has exactly one real solution for $(x, y)$ , what is the value of $c$ ?

Show answer & explanation

Answer: 25

Answer

The value of

c

is

25

.

The correct answer is

25

. The equation

x^2 + y^2 = 25

represents a circle centered at

(0, 0)

with a radius of

5

. For the system of equations to have exactly one real solution, the line

3x - 4y = c

must be tangent to the circle. The perpendicular distance from the center

(0,0)

to the line

3x - 4y - c = 0

is given by

\frac{|3(0) - 4(0) - c|}{\sqrt{3^2 + (-4)^2}} = \frac{|c|}{5}

. Setting this distance equal to the radius of the circle yields

\frac{|c|}{5} = 5

, which gives

|c| = 25

. Since

c

is specified as a positive constant,

c

must be

25

.

Step-by-Step Solution

1

Find the center and radius of the circle.

Center is

(0, 0)

and radius is

r = 5

.

The circle equation

x^2 + y^2 = 25

is in the standard form

x^2 + y^2 = r^2

centered at the origin with radius

r = \sqrt{25} = 5

.

2

Set up the condition for tangency (exactly one real solution).

The perpendicular distance from the center

(0,0)

to the line

3x - 4y - c = 0

must equal the radius

5

.

A line intersects a circle at exactly one point if and only if the line is tangent to the circle.

3

Apply the point-to-line distance formula.

Distance

d = \frac{|3(0) - 4(0) - c|}{\sqrt{3^2 + (-4)^2}} = \frac{|c|}{5}

.

The formula for the distance from

(x_0, y_0)

to the line

Ax + By + C = 0

is

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

.

4

Solve for the positive constant

c

.

c = 25

Setting the distance

\frac{|c|}{5}

equal to the radius

5

gives

|c| = 25

. Since

c

is a positive constant,

c = 25

.

Key Concept

Determining conditions for tangency in a system of linear and circular equations.

Alternative Method

Instead of using the geometric distance formula, the system can be solved algebraically by substitution. Express

y

in terms of

x

from the linear equation:

y = \frac{3x - c}{4}

. Substitute this expression into the circle's equation:

x^2 + \left(\frac{3x - c}{4}\right)^2 = 25

. Expand the terms and multiply by

16

to clear the denominator:

16x^2 + 9x^2 - 6cx + c^2 = 400

, which simplifies to the quadratic equation

25x^2 - 6cx + (c^2 - 400) = 0

. For the system to have exactly one solution, this quadratic equation must have a discriminant equal to zero. Calculate the discriminant:

D = (-6c)^2 - 4(25)(c^2 - 400) = 36c^2 - 100c^2 + 40000 = -64c^2 + 40000 = 0

. Solving for

c

yields

64c^2 = 40000 \implies c^2 = 625

. Since

c

is a positive constant,

c = 25

.

Estimated Time:1m 30s

Intermediate Algebra

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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Alternative Method

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Common Mistakes

Alternative Method

Answer

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Key Concept

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