All practice questions

Question 1661Question →

In the $xy$ -plane, line $k$ has a slope of $-\frac{3}{4}$ and intersects the positive $x$ -axis at $(a, 0)$ and the positive $y$ -axis at $(0, b)$ . If the distance between the two intercept points $(a, 0)$ and $(0, b)$ is $15$ , what is the value of $a$ ?

9

12

15

16

20

Show answer & explanation

Answer: $12$

Answer

The value of

a

is

12

.

Using the slope formula between

(a,0)

and

(0,b)

, we find

m = -\frac{b}{a} = -\frac{3}{4}

, giving

b = \frac{3}{4}a

. Applying the distance formula yields

\sqrt{a^2 + b^2} = 15

, or

a^2 + b^2 = 225

. Substituting

b = \frac{3}{4}a

produces

a^2 + \frac{9}{16}a^2 = 225

, which simplifies to

\frac{25}{16}a^2 = 225

. Multiplying by 16 and dividing by 25 yields

a^2 = 144

, giving

a = 12

since

a > 0

.

Step-by-Step Solution

1

Express

b

in terms of

a

using the slope formula.

The slope of line

k

passing through

(a, 0)

and

(0, b)

is

m = \frac{b - 0}{0 - a} = -\frac{b}{a} = -\frac{3}{4}

, which simplifies to

b = \frac{3}{4}a

.

The slope of a line through two points

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Set up the distance equation between

(a, 0)

and

(0, b)

.

\sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} = 15

, so

a^2 + b^2 = 225

.

The distance formula between two points in the coordinate plane is derived from the Pythagorean theorem.

3

Substitute

b = \frac{3}{4}a

into the distance equation and solve for

a

.

a^2 + \left(\frac{3}{4}a\right)^2 = 225 \implies a^2 + \frac{9}{16}a^2 = 225 \implies \frac{25}{16}a^2 = 225 \implies a^2 = 144 \implies a = 12

.

Since

(a, 0)

is on the positive

x

-axis,

a

must be positive.

Key Concept

Slope and Distance in Coordinate Geometry

Estimated Time:1m 30s

Question 1662Question →

A rectangular plot of land has a perimeter of 56 meters and an area of 180 square meters. A border of uniform width $x$ meters is constructed around the inside edge of the plot, reducing the remaining inner area to 84 square meters. What is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The border width x is 2 meters.

Solving the system

L + W = 28

and

LW = 180

gives outer dimensions of 18 and 10 meters. The inner area equation

(18 - 2x)(10 - 2x) = 84

simplifies to

x^2 - 14x + 24 = 0

, which factors into

(x - 2)(x - 12) = 0

. Since

2x

must be less than the total width of 10 meters (

x < 5

), the extraneous root

x = 12

is discarded, leaving

x = 2

.

Step-by-Step Solution

1

Determine the outer dimensions of the plot.

Length = 18 meters, Width = 10 meters

The semi-perimeter is L + W = 56 / 2 = 28, and the area is L * W = 180. Forming the quadratic equation t^2 - 28t + 180 = 0 yields roots t = 18 and t = 10.

2

Write the quadratic expression for the inner area in terms of x.

(18 - 2x)(10 - 2x) = 84

A uniform border of width x along all four sides reduces both the length and width by 2x.

3

Expand and simplify the quadratic equation into standard form.

x^2 - 14x + 24 = 0

Expanding yields 180 - 56x + 4x^2 = 84. Subtracting 84 gives 4x^2 - 56x + 96 = 0, which divides by 4 to give x^2 - 14x + 24 = 0.

4

Factor the quadratic equation to solve for possible values of x.

x = 2 or x = 12

Factoring x^2 - 14x + 24 = 0 gives (x - 2)(x - 12) = 0.

5

Eliminate the extraneous root based on physical geometry constraints.

x = 2

The total width is 10 meters, so 2x must be less than 10 (x < 5). Thus, x = 12 is physically impossible and discarded.

Key Concept

Quadratic Equations, Factoring, and Extraneous Solutions

Question 1663Question →

Dataset

S

consists of

4n

distinct real numbers arranged in ascending order, where

n \ge 5

. The interquartile range of Dataset

S

is

IQR_S

, and its standard deviation is

\sigma_S

. A new dataset,

T

, is formed by transforming each value

x

in Dataset

S

into a corresponding value

y

as follows:

y = \begin{cases} x + k & \text{if } x \le Q_3 \\ x + 2k & \text{if } x > Q_3 \end{cases}

where

Q_3

is the third quartile (75th percentile) of Dataset

S

, and

k

is a positive constant. Which of the following statements must be true regarding the interquartile range

IQR_T

and standard deviation

\sigma_T

of Dataset

T

relative to Dataset

S

?

IQR_T = IQR_S

and

\sigma_T > \sigma_S

IQR_T > IQR_S

and

\sigma_T > \sigma_S

IQR_T = IQR_S

and

\sigma_T = \sigma_S

IQR_T > IQR_S

and

\sigma_T = \sigma_S

IQR_T < IQR_S

and

\sigma_T > \sigma_S

Show answer & explanation

Answer: $IQR_T = IQR_S$ and $\sigma_T > \sigma_S$

Answer

The interquartile range remains unchanged (

IQR_T = IQR_S

) while the standard deviation strictly increases (

\sigma_T > \sigma_S

).

The statement asserting that

IQR_T = IQR_S

and

\sigma_T > \sigma_S

is correct. Both

Q_1

and

Q_3

belong to the condition

x \le Q_3

, meaning both quartile values increase by exactly

k

. Thus,

IQR_T = (Q_3 + k) - (Q_1 + k) = Q_3 - Q_1 = IQR_S

. Meanwhile, the highest 25% of data values are shifted by an additional distance of

k

, increasing the overall spread of values around the mean, which strictly increases the standard deviation.

Step-by-Step Solution

1

Analyze the impact of the transformation on the first quartile (

Q_1

) and third quartile (

Q_3

).

Since

Q_1 < Q_3

, the value corresponding to

Q_1

is less than or equal to

Q_3

, so it is shifted to

Q_1 + k

. The value corresponding to

Q_3

satisfies

x \le Q_3

, so it is also shifted to

Q_3 + k

.

The definition of the piecewise rule adds

k

to all values less than or equal to

Q_3

.

2

Calculate the new interquartile range

IQR_T

.

IQR_T = (Q_3 + k) - (Q_1 + k) = Q_3 - Q_1 = IQR_S

.

The constant shift

k

cancels out when taking the difference between the upper and lower quartiles.

3

Analyze the impact on standard deviation

\sigma_T

.

The lower 75% of elements are shifted by

+k

, while the upper 25% of elements are shifted further by

+2k

. This increases the relative distance between upper-tail data points and the rest of the dataset.

A non-uniform shift that spreads the upper tail farther from the rest of the distribution increases total variation around the mean, resulting in

\sigma_T > \sigma_S

.

Key Concept

Effect of non-linear piecewise transformations on dispersion measures (IQR invariance under equal quartile shifts vs. standard deviation sensitivity to upper-tail displacement)

Question 1664Question →

In the $xy$ -plane, line $L$ passes through the points $(t, t+2)$ and $(3t-1, 2t+7)$ , where $t$ is a constant. Line $M$ is perpendicular to line $L$ and passes through the point $(4, -1)$ . If the $y$ -intercept of line $M$ is $11$ , what is the value of $t$ ?

-16

-\frac{14}{5}

\frac{8}{5}

-\frac{2}{7}

14

Show answer & explanation

Answer: $-16$

Answer

-16

The slope of line M is computed from its

y

-intercept

(0, 11)

and the point

(4, -1)

as

\frac{11 - (-1)}{0 - 4} = -3

. Because line L is perpendicular to line M, the slope of line L is the negative reciprocal of

-3

, which is

\frac{1}{3}

. Calculating the slope of line L using points

(t, t+2)

and

(3t-1, 2t+7)

gives

\frac{(2t+7)-(t+2)}{(3t-1)-t} = \frac{t+5}{2t-1}

. Setting

\frac{t+5}{2t-1} = \frac{1}{3}

leads to

3t + 15 = 2t - 1

, which simplifies to

t = -16

.

Step-by-Step Solution

1

Determine the slope of line M using its given points.

Line M passes through

(4, -1)

and its

y

-intercept

(0, 11)

. The slope

m_M = \frac{11 - (-1)}{0 - 4} = \frac{12}{-4} = -3

.

The slope of a line through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Determine the required slope of line L.

Since line L is perpendicular to line M, its slope

m_L = -\frac{1}{m_M} = -\frac{1}{-3} = \frac{1}{3}

.

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

3

Express the slope of line L in terms of

t

and solve for

t

.

Using points

(t, t+2)

and

(3t-1, 2t+7)

,

m_L = \frac{(2t+7) - (t+2)}{(3t-1) - t} = \frac{t+5}{2t-1}

. Setting

\frac{t+5}{2t-1} = \frac{1}{3}

yields

3(t+5) = 1(2t-1) \implies 3t + 15 = 2t - 1 \implies t = -16

.

Equating the algebraic slope expression to the numerical slope allows solving for the unknown parameter

t

.

Key Concept

Perpendicular Slopes and Coordinate Line Equations

Estimated Time:2m 0s

Question 1665Question →

Which of the following values of $x$ satisfy the inequality $|6 - 3x| > 9$ ? Select all that apply.

Select all that apply

-4

-2

1

3

6

Show answer & explanation

Answer: $-4$ ; $-2$ ; $6$

Answer

The values

-4

,

-2

, and

6

satisfy the inequality

|6 - 3x| > 9

.

The absolute value inequality

|6 - 3x| > 9

is equivalent to

6 - 3x > 9

or

6 - 3x < -9

. Solving these yields

x < -1

or

x > 5

. Among the given choices,

-4

,

-2

, and

6

fall within these solution ranges.

Step-by-Step Solution

1

Set up the compound linear inequalities from the absolute value inequality

|6 - 3x| > 9

.

6 - 3x > 9

or

6 - 3x < -9

An absolute value inequality of the form

|u| > c

(where

c > 0

) splits into

u > c

or

u < -c

.

2

Solve the first inequality

6 - 3x > 9

.

-3x > 3 \implies x < -1

Subtracting

6

gives

-3x > 3

. Dividing both sides by

-3

reverses the inequality sign to yield

x < -1

.

3

Solve the second inequality

6 - 3x < -9

.

-3x < -15 \implies x > 5

Subtracting

6

gives

-3x < -15

. Dividing both sides by

-3

reverses the inequality sign to yield

x > 5

.

4

Combine the solution sets and evaluate the given options.

The solution set consists of all real numbers where

x < -1

or

x > 5

. Therefore,

-4

,

-2

, and

6

are valid solutions.

Values

-4

and

-2

are strictly less than

-1

, while

6

is strictly greater than

5

.

Key Concept

Solving absolute value inequalities and reversing inequality signs when multiplying or dividing by negative numbers.

Question 1666Question →

A survey of $500$ university researchers evaluated their usage of three high-performance computing resources: Cloud Containers ( $C$ ), GPU Accelerators ( $G$ ), and Distributed Storage ( $D$ ). Exactly $50$ researchers use none of these three resources. The survey revealed that equal numbers of researchers use Cloud Containers and GPU Accelerators ( $|C| = |G| = 240$ ), while $210$ researchers use Distributed Storage ( $|D| = 210$ ). Exactly $40$ researchers use all three resources. Furthermore, the number of researchers who use both $C$ and $G$ but not $D$ is equal to the number who use both $G$ and $D$ but not $C$ , and this quantity is exactly twice the number of researchers who use both $C$ and $D$ but not $G$ .

How many researchers use GPU Accelerators ( $G$ ) ONLY?

32

72

104

112

144

Show answer & explanation

Answer: 72

Answer

72 researchers use GPU Accelerators (

G

) only.

The total number of researchers using at least one resource is

500 - 50 = 450

. Assigning

x

to the region using Cloud Containers and Distributed Storage only, the regions for C-and-G-only and G-and-D-only are each

2x

. Applying the three-set inclusion-exclusion principle gives

450 = 240 + 240 + 210 - (5x + 120) + 40

, which simplifies to

5x = 160

, so

x = 32

. The exclusive overlapping regions containing GPU Accelerators are

2(32) = 64

and

2(32) = 64

. Subtracting these overlaps along with the triple intersection (

40

) from

|G| = 240

yields

240 - (64 + 64 + 40) = 72

.

Step-by-Step Solution

1

Determine the total number of researchers using at least one resource.

|C \cup G \cup D| = 500 - 50 = 450

.

Subtract researchers who use none of the resources from the total surveyed.

2

Define variables for the two-set exclusive intersection regions.

Let

|C \cap D \cap G^c| = x

. Then

|C \cap G \cap D^c| = 2x

and

|G \cap D \cap C^c| = 2x

.

The problem states that the C-and-D-only region is half of the other two exclusive two-set intersection regions.

3

Express the full pairwise intersections including the triple intersection (

|C \cap G \cap D| = 40

).

|C \cap G| = 2x + 40

,

|G \cap D| = 2x + 40

, and

|C \cap D| = x + 40

.

Each pairwise intersection is the sum of its exclusive two-set intersection and the three-set intersection.

4

Apply the 3-Set Principle of Inclusion-Exclusion.

450 = 240 + 240 + 210 - [(2x + 40) + (2x + 40) + (x + 40)] + 40 \implies 450 = 690 - (5x + 120) + 40 \implies 450 = 610 - 5x \implies 5x = 160 \implies x = 32

.

Inclusion-exclusion formula:

|C \cup G \cup D| = |C| + |G| + |D| - (|C \cap G| + |G \cap D| + |C \cap D|) + |C \cap G \cap D|

.

5

Calculate the number of researchers using GPU Accelerators (

G

) only.

Exclusive

G

-only

= |G| - (|C \cap G \cap D^c| + |G \cap D \cap C^c| + |C \cap G \cap D|) = 240 - (64 + 64 + 40) = 240 - 168 = 72

.

Subtract all overlapping regions within set

G

from the total size of set

G

.

Key Concept

3-Set Inclusion-Exclusion Principle and Venn Diagram Region Partitioning

Question 1667Question →

A statistics instructor compiles the exam scores of 50 students in Dataset $S$ . The dataset has a range $R$ , an interquartile range $IQR$ , and a standard deviation $s$ . A revised dataset $T$ is created by multiplying each score in Dataset $S$ by $-1.5$ and then adding $25$ to each result. Which of the following expressions correctly states the measures of dispersion for Dataset $T$ in terms of the measures of dispersion for Dataset $S$ ?

\text{Range}_T = -1.5R + 25

,

IQR_T = -1.5 IQR + 25

, and

s_T = -1.5s + 25

\text{Range}_T = 1.5R + 25

,

IQR_T = 1.5 IQR + 25

, and

s_T = 1.5s + 25

\text{Range}_T = 1.5R

,

IQR_T = 1.5 IQR

, and

s_T = 1.5s

\text{Range}_T = 1.5R

,

IQR_T = 1.5 IQR

, and

s_T = 2.25s

\text{Range}_T = 1.5R

,

IQR_T = IQR

, and

s_T = 1.5s

Show answer & explanation

Answer: $\text{Range}_T = 1.5R$ , $IQR_T = 1.5 IQR$ , and $s_T = 1.5s$

Answer

The measures of dispersion for Dataset T are given by Range_T = 1.5R, IQR_T = 1.5 IQR, and s_T = 1.5s.

When a linear transformation

y = ax + b

is applied to a dataset, any measure of dispersion

D

(such as range, interquartile range, or standard deviation) transforms according to

D_y = |a| D_x

. The additive constant

b = 25

shifts all values equally and does not change the distances between data points, so it has no effect on dispersion. The multiplicative factor

a = -1.5

scales all distances by

|-1.5| = 1.5

. Therefore,

\text{Range}_T = 1.5R

,

IQR_T = 1.5 IQR

, and

s_T = 1.5s

.

Step-by-Step Solution

1

Analyze the general effect of a linear transformation y = ax + b on measures of dispersion.

Measures of dispersion (range, interquartile range, standard deviation) quantify spread and distance between data points. Adding a constant b shifts the whole distribution without altering distance between points, so b has zero impact on spread.

Additive constants shift position, not dispersion.

2

Determine the effect of multiplying by scale factor a = -1.5 on range, IQR, and standard deviation.

Multiplying data by a constant scale factor a scales all distance-based measures by |a|. Since dispersion measures must be non-negative, the scale factor used is |-1.5| = 1.5.

Distances between values scale by the absolute value of the multiplicative factor.

3

Combine the results to state Range_T, IQR_T, and s_T in terms of R, IQR, and s.

Range_T = 1.5R, IQR_T = 1.5 IQR, and s_T = 1.5s.

All three metrics scale by 1.5 and are unaffected by the +25 term.

Key Concept

Linear Transformations of Dispersion Metrics

Estimated Time:2m 0s

Question 1668Question →

A logistics company uses two delivery vehicles, Vehicle P and Vehicle Q, to transport cargo between two warehouses that are $240$ miles apart. Vehicle P travels at a constant average speed of $r$ miles per hour ( $r > 0$ ). Vehicle Q travels at a constant average speed that is $20$ miles per hour faster than that of Vehicle P. In addition to driving time, Vehicle P requires a flat setup time of $1$ hour before departure, while Vehicle Q requires a flat setup time of $2$ hours before departure.

Let $T_P(r)$ and $T_Q(r)$ represent the total elapsed time in hours (including setup time) required for Vehicle P and Vehicle Q to complete the trip, respectively.

Which of the following statements are true? Select all such statements.

Select all that apply

The total elapsed time for Vehicle P, in hours, as a function of its speed

r

, is given by

T_P(r) = \frac{240 + r}{r}

.

Vehicle P and Vehicle Q take the exact same total elapsed time to complete the trip when

r = 60

miles per hour.

For all positive speeds

r > 0

, Vehicle Q always requires strictly less total elapsed time than Vehicle P to complete the trip.

If

r = 30

miles per hour, the travel time (excluding setup time) for Vehicle Q is half the travel time (excluding setup time) for Vehicle P.

Doubling the speed of Vehicle P from

r = 40

miles per hour to

r = 80

miles per hour reduces its total elapsed time

T_P(r)

by

50\%

.

Show answer & explanation

Answer: The total elapsed time for Vehicle P, in hours, as a function of its speed $r$ , is given by $T_P(r) = \frac{240 + r}{r}$ .; Vehicle P and Vehicle Q take the exact same total elapsed time to complete the trip when $r = 60$ miles per hour.

Answer

The correct statements are the algebraic expression for Vehicle P's total time as a function of speed and the equal total elapsed time condition at a speed of 60 miles per hour.

The expression for Vehicle P's total time accurately combines

1

hour of setup time with

\frac{240}{r}

driving hours to get

\frac{240+r}{r}

. Setting

T_P(r) = T_Q(r)

yields the quadratic equation

r^2 + 20r - 4800 = 0

, which correctly solves to

r = 60

mph, at which point both vehicles require exactly

5

hours total.

Step-by-Step Solution

1

Model total elapsed time functions

T_P(r)

and

T_Q(r)

using setup time plus travel time.

T_P(r) = 1 + \frac{240}{r} = \frac{240 + r}{r}

and

T_Q(r) = 2 + \frac{240}{r + 20}

.

Total elapsed time is the sum of fixed pre-departure setup overhead and variable driving time.

2

Determine the speed

r

where total elapsed times are equal by setting

T_P(r) = T_Q(r)

.

1 + \frac{240}{r} = 2 + \frac{240}{r + 20} \implies \frac{240}{r} - \frac{240}{r + 20} = 1 \implies 240 \cdot 20 = r(r + 20) \implies r^2 + 20r - 4800 = 0

. Factoring gives

(r - 60)(r + 80) = 0

, so

r = 60

mph.

Solving the rational equation identifies the exact breakeven speed where higher travel efficiency balances extra setup time.

3

Evaluate the remaining candidate assertions against the derived model.

For

r > 60

,

T_Q(r) > T_P(r)

, invalidating the claim that Vehicle Q is always faster. At

r = 40

,

T_P(40) = 7

hours and

T_P(80) = 4

hours, giving a decrease of

\frac{3}{7} \approx 42.86\%

, invalidating the

50\%

reduction claim.

Fixed setup costs distort simple constant-proportion and percentage changes in total time.

Key Concept

Algebraic Modeling of Combined Time with Fixed Overhead and Variable Rates

Question 1669Question →

In $\triangle ABC$ , the length of side $AB$ is $13$ units, the length of side $BC$ is $21$ units, and the area of $\triangle ABC$ is $126$ square units. If $\angle ABC$ is an acute angle, what is the length of side $AC$ ?

15

18

20

22

25

Show answer & explanation

Answer: 20

Answer

20

The area of triangle

ABC

is

\frac{1}{2} \times 21 \times h = 126

, which yields an altitude

AH = 12

perpendicular to side

BC

. In right triangle

ABH

, the base segment

BH = \sqrt{13^2 - 12^2} = 5

. Because angle

ABC

is acute,

H

falls between

B

and

C

, making

HC = 21 - 5 = 16

. Finally, in right triangle

AHC

,

AC = \sqrt{12^2 + 16^2} = \sqrt{400} = 20

. Therefore, 20 is the correct answer.

Step-by-Step Solution

1

Calculate the altitude

h

from vertex

A

to base

BC

.

h = 12

units.

The area formula for a triangle is

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

. Substituting

BC = 21

and

\text{Area} = 126

yields

126 = \frac{1}{2} \times 21 \times h \implies h = 12

.

2

Find the length of segment

BH

where

H

is the foot of the altitude on

BC

.

BH = 5

units.

In right triangle

\triangle ABH

,

AB = 13

and

AH = 12

. By the Pythagorean theorem,

BH = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5

.

3

Determine the length of segment

HC

.

HC = 16

units.

Since

\angle ABC

is an acute angle, point

H

lies on segment

BC

. Therefore,

HC = BC - BH = 21 - 5 = 16

.

4

Calculate the length of side

AC

.

AC = 20

units.

In right triangle

\triangle AHC

,

AH = 12

and

HC = 16

. By the Pythagorean theorem,

AC = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20

.

Key Concept

Triangles: Altitude, Area, and Pythagorean Theorem

Estimated Time:1m 30s

Question 1670Question →

A market research firm surveyed a group of $200$ consumers regarding their active subscriptions to three digital services: FilmStream ( $F$ ), AudioVibe ( $A$ ), and PrintPlus ( $P$ ). Every surveyed consumer subscribed to at least one of the three services. The survey gathered the following information:
- $120$ consumers subscribed to FilmStream
- $100$ consumers subscribed to AudioVibe
- $75$ consumers subscribed to PrintPlus
- $45$ consumers subscribed to both FilmStream and AudioVibe
- $35$ consumers subscribed to both AudioVibe and PrintPlus
- $30$ consumers subscribed to both FilmStream and PrintPlus

How many consumers subscribed to all three digital services?

15

25

35

55

95

Show answer & explanation

Answer: $15$

Answer

15 consumers subscribed to all three digital services.

According to the 3-set inclusion-exclusion principle,

|F \cup A \cup P| = |F| + |A| + |P| - (|F \cap A| + |A \cap P| + |F \cap P|) + |F \cap A \cap P|

. Substituting the given values gives

200 = 120 + 100 + 75 - (45 + 35 + 30) + |F \cap A \cap P|

, which simplifies to

200 = 185 + |F \cap A \cap P|

. Subtracting

185

from

200

yields

15

.

Step-by-Step Solution

1

Apply the Principle of Inclusion-Exclusion for three sets.

|F \cup A \cup P| = |F| + |A| + |P| - (|F \cap A| + |A \cap P| + |F \cap P|) + |F \cap A \cap P|

This formula accounts for elements counted multiple times across overlapping sets.

2

Substitute the known values into the equation.

200 = 120 + 100 + 75 - (45 + 35 + 30) + |F \cap A \cap P|

Every consumer subscribes to at least one service, so

|F \cup A \cup P| = 200

.

3

Simplify the numerical terms on the right side of the equation.

200 = 295 - 110 + |F \cap A \cap P| \implies 200 = 185 + |F \cap A \cap P|

Sum of individual sets is

295

, and sum of pairwise intersections is

110

.

4

Solve for the target three-set intersection

|F \cap A \cap P|

.

|F \cap A \cap P| = 200 - 185 = 15

Isolating the variable gives the number of consumers in all three sets.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Estimated Time:1m 30s

Question 1671Question →

In the $xy$ -coordinate plane, point $P$ lies in the first quadrant such that the distance from the origin $O(0,0)$ to $P$ is $4\sqrt{3}$ , and line segment $OP$ forms a $30^\circ$ angle with the positive $x$ -axis. A circle centered at point $P$ with radius $10$ intersects the $y$ -axis at two distinct points. Which of the following statements are true? Select all such statements.

Select all that apply

The

x

-coordinate of point

P

is

6

.

The length of the chord formed by the circle on the

y

-axis is

16

.

The

x

-coordinate of point

P

is

2\sqrt{3}

.

The length of the chord formed by the circle on the

y

-axis is

4\sqrt{34}

.

The perpendicular distance from point

P

to the

y

-axis is

2\sqrt{3}

.

Show answer & explanation

Answer: The $x$ -coordinate of point $P$ is $6$ .; The length of the chord formed by the circle on the $y$ -axis is $16$ .

Answer

The correct statements are that the

x

-coordinate of point

P

is

6

and the length of the chord formed by the circle on the

y

-axis is

16

.

The horizontal leg of the

30^\circ-60^\circ-90^\circ

triangle gives the

x

-coordinate

x_P = 4\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6

. The perpendicular distance from center

P

to the

y

-axis is

6

. Using the Pythagorean theorem with circle radius

10

and distance

6

, half the chord length is

\sqrt{10^2 - 6^2} = 8

, yielding a full chord length of

16

.

Step-by-Step Solution

1

Determine the coordinates of point P using special right triangle ratio (30-60-90)

Point

P = (6, 2\sqrt{3})

Segment

OP

forms the hypotenuse of length

4\sqrt{3}

in a right triangle with the

x

-axis. The horizontal component is

x_P = 4\sqrt{3} \cos(30^\circ) = 4\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6

. The vertical component is

y_P = 4\sqrt{3} \sin(30^\circ) = 4\sqrt{3} \cdot \frac{1}{2} = 2\sqrt{3}

.

2

Find the perpendicular distance from P to the y-axis

Perpendicular distance

= 6

The

y

-axis is defined by the line equation

x = 0

. The perpendicular distance from

P(6, 2\sqrt{3})

to

x = 0

is

|6 - 0| = 6

.

3

Apply the Pythagorean theorem to calculate the chord length on the y-axis

Total chord length

= 16

A perpendicular segment from center

P

to the chord bisects the chord at midpoint

M(0, 2\sqrt{3})

. In right triangle

\triangle PMQ

formed by radius

PQ = 10

, distance

PM = 6

, and half-chord

MQ = d

:

d^2 + PM^2 = PQ^2 \implies d^2 + 6^2 = 10^2 \implies d^2 = 64 \implies d = 8

. The full chord length is

2d = 16

.

Key Concept

Properties of 30-60-90 special right triangles and application of the Pythagorean theorem to circle chord geometry

Question 1672Question →

A biotechnology consortium surveyed $180$ research laboratories regarding their implementation of three diagnostic platforms: Platform $A$ , Platform $B$ , and Platform $C$ .

- $95$ laboratories use Platform $A$ .
- $85$ laboratories use Platform $B$ .
- $80$ laboratories use Platform $C$ .
- $15$ laboratories use all three platforms.
- $20$ laboratories use none of the three platforms.
- The number of laboratories that use Platform $A$ and Platform $B$ but NOT Platform $C$ is equal to the number of laboratories that use Platform $B$ and Platform $C$ but NOT Platform $A$ .
- $30$ laboratories use Platform $A$ and Platform $C$ but NOT Platform $B$ .

How many laboratories use Platform $A$ ONLY?

15

20

30

45

50

Show answer & explanation

Answer: 30

Answer

30 laboratories use Platform A only.

Subtracting the 20 laboratories that use none of the platforms from the total population of 180 gives a union size of 160. Applying the inclusion-exclusion formula yields

160 = (95 + 85 + 80) - (x + 15 + x + 15 + 30 + 15) + 15

, which resolves to

x = 20

laboratories that use Platforms A and B only. Subtracting all overlap regions from Platform A's total gives

95 - 20 - 30 - 15 = 30

laboratories that use Platform A only.

Step-by-Step Solution

1

Calculate the total number of laboratories that use at least one platform.

|A \cup B \cup C| = 180 - 20 = 160

Laboratories using at least one platform equal the total surveyed minus those using none.

2

Set up the Principle of Inclusion-Exclusion for three sets.

|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |A \cap C|) + |A \cap B \cap C|

Standard formula relating total union size to individual set sizes and intersections.

3

Express pairwise intersections in terms of non-overlapping regions and solve for the unknown region

x

.

Let

x

be the number of labs using Platform

A

and

B

only. Given

A \cap C \text{ only} = 30

and

A \cap B \cap C = 15

, we have:

160 = (95 + 85 + 80) - [(x + 15) + (x + 15) + (30 + 15)] + 15

160 = 260 - (2x + 75) + 15 = 200 - 2x \implies 2x = 40 \implies x = 20

Substituting the given equality of regions allows finding

x = 20

.

4

Calculate the number of laboratories that use Platform

A

only.

\text{Platform } A \text{ only} = |A| - (A \cap B \text{ only}) - (A \cap C \text{ only}) - (A \cap B \cap C) = 95 - 20 - 30 - 15 = 30

Subtracting all overlap regions containing Platform A from the total Platform A count gives the single-region count.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Region Partitioning

Question 1673Question →

If $x$ is an integer that satisfies both $|2x - 1| \le 9$ and $\frac{x + 2}{-3} > -2$ , how many possible values of $x$ exist?

1

4

7

8

9

Show answer & explanation

Answer: 8

Answer

There are 8 possible integer values for x.

Solving

|2x - 1| \le 9

yields

-4 \le x \le 5

. Solving

\frac{x + 2}{-3} > -2

requires flipping the inequality sign when multiplying by

-3

, which gives

x < 4

. Combining both inequalities yields

-4 \le x < 4

. The integers satisfying this compound inequality are

-4, -3, -2, -1, 0, 1, 2, 3

, making a total of 8 possible integer values.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x - 1| \le 9

-9 \le 2x - 1 \le 9 \implies -8 \le 2x \le 10 \implies -4 \le x \le 5

An absolute value inequality of the form

|u| \le k

expands to

-k \le u \le k

.

2

Solve the linear inequality

\frac{x + 2}{-3} > -2

x + 2 < 6 \implies x < 4

Multiplying both sides of an inequality by a negative quantity (

-3

) requires reversing the inequality sign from

>

to

<

.

3

Find the intersection of the two solution sets

-4 \le x < 4

The integer

x

must satisfy both

-4 \le x \le 5

and

x < 4

simultaneously.

4

Count the integer values satisfying

-4 \le x < 4

The integers are

-4, -3, -2, -1, 0, 1, 2, 3

, which totals 8 values.

Counting all integers from

-4

up to (but not including)

4

gives 8 valid integers.

Key Concept

Solving systems of linear inequalities involving absolute values and negative multipliers

Estimated Time:1m 30s

Question 1674Question →

In the $xy$ -plane, line segment $AB$ has endpoints $A(-3, 1)$ and $B(1, 9)$ . Line $N$ is the perpendicular bisector of segment $AB$ . Which of the following statements must be true? Select all that apply.

Select all that apply

Line

N

has a slope of

-\frac{1}{2}

.

Line

N

passes through the point

(3, 3)

.

Line

N

has a

y

-intercept of

(0, 5)

.

The length of line segment

AB

is

10

.

The midpoint of line segment

AB

lies in Quadrant I.

Show answer & explanation

Answer: Line $N$ has a slope of $-\frac{1}{2}$ .; Line $N$ passes through the point $(3, 3)$ .

Answer

The statement that Line

N

has a slope of

-\frac{1}{2}

and the statement that Line

N

passes through the point

(3, 3)

are both true.

The slope of segment

AB

is

2

, making the perpendicular slope

-\frac{1}{2}

. The midpoint of

AB

is

(-1, 5)

. Using point-slope form with

M(-1, 5)

and slope

-\frac{1}{2}

, Line

N

has equation

y = -\frac{1}{2}x + \frac{9}{2}

. Evaluating at

x = 3

gives

y = 3

, so Line

N

passes through

(3, 3)

.

Step-by-Step Solution

1

Calculate the slope of line segment

AB

slope

m_{AB} = \frac{9 - 1}{1 - (-3)} = \frac{8}{4} = 2

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Determine the slope of the perpendicular bisector (Line

N

)

slope

m_N = -\frac{1}{m_{AB}} = -\frac{1}{2}

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Find the midpoint of line segment

AB

Midpoint

M = \left(\frac{-3 + 1}{2}, \frac{1 + 9}{2}\right) = (-1, 5)

A perpendicular bisector must pass through the midpoint of the segment it bisects.

4

Find the equation of Line

N

and test given points

Line

N

:

y - 5 = -\frac{1}{2}(x - (-1)) \implies y = -\frac{1}{2}x + \frac{9}{2}

. Testing

(3, 3)

:

3 = -\frac{1}{2}(3) + \frac{9}{2} = \frac{6}{2} = 3

(True). Testing

y

-intercept (

x=0

):

y = 4.5 \neq 5

.

Point-slope form

y - y_1 = m(x - x_1)

defines the line.

5

Calculate the length of segment

AB

AB = \sqrt{(1 - (-3))^2 + (9 - 1)^2} = \sqrt{4^2 + 8^2} = \sqrt{80} = 4\sqrt{5} \approx 8.94

The distance formula between two points is

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

.

Key Concept

Coordinate Geometry: Perpendicular Bisectors, Slopes, Midpoints, and Distance

Estimated Time:1m 45s

Question 1675Question →

A committee is to be selected from a group of $5$ distinct people: $P$ , $Q$ , $R$ , $S$ , and $T$ . Which of the following statements regarding the possible selections or arrangements of people from this group are true? Select all that apply.

Select all that apply

The number of different

2

-person committees that can be formed from the group is

10

.

The number of different

3

-person committees that can be formed from the group is

10

.

The number of different ways to assign

2

distinct officer roles (President and Vice President) from the group is

10

.

The total number of different ways to arrange all

5

people in a single line is

25

.

The number of different ways to select and arrange

3

of the

5

people in a line is

60

.

Show answer & explanation

Answer: The number of different $2$ -person committees that can be formed from the group is $10$ .; The number of different $3$ -person committees that can be formed from the group is $10$ .; The number of different ways to select and arrange $3$ of the $5$ people in a line is $60$ .

Answer

The statements confirming that

10

different

2

-person committees can be formed,

10

different

3

-person committees can be formed, and

60

different

3

-person linear arrangements can be formed are all correct.

Selecting committees without specific roles requires combinations

\binom{n}{k}

, giving

\binom{5}{2} = 10

and

\binom{5}{3} = 10

. Arranging 3 people in ordered positions requires permutations

P(5,3) = 5 \times 4 \times 3 = 60

. Thus, all three corresponding statements are correct.

Step-by-Step Solution

1

Evaluate the 2-person committee selection statement.

\binom{5}{2} = \frac{5 \times 4}{2} = 10

Selection of a committee without specific roles is an unordered combination.

2

Evaluate the 3-person committee selection statement.

\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10

Choosing 3 items out of 5 yields the same number of outcomes as choosing 2 items out of 5.

3

Evaluate the 2-person officer assignment statement.

P(5,2) = 5 \times 4 = 20

Assigning distinct officer positions means order matters, requiring permutations rather than combinations.

4

Evaluate the 5-person line arrangement statement.

5! = 120

The total number of linear arrangements of 5 distinct objects is given by 5 factorial.

5

Evaluate the 3-person line arrangement statement.

P(5,3) = 5 \times 4 \times 3 = 60

Ordering 3 out of 5 people in a line uses the fundamental counting principle with decreasing choices per slot.

Key Concept

Distinguishing between combinations (where selection order does not matter) and permutations (where selection or position order does matter).

Question 1676Question →

A dataset $S$ consists of $7$ distinct positive integers arranged in ascending order: $x_1, x_2, x_3, x_4, x_5, x_6, x_7$ . The arithmetic mean of the entire dataset is $28$ , and the median is $25$ . The arithmetic mean of the $3$ smallest integers in $S$ is $12$ . If $M$ is the maximum possible value of $x_7$ and $m$ is the minimum possible value of $x_7$ , what is the value of $M - m$ ?

32

36

37

40

52

Show answer & explanation

Answer: 36

Answer

The value of

M - m

is

36

.

The sum of all

7

distinct positive integers is

7 \times 28 = 196

. Since the dataset is ordered and has

7

elements, the median is

x_4 = 25

. The sum of the smallest

3

integers is

3 \times 12 = 36

. Therefore, the sum of the remaining three integers

x_5 + x_6 + x_7 = 196 - 36 - 25 = 135

.

To maximize

x_7

,

x_5

and

x_6

must be as small as possible. Since all elements are distinct integers greater than

x_4 = 25

, the smallest possible values are

x_5 = 26

and

x_6 = 27

. Thus,

M = 135 - 26 - 27 = 82

.

To minimize

x_7

,

x_5

,

x_6

, and

x_7

must be as close together as possible while preserving

25 < x_5 < x_6 < x_7

. Dividing

135

by

3

gives

45

. The consecutive integers centered around

45

are

44, 45, 46

, which sum to

135

and satisfy all inequalities. Thus,

m = 46

.

The difference

M - m = 82 - 46 = 36

, which corresponds to the value

36

.

Step-by-Step Solution

1

Calculate the total sum of all 7 integers in dataset S.

Total sum =

7 \times 28 = 196

.

The arithmetic mean of

n

numbers is the total sum divided by

n

.

2

Identify the median value and the sum of the smallest 3 integers.

Median

x_4 = 25

, and

x_1 + x_2 + x_3 = 3 \times 12 = 36

.

For an odd number of ordered elements (

7

), the middle term

x_4

is the median. The mean of the first 3 terms gives their sum.

3

Determine the sum of the top 3 integers

(x_5 + x_6 + x_7)

.

x_5 + x_6 + x_7 = 196 - 36 - 25 = 135

.

Subtracting

x_1 + x_2 + x_3

and

x_4

from the total sum leaves the sum of the remaining three elements.

4

Calculate the maximum possible value

M

of

x_7

.

M = 82

.

To maximize

x_7

, minimize

x_5

and

x_6

. Since elements are distinct integers and

x_4 = 25

, the minimum values are

x_5 = 26

and

x_6 = 27

. Thus

x_7 = 135 - 26 - 27 = 82

.

5

Calculate the minimum possible value

m

of

x_7

.

m = 46

.

To minimize

x_7

, maximize

x_5

and

x_6

such that

25 < x_5 < x_6 < x_7

and

x_5 + x_6 + x_7 = 135

. Setting

x_5 = 44, x_6 = 45, x_7 = 46

gives

44 + 45 + 46 = 135

, maintaining strict inequalities.

6

Compute

M - m

.

M - m = 82 - 46 = 36

.

Subtract the minimum possible value of

x_7

from its maximum possible value.

Key Concept

Measures of Central Tendency with Extreme Value Optimization

Estimated Time:2m 30s

Question 1677Question →

A water purification facility uses a primary filtration system and a secondary filtration system to process untreated water. The primary system operates at a constant rate that is $25\%$ faster than the secondary system. Working together at their normal constant rates, both systems can process a full reservoir of $36,000$ gallons in $8$ hours.

On a day when the primary system operates at only $80\%$ of its normal rate due to maintenance while the secondary system operates at its normal rate, both systems work together for $6$ hours. At that point, the primary system is shut down completely. How many additional hours will it take the secondary system, working alone at its normal rate, to process the remainder of the reservoir?

4.5

hours

5.0

hours

6.0

hours

7.2

hours

8.0

hours

Show answer & explanation

Answer: $6.0$ hours

Answer

6.0

hours

The correct answer of

6.0

hours is derived by establishing that the combined normal processing rate is

4,500

gal/hr. Since the primary system is

25\%

faster than the secondary system (

P = 1.25S

), the secondary rate is

2,000

gal/hr and the primary rate is

2,500

gal/hr. At

80\%

efficiency, the primary system operates at

2,000

gal/hr, making the joint rate

4,000

gal/hr. In

6

hours,

24,000

gallons are processed, leaving

12,000

gallons. The secondary system working alone at

2,000

gal/hr processes the remaining volume in exactly

6.0

hours.

Step-by-Step Solution

1

Determine the combined normal operating rate and set up individual rates.

Combined rate =

4,500

gal/hr; Secondary rate =

2,000

gal/hr; Primary rate =

2,500

gal/hr.

The combined rate is

\frac{36,000\text{ gallons}}{8\text{ hours}} = 4,500\text{ gal/hr}

. Let

S

be the secondary rate. The primary rate is

1.25S

. Thus,

S + 1.25S = 2.25S = 4,500

, yielding

S = 2,000\text{ gal/hr}

and

P = 2,500\text{ gal/hr}

.

2

Calculate the reduced primary rate and the total water processed in the first 6 hours.

24,000

gallons processed in the first 6 hours.

During maintenance, the primary system operates at

80\%

of

2,500\text{ gal/hr}

, which is

0.80 \times 2,500 = 2,000\text{ gal/hr}

. The combined rate during this period is

2,000 + 2,000 = 4,000\text{ gal/hr}

. Over 6 hours, the volume processed is

4,000 \times 6 = 24,000\text{ gallons}

.

3

Find the remaining volume of water to process.

12,000

gallons remaining.

Subtract the volume processed from the total reservoir capacity:

36,000 - 24,000 = 12,000\text{ gallons}

.

4

Calculate the time required for the secondary system alone to process the remainder.

6.0

hours.

Divide the remaining volume by the normal secondary system rate

S = 2,000\text{ gal/hr}

:

\frac{12,000\text{ gallons}}{2,000\text{ gal/hr}} = 6.0\text{ hours}

.

Key Concept

Linear rate modeling and percentage rate adjustment in combined work problems

Estimated Time:2m 30s

Question 1678Question →

In the coordinate plane, triangle $JKL$ has vertices $J(0,0)$ , $K(14,0)$ , and $L(x, 12)$ , where $x > 0$ . If the perimeter of triangle $JKL$ is $42$ units and the length of side $JL$ is less than the length of side $KL$ , what is the value of $x$ ?

4

5

7

9

12

Show answer & explanation

Answer: 5

Answer

5

The correct answer is 5. Using the distance formula, the base length

JK = 14

. Expressing the side lengths as

JL = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + 144}

, setting the perimeter

JK + JL + KL = 42

leads to the quadratic equation

x^2 - 14x + 45 = 0

. This yields

x = 5

or

x = 9

. Evaluating the sides for

x = 5

gives

JL = 13

and

KL = 15

, which satisfies the problem condition

JL < KL

.

Step-by-Step Solution

1

Calculate the length of base

JK

using the distance formula.

The distance between

J(0,0)

and

K(14,0)

is

14 - 0 = 14

units.

Base

JK

lies along the horizontal

x

-axis.

2

Express side lengths

JL

and

KL

in terms of

x

.

JL = \sqrt{(x-0)^2 + (12-0)^2} = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + (12-0)^2} = \sqrt{(14-x)^2 + 144}

.

Apply the distance formula between coordinates

L(x,12)

and vertices

J

and

K

.

3

Set up and solve the perimeter equation.

14 + \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 42 \implies \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 28

. Squaring both sides systematically yields

x^2 - 14x + 45 = 0

, giving roots

x = 5

and

x = 9

.

The total perimeter is given as 42 units.

4

Apply the constraint

JL < KL

to choose the valid root.

For

x = 5

,

JL = \sqrt{25 + 144} = 13

and

KL = \sqrt{81 + 144} = 15

, satisfying

JL < KL

.

For

x = 9

,

JL = 15

and

KL = 13

, which violates

JL < KL

.

Key Concept

Coordinate Geometry and Triangle Side Length Constraints

Estimated Time:1m 30s

Question 1679Question →

A security code consists of three distinct digits chosen from the non-zero digits $1$ through $9$ . If the first digit must be odd and the third digit must be even, how many such three-digit security codes can be formed?

16

70

140

160

180

Show answer & explanation

Answer: 140

Answer

140

To find the number of three-digit codes with distinct digits from

1

through

9

satisfying the constraints, count the options for each slot: the first position has

5

odd options (

1, 3, 5, 7, 9

), the third position has

4

even options (

2, 4, 6, 8

), and the middle position has

9 - 2 = 7

remaining options. By the Fundamental Counting Principle, multiplying these options gives

5 \times 7 \times 4 = 140

.

Step-by-Step Solution

1

Determine the number of possibilities for the first digit.

5 choices (the odd digits:

1, 3, 5, 7, 9

).

The question specifies that the first digit must be odd.

2

Determine the number of possibilities for the third digit.

4 choices (the even digits:

2, 4, 6, 8

).

The question specifies that the third digit must be even.

3

Determine the number of possibilities for the middle (second) digit.

7 choices.

There are

9

total non-zero digits (

1

through

9

). Since

2

distinct digits have already been used for the first and third positions,

9 - 2 = 7

digits remain available for the middle position.

4

Apply the Fundamental Counting Principle to find the total number of codes.

5 \times 7 \times 4 = 140

.

The total number of sequential independent choices is found by multiplying the number of options for each position.

Key Concept

Fundamental Counting Principle with Restricted Positions and Distinct Elements

Estimated Time:45s

Question 1680Question →

Consider the quadratic equation $x^2 - 2(k + 1)x + (k^2 + 4k + 1) = 0$ , where $k$ is a real constant. Let $\alpha$ and $\beta$ be the real roots of this equation. If $\alpha^2 + \beta^2 = 18$ , what is the sum of all possible values of $k$ ?

-2\sqrt{2}

0

2\sqrt{2}

2 - 2\sqrt{3}

-4\sqrt{2}

Show answer & explanation

Answer: $-2\sqrt{2}$

Answer

The sum of all possible real values of

k

is

-2\sqrt{2}

.

Using Vieta's formulas,

\alpha + \beta = 2(k + 1)

and

\alpha\beta = k^2 + 4k + 1

. Expanding

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

gives

2k^2 + 2

. Setting

2k^2 + 2 = 18

yields

k^2 = 8

, so

k = \pm 2\sqrt{2}

. Furthermore, for the roots

\alpha

and

\beta

to be real numbers, the discriminant

\Delta = [-2(k+1)]^2 - 4(k^2+4k+1) = -8k

must be non-negative, requiring

k \le 0

. Only

k = -2\sqrt{2}

satisfies this requirement. Hence, the sum of all valid values of

k

is

-2\sqrt{2}

.

Step-by-Step Solution

1

Apply Vieta's formulas to express the sum and product of roots in terms of

k

.

\alpha + \beta = 2(k + 1)

and

\alpha\beta = k^2 + 4k + 1

.

For any quadratic equation

x^2 - bx + c = 0

, the sum of roots is

b

and the product of roots is

c

.

2

Express

\alpha^2 + \beta^2

in terms of

k

using the algebraic identity

(\alpha + \beta)^2 - 2\alpha\beta

.

\alpha^2 + \beta^2 = [2(k + 1)]^2 - 2(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 2k^2 - 8k - 2 = 2k^2 + 2

.

Expanding and simplifying allows us to relate the given sum of squares to the parameter

k

.

3

Set the simplified expression equal to

18

and solve for

k

.

2k^2 + 2 = 18 \implies 2k^2 = 16 \implies k^2 = 8 \implies k = \pm 2\sqrt{2}

.

This yields two candidate real values for the parameter

k

.

4

Analyze the discriminant

\Delta

to enforce the condition that roots

\alpha

and

\beta

are real numbers.

\Delta = [-2(k + 1)]^2 - 4(1)(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 4(k^2 + 4k + 1) = -8k

. For real roots,

\Delta \ge 0 \implies -8k \ge 0 \implies k \le 0

.

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

5

Filter candidate values of

k

against the condition

k \le 0

.

k = -2\sqrt{2}

satisfies

k \le 0

(yielding

\Delta = 16\sqrt{2} > 0

), while

k = 2\sqrt{2}

fails

k \le 0

(yielding

\Delta = -16\sqrt{2} < 0

). Thus,

k = -2\sqrt{2}

is the unique valid value.

Values of

k

that make the discriminant negative produce non-real complex roots and must be discarded as extraneous.

Key Concept

Quadratic Vieta's formulas combined with discriminant non-negativity constraint for real roots

Estimated Time:2m 30s

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