Algebra and Functions Practice Questions — GMAT — Page 8

Question 141Question →

The roots of the quadratic polynomial $P(x) = x^2 + bx + c$ are $r_1$ and $r_2$ , where $b$ and $c$ are constants. If $r_1 + 2$ and $r_2 + 2$ are the roots of the quadratic equation $x^2 - 10x + 21 = 0$ , what is the value of $c$ ?

Show answer & explanation

Answer: 5

Answer

The value of

c

is 5.

Factoring

x^2 - 10x + 21 = 0

yields roots

3

and

7

. Because these roots represent

r_1 + 2

and

r_2 + 2

, subtracting

2

from each root gives the original roots

r_1 = 1

and

r_2 = 5

. In the monic polynomial

P(x) = x^2 + bx + c

, the constant coefficient

c

equals the product of the roots

r_1 \times r_2 = 1 \times 5 = 5

.

Step-by-Step Solution

1

Solve for the roots of the given equation

x^2 - 10x + 21 = 0

The roots are

3

and

7

.

Factoring the quadratic expression yields

(x - 3)(x - 7) = 0

, so

x = 3

or

x = 7

.

2

Determine the roots

r_1

and

r_2

of

P(x)

r_1 = 1

and

r_2 = 5

.

Since the roots of the second equation are shifted by

+2

, we set

r_1 + 2 = 3 \implies r_1 = 1

and

r_2 + 2 = 7 \implies r_2 = 5

.

3

Calculate the constant term

c

c = 5

.

By Vieta's formulas, for any monic quadratic polynomial

x^2 + bx + c

, the constant term

c

is the product of the roots

r_1 r_2 = 1 \times 5 = 5

.

Key Concept

Root transformation of quadratic equations and relationship between roots and coefficients via Vieta's formulas.

Question 142Question →

A commercial bakery operates two automated production lines, Line A and Line B, to fulfill a production order for $B$ boxes of goods:

- Line A operates at a constant rate of $r$ boxes per hour ( $r > 0$ ) for $t$ hours.
- Line B operates at a constant rate that is $25\%$ higher than Line A's rate and works for $2$ hours longer than Line A.
- Line B produces twice as many boxes as Line A during their respective operational periods.
- The total contract volume $B$ is the sum of the boxes produced by Line A and Line B.

Which of the following statements MUST be true regarding this algebraic model? (Select ALL that apply.)

Select all that apply

Line A operates for

t = \frac{10}{3}

hours (3 hours and 20 minutes).

The total contract volume

B

is equal to

10r

boxes.

The overall average production rate over the combined total operating time of both lines is

1.125r

boxes per hour.

If both lines were to work together simultaneously to complete the total order of

B

boxes, the required time would equal the sum of their individual operating times (

\frac{26}{3}

hours).

Dividing both sides of the equation

1.25r(t + 2) = 2rt

by

r

loses

r = 0

as a physically valid non-zero rate solution.

Show answer & explanation

Answer: Line A operates for $t = \frac{10}{3}$ hours (3 hours and 20 minutes).; The total contract volume $B$ is equal to $10r$ boxes.

Answer

The correct statements are the one establishing Line A's operating time as 10/3 hours and the one establishing the total volume as 10r boxes.

The model yields

t = \frac{10}{3}

hours by equating

1.25r(t + 2) = 2rt

and dividing by

r

. Substituting

t = \frac{10}{3}

into the total output expression

B = rt + 1.25r(t + 2)

gives

B = \frac{10}{3}r + \frac{20}{3}r = 10r

. Thus, both the operating time statement for Line A and the total contract volume statement are mathematically true.

Step-by-Step Solution

1

Formulate algebraic expressions for the output of each production line.

Line A output:

Q_A = r \cdot t

. Line B rate is

1.25r

and operating time is

t + 2

, so Line B output:

Q_B = 1.25r(t + 2)

.

Word problem translation requires modeling quantities produced as rate multiplied by time.

2

Set up the relational equation based on the condition that Line B produces twice as much as Line A.

1.25r(t + 2) = 2(r \cdot t) \implies 1.25r(t + 2) = 2rt

.

Translates the given relationship

Q_B = 2 Q_A

into a single variable equation for

t

.

3

Solve for

t

by dividing out non-zero

r

.

1.25(t + 2) = 2t \implies 1.25t + 2.5 = 2t \implies 0.75t = 2.5 \implies t = \frac{2.5}{0.75} = \frac{10}{3}

hours.

Since rate

r > 0

, dividing by

r

isolates

t

.

4

Calculate individual outputs and total contract volume

B

.

Q_A = r \cdot \frac{10}{3} = \frac{10}{3}r

.

Q_B = 2 Q_A = \frac{20}{3}r

. Total volume

B = Q_A + Q_B = \frac{10}{3}r + \frac{20}{3}r = 10r

.

Summing the output of both lines gives the exact total contract volume in terms of rate

r

.

Key Concept

Algebraic Modeling of Multi-Rate Work Systems

Question 143Question →

A software consulting firm calculates the total cost of a project using a fixed administrative fee plus a constant hourly rate for each hour worked. For a project requiring $H$ hours, the firm originally charged a total of $\$2,400$ , where the fixed administrative fee was equal to the total hourly charges. Under a new pricing model, the fixed administrative fee is reduced by $25\%$ , while the hourly rate is increased by $20\%$ . Under this new model, a project requiring $H + 15$ hours costs a total of $\$2,880$ . What is the value of $H$ ?

25

30

40

50

60

Show answer & explanation

Answer: 40

Answer

The value of

H

is 40.

The original cost model breaks into

F = \$1,200

and

R \cdot H = \$1,200

. Reducing

F

by

25\%

gives

\$900

. Increasing

R

by

20\%

gives a new hourly charge of

1.20 R(H + 15) = 1.20 RH + 18 R

. Substituting

RH = 1,200

yields

1,440 + 18 R = 1,980

, solving to

R = 30

. Dividing

1,200

by

30

gives

H = 40

.

Step-by-Step Solution

1

Formulate equations for the original pricing structure.

Let

F

be the fixed administrative fee and

R

be the original hourly rate. The total original cost is given by

F + R \cdot H = 2,400

. Since

F

equals the total hourly charges (

R \cdot H

), we have

F = R \cdot H = 1,200

.

Splitting the total cost of

\$2,400

into two equal parts establishes the exact baseline values for

F

and

R \cdot H

.

2

Formulate the equation under the new pricing model.

The new fixed fee is

0.75 \times 1,200 = 900

. The new hourly rate is

1.20 R

. The new total cost equation for

H + 15

hours is

900 + 1.20 R(H + 15) = 2,880

.

Applies the specified percentage changes to the fixed fee and hourly rate individually.

3

Solve for the hourly rate

R

.

Subtract

900

from both sides:

1.20 R(H + 15) = 1,980 \implies 1.20 RH + 18 R = 1,980

. Substituting

RH = 1,200

gives

1.20(1,200) + 18 R = 1,980 \implies 1,440 + 18 R = 1,980 \implies 18 R = 540 \implies R = 30

.

Expands the algebraic expression and uses the substitution

RH = 1,200

to isolate

R

.

4

Calculate the value of

H

.

Since

R \cdot H = 1,200

and

R = 30

, we find

H = \frac{1,200}{30} = 40

.

Divides the total hourly charge by the hourly rate to obtain the number of hours.

Key Concept

Linear Equation Modeling with Percentage Adjustments

Estimated Time:2m 0s

Question 144Question →

What is the numerical value of $\frac{2^8 \cdot 3^4}{6^4}$ ?

Show answer & explanation

Answer: 16

Answer

16

Rewriting the base

6

as

2 \cdot 3

gives

6^4 = 2^4 \cdot 3^4

. Substituting this into the original expression yields

\frac{2^8 \cdot 3^4}{2^4 \cdot 3^4}

. Cancelling

3^4

leaves

\frac{2^8}{2^4} = 2^{8-4} = 2^4 = 16

.

Step-by-Step Solution

1

Rewrite composite bases into prime factors

6^4 = (2 \cdot 3)^4 = 2^4 \cdot 3^4

Applying the power of a product rule

(ab)^n = a^n b^n

allows base matching.

2

Simplify the quotient using exponent subtraction

\frac{2^8 \cdot 3^4}{2^4 \cdot 3^4} = 2^{8-4} \cdot 3^{4-4} = 2^4 \cdot 3^0 = 16 \cdot 1 = 16

Applying the quotient rule

\frac{a^m}{a^n} = a^{m-n}

for powers with equal bases.

Key Concept

Exponent rules: Power of a Product

(ab)^n = a^n b^n

and Quotient of Powers

\frac{a^m}{a^n} = a^{m-n}

Question 145Question →

If $x$ is a real number satisfying the equation $(x^2 - 2x)^2 - (x^2 - 2x) - 12 = 0$ , what is the product of all distinct real roots of the equation?

-12

-4

-3

2

12

Show answer & explanation

Answer: $-4$

Answer

The product of all distinct real roots of the equation is

-4

.

Substituting

u = x^2 - 2x

simplifies the given equation to

u^2 - u - 12 = 0

, which factors as

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

. Setting

x^2 - 2x = 4

yields

x^2 - 2x - 4 = 0

, which has two real roots (since the discriminant is

20 > 0

) with product

-4

. Setting

x^2 - 2x = -3

yields

x^2 - 2x + 3 = 0

, which has a negative discriminant (

-8

) and therefore no real solutions. Thus, the product of all distinct real roots is

-4

.

Step-by-Step Solution

1

Use algebraic substitution to simplify the disguised quadratic equation.

Let

u = x^2 - 2x

. Substituting

u

into

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

gives

u^2 - u - 12 = 0

.

Recognizing the repeated expression

x^2 - 2x

reduces a 4th-degree polynomial into a standard 2nd-degree quadratic equation in terms of

u

.

2

Solve for

u

by factoring.

u^2 - u - 12 = (u - 4)(u + 3) = 0

, so

u = 4

or

u = -3

.

Factoring provides the possible numerical values for the expression

x^2 - 2x

.

3

Analyze each case for

x

and check for real roots using the discriminant.

Case 1:

x^2 - 2x = 4 \implies x^2 - 2x - 4 = 0

. Discriminant

D_1 = (-2)^2 - 4(1)(-4) = 20 > 0

. Since

D_1 > 0

, this quadratic has 2 distinct real roots. By Vieta's formula, the product of these two real roots is

\frac{c}{a} = \frac{-4}{1} = -4

.
Case 2:

x^2 - 2x = -3 \implies x^2 - 2x + 3 = 0

. Discriminant

D_2 = (-2)^2 - 4(1)(3) = -8 < 0

. Since

D_2 < 0

, this quadratic has no real roots.

Only roots with a non-negative discriminant belong to the real number domain and contribute to the product of real roots.

4

Determine the product of all distinct real roots.

The total product of all distinct real roots is

-4

.

Since Case 1 produces the only real roots of the original equation, their product

-4

is the complete product of all real solutions.

Key Concept

Disguised Quadratic Equations and Discriminant Real Root Filtering

Question 146Question →

What is the real solution to the equation $|x - 5| = 2x + 1$ ?

x = \frac{4}{3}

x = -6

x = -6

and

x = \frac{4}{3}

x = -\frac{4}{3}

x = 6

Show answer & explanation

Answer: $x = \frac{4}{3}$

Answer

x = \frac{4}{3}

The correct answer is

x = \frac{4}{3}

. Setting

x - 5 = -(2x + 1)

gives

x - 5 = -2x - 1

, which solves to

3x = 4

or

x = \frac{4}{3}

. Plugging

x = \frac{4}{3}

into both sides of the original equation yields

\frac{11}{3} = \frac{11}{3}

, satisfying the equation.

Step-by-Step Solution

1

Set up the two cases for the absolute value equation

Case 1:

x - 5 = 2x + 1

; Case 2:

x - 5 = -(2x + 1)

By definition,

|A| = B

implies

A = B

or

A = -B

, provided

B \geq 0

.

2

Solve Case 1 algebraically

x - 5 = 2x + 1 \implies x = -6

Subtracting

x

and

1

from both sides isolates

x

.

3

Solve Case 2 algebraically

x - 5 = -2x - 1 \implies 3x = 4 \implies x = \frac{4}{3}

Distributing the negative sign and adding

2x

and

5

to both sides isolates

x

.

4

Check for extraneous solutions in the original equation

|x - 5| = 2x + 1

For

x = -6

:

|-6 - 5| = |-11| = 11

, but

2(-6) + 1 = -11 \neq 11

(extraneous). For

x = \frac{4}{3}

:

|\frac{4}{3} - 5| = |-\frac{11}{3}| = \frac{11}{3}

, and

2(\frac{4}{3}) + 1 = \frac{11}{3}

(valid).

An absolute value expression cannot equal a negative number.

Key Concept

Solving absolute value equations with a variable on the right-hand side requires checking for extraneous solutions.

Question 147Question →

For all real numbers $x$ and $y$ , the custom binary operator $\odot$ is defined by $x \odot y = x^2 y - x y^2$ . The function $f$ is defined by $f(x) = x \odot 3$ . What is the value of $f(f(2))$ ?

Show answer & explanation

Answer: 162

Answer

The value of

f(f(2))

is 162.

To evaluate the nested function

f(f(2))

, first determine

f(2)

using the definition

f(x) = x \odot 3

. Applying the definition of the custom operator

x \odot y = x^2 y - x y^2

with

x = 2

and

y = 3

yields

2 \odot 3 = (2)^2(3) - (2)(3)^2 = 12 - 18 = -6

. Then, evaluate

f(-6) = -6 \odot 3 = (-6)^2(3) - (-6)(3)^2 = 108 - (-54) = 162

.

Step-by-Step Solution

1

Evaluate the inner function expression f(2)

f(2) = 2 \odot 3 = (2)^2(3) - (2)(3)^2 = 12 - 18 = -6

By definition, f(x) = x \odot 3. Substituting x = 2 gives 2 \odot 3, which applies the custom operator definition x^2 y - x y^2.

2

Evaluate the outer function expression f(f(2)) = f(-6)

f(-6) = -6 \odot 3 = (-6)^2(3) - (-6)(3)^2 = 36(3) - (-54) = 108 + 54 = 162

Substitute the inner result -6 into the function f(x).

Key Concept

Custom Binary Operators and Nested Function Evaluation

Question 148Question →

What is the value of the expression $\frac{3^8 + 3^8 + 3^8}{3^4}$ ?

3^4

3^5

3^8

3^{12}

3^{20}

Show answer & explanation

Answer: $3^5$

Answer

3^5

The correct answer is

3^5

. Adding three identical values

3^8 + 3^8 + 3^8

is equivalent to

3 \times 3^8

, which equals

3^9

. Dividing

3^9

by

3^4

gives

3^{9-4} = 3^5

.

Step-by-Step Solution

1

Simplify the numerator by combining like terms

3^8 + 3^8 + 3^8 = 3 \cdot 3^8 = 3^1 \cdot 3^8 = 3^{1+8} = 3^9

Repeated addition of

3

identical terms is equivalent to multiplying one term by

3

.

2

Apply the quotient rule for exponents

\frac{3^9}{3^4} = 3^{9-4} = 3^5

When dividing exponential terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.

Key Concept

Combining like exponential terms using multiplication rules and applying the quotient rule for exponents

Question 149Question →

For how many integer values of $k$ does the equation $|x - 2| + |x + 4| = kx + 1$ have no real solutions for $x$ ?

Show answer & explanation

Answer: 4

Answer

4

The equation has no real solutions when the line y = kx + 1 lies completely below the graph of f(x) = |x - 2| + |x + 4|. Analyzing the three piecewise regions of f(x) shows that f(x) = 6 on [-4, 2], with linear rays extending upward outside this interval. Setting the line to remain below f(x) forces -1.25 < k <= 2. The four integers in this range are -1, 0, 1, and 2.

Step-by-Step Solution

1

Decompose the sum of absolute values into a piecewise linear function

f(x) = -2x - 2 for x < -4; f(x) = 6 for -4 <= x < 2; f(x) = 2x + 2 for x >= 2

The critical points x = -4 and x = 2 split the real number line into three intervals where each absolute value expression maintains a constant sign.

2

Set up conditions for zero intersections with the line g(x) = kx + 1

g(x) must remain strictly below f(x) for all x

Any intersection point between y = f(x) and y = g(x) corresponds to a real solution of the equation.

3

Evaluate boundary points and slope constraints for each interval

From the middle interval and right ray: k <= 2. From the left ray: k > -1.25.

For x >= 2, the line slope k cannot exceed the ray slope of 2. For x <= -4, g(-4) < 6 requires -4k + 1 < 6, which yields k > -1.25.

4

Determine the allowable interval for k and count integer values

-1.25 < k <= 2, giving integer values k in {-1, 0, 1, 2}

The integer values strictly inside (-1.25, 2] are -1, 0, 1, and 2, making a total of 4 integer values.

Key Concept

Piecewise analysis of absolute value functions and linear line intersection conditions

Question 150Question →

For how many integer values of $k$ does the equation $|x - 4| + |x + 2| = k$ have at least one real solution $x$ satisfying $|x| \leq 5$ ?

5

6

7

8

12

Show answer & explanation

Answer: 7

Answer

7 integer values of k satisfy the equation.

The sum of absolute values

f(x) = |x - 4| + |x + 2|

represents the total distance from

x

to

4

and

-2

on the number line. On the interval

[-2, 4]

, this total distance is constant and equal to

6

, which is the minimum possible value. On the restricted domain

[-5, 5]

, the maximum distance occurs at the boundary point

x = -5

, where

f(-5) = |-9| + |-3| = 12

. Because

f(x)

is continuous, it takes every real value from

6

to

12

inclusive. The integer values that

k

can assume are

6, 7, 8, 9, 10, 11,

and

12

, yielding a total of 7 integer values.

Step-by-Step Solution

1

Define the piecewise absolute value function

f(x) = |x - 4| + |x + 2|

and analyze its critical points.

Critical points occur at

x = -2

and

x = 4

.

Absolute value expressions change definition where their inner terms equal zero.

2

Evaluate

f(x)

across the three intervals defining the domain

[-5, 5]

.

For

x < -2

,

f(x) = (4 - x) - (x + 2) = 2 - 2x

. For

-2 \leq x \leq 4

,

f(x) = (4 - x) + (x + 2) = 6

. For

x > 4

,

f(x) = (x - 4) + (x + 2) = 2x - 2

.

Removing absolute value bars according to domain regions gives explicit linear formulas.

3

Find the minimum and maximum values of

f(x)

on the domain

|x| \leq 5

(i.e.,

-5 \leq x \leq 5

).

Minimum value is

f(x) = 6

for

x \in [-2, 4]

. Maximum value occurs at

x = -5

, giving

f(-5) = 2 - 2(-5) = 12

. At

x = 5

,

f(5) = 2(5) - 2 = 8

. Range of

f(x)

is

[6, 12]

.

Continuous piecewise functions achieve extreme values at domain boundaries or constant minimum intervals.

4

Count the integer values of

k

in the closed interval

[6, 12]

.

The integers are

6, 7, 8, 9, 10, 11, 12

, giving

12 - 6 + 1 = 7

integers.

The number of integers in an inclusive range

[a, b]

is given by

b - a + 1

.

Key Concept

Absolute Value Equations and Piecewise Function Ranges

Estimated Time:2m 0s

Question 151Question →

How many integer values of $x$ satisfy the inequality $\big||x + 1| - 4\big| \leq 3 - |x|$ ?

2

3

4

5

7

Show answer & explanation

Answer: 4

Answer

4

The correct answer is 4. The inequality requires

3 - |x| \geq 0

, restricting candidate solutions to integers in the range

[-3, 3]

. Testing each integer directly reveals that only

x = 0, 1, 2, 3

satisfy the inequality, yielding a total of 4 valid integers.

Step-by-Step Solution

1

Establish the domain restriction from the non-negativity of the absolute value.

Since the left-hand side

\big||x + 1| - 4\big| \geq 0

for all real numbers, the right-hand side must also be non-negative:

3 - |x| \geq 0 \implies |x| \leq 3 \implies -3 \leq x \leq 3

.

An absolute value expression cannot be less than a negative number.

2

Test non-negative integer candidates in the domain

[-3, 3]

.

For

x = 3

:

\big|4 - 4\big| = 0 \leq 3 - 3 = 0

(True);
For

x = 2

:

\big|3 - 4\big| = 1 \leq 3 - 2 = 1

(True);
For

x = 1

:

\big|2 - 4\big| = 2 \leq 3 - 1 = 2

(True);
For

x = 0

:

\big|1 - 4\big| = 3 \leq 3 - 0 = 3

(True).

Substituting non-negative integers verifies that

x \in \{0, 1, 2, 3\}

satisfy the inequality.

3

Test negative integer candidates in the domain

[-3, 3]

.

For

x = -1

:

\big|0 - 4\big| = 4 \leq 3 - 1 = 2

(False);
For

x = -2

:

\big|-1 - 4\big| = 5 \leq 3 - 2 = 1

(False);
For

x = -3

:

\big|-2 - 4\big| = 6 \leq 3 - 3 = 0

(False).

Evaluating negative integers shows no negative integer satisfies the inequality.

4

Count the total number of valid integer solutions.

The valid integer solutions are

x = 0, 1, 2, 3

, giving a total count of 4.

Combining all valid cases yields exactly 4 integer solutions.

Key Concept

Solving nested absolute value inequalities using domain constraints and piecewise sign analysis.

Question 152Question →

A boutique marketing agency charges a corporate client a fixed monthly retainer fee of $\$1,600$ plus a uniform monthly fee for each active digital ad campaign managed. The client has a total monthly ad management budget of $\$4,000$ . If the agency were to increase the fee per campaign by $\$40$ , the number of ad campaigns the client could afford under the same total budget would decrease by $2$ . What is the original monthly fee charged per ad campaign?

\$160

\$180

\$200

\$240

\$250

Show answer & explanation

Answer: $\$200$

Answer

The original monthly fee charged per ad campaign is

\$200

.

Subtracting the

\$1,600

retainer from the

\$4,000

total budget leaves

\$2,400

for campaign fees. Letting

p

represent the original price per campaign, the original number of campaigns is

\frac{2,400}{p}

and the new number of campaigns at price

p+40

is

\frac{2,400}{p+40}

. The condition that the client can afford

2

fewer campaigns yields the equation

\frac{2,400}{p} - \frac{2,400}{p+40} = 2

. Simplifying leads to

p^2 + 40p - 48,000 = 0

, which factors as

(p - 200)(p + 240) = 0

. Since price must be positive, the original fee per campaign is

\$200

.

Step-by-Step Solution

1

Determine the portion of the budget available for ad campaigns.

Variable Budget = Total Budget - Fixed Retainer =

\$4,000 - \$1,600 = \$2,400

.

The fixed retainer fee must be subtracted before calculating how many campaigns can be purchased.

2

Define variables and set up the equation for the number of campaigns.

Let

p

be the original fee per campaign. The original number of campaigns is

\frac{2,400}{p}

. With the fee increase, the new number of campaigns is

\frac{2,400}{p + 40}

. The difference is given as

2

, so

\frac{2,400}{p} - \frac{2,400}{p + 40} = 2

.

Model the real-world reduction in campaigns using algebraic expressions.

3

Simplify the equation and form a standard quadratic equation.

Divide the entire equation by

2

:

\frac{1,200}{p} - \frac{1,200}{p + 40} = 1

. Multiply by

p(p + 40)

:

1,200(p + 40) - 1,200p = p(p + 40) \implies 48,000 = p^2 + 40p \implies p^2 + 40p - 48,000 = 0

.

Clear fractions to transform the rational equation into quadratic form.

4

Solve the quadratic equation for

p

.

(p + 240)(p - 200) = 0

. Since price must be positive,

p = 200

.

Factoring gives two roots (

200

and

-240

); reject the negative root in physical pricing context.

Key Concept

Algebraic Word Problems and Equation Modeling

Estimated Time:2m 0s

Question 153Question →

For all real numbers $a$ and $b$ , the custom binary operator $\star$ is defined by $a \star b = a^2 - b^2$ . The function $f$ is defined by $f(x) = x \star 3$ . If $f(x) = 4x - 12$ , what is the sum of all real values of $x$ that satisfy this equation?

4

3

-4

-3

12

Show answer & explanation

Answer: 4

Answer

The sum of all real values of x that satisfy the equation is 4.

Applying the custom operator definition gives f(x) = x^2 - 9. Setting this equal to 4x - 12 produces the quadratic equation x^2 - 4x + 3 = 0. Factoring yields (x - 1)(x - 3) = 0, which has solutions x = 1 and x = 3. The sum of these solutions is 1 + 3 = 4.

Step-by-Step Solution

1

Evaluate the custom operator expression f(x) = x * 3.

f(x) = x^2 - 3^2 = x^2 - 9

Applying the definition of the custom operator a * b = a^2 - b^2 with a = x and b = 3.

2

Set f(x) equal to 4x - 12 and rearrange into standard quadratic form.

x^2 - 9 = 4x - 12 => x^2 - 4x + 3 = 0

Equating the function expression to the given linear expression.

3

Factor the quadratic equation to find the roots.

(x - 1)(x - 3) = 0 => x = 1 or x = 3

Finding the values of x that satisfy the quadratic equation.

4

Calculate the sum of the roots.

1 + 3 = 4

Summing all real solutions to satisfy the prompt requirement.

Key Concept

Custom Operator Evaluation and Quadratic Equation Solving

Estimated Time:1m 30s

Question 154Question →

If $x$ and $y$ are non-zero real numbers such that $x^2 - 5xy + 6y^2 = 0$ , what is the sum of all possible values of $\frac{x^2 + y^2}{xy}$ ?

\frac{5}{2}

\frac{35}{6}

\frac{59}{6}

-\frac{5}{6}

5

Show answer & explanation

Answer: $\frac{35}{6}$

Answer

The sum of all possible values of

\frac{x^2 + y^2}{xy}

is

\frac{35}{6}

.

Factoring

x^2 - 5xy + 6y^2 = 0

yields

(x - 2y)(x - 3y) = 0

. Thus,

\frac{x}{y} = 2

or

\frac{x}{y} = 3

. Simplifying

\frac{x^2 + y^2}{xy}

gives

\frac{x}{y} + \frac{y}{x}

. Substituting

2

yields

2 + \frac{1}{2} = \frac{5}{2}

, and substituting

3

yields

3 + \frac{1}{3} = \frac{10}{3}

. The sum of these values is

\frac{5}{2} + \frac{10}{3} = \frac{35}{6}

.

Step-by-Step Solution

1

Factor the quadratic equation

x^2 - 5xy + 6y^2 = 0

in terms of

x

and

y

.

(x - 2y)(x - 3y) = 0

Factoring determines the proportional relationships between

x

and

y

.

2

Solve for the possible ratio values of

\frac{x}{y}

.

Either

x = 2y \implies \frac{x}{y} = 2

, or

x = 3y \implies \frac{x}{y} = 3

.

Since

y \neq 0

, dividing by

y

gives the linear root ratios.

3

Rewrite the target expression

\frac{x^2 + y^2}{xy}

as a sum of ratios.

\frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}

Splitting the numerator simplifies substitution of

\frac{x}{y}

.

4

Evaluate the expression for each possible ratio.

Case 1 (

x/y = 2

):

2 + \frac{1}{2} = \frac{5}{2}

. Case 2 (

x/y = 3

):

3 + \frac{1}{3} = \frac{10}{3}

.

Each root ratio produces a distinct value for the target expression.

5

Sum the two values obtained.

\frac{5}{2} + \frac{10}{3} = \frac{15 + 20}{6} = \frac{35}{6}

The question asks for the sum of all possible values of the expression.

Key Concept

Homogeneous Quadratic Equations and Algebraic Factoring

Estimated Time:2m 0s

Question 155Question →

What is the sum of all real solutions to the equation $\sqrt{2x + 15} = x$ ?

5

2

-3

8

15

Show answer & explanation

Answer: 5

Answer

The sum of all real solutions to the equation is 5.

Squaring both sides of

\sqrt{2x + 15} = x

yields

2x + 15 = x^2

, which rearranges to

x^2 - 2x - 15 = 0

. Factoring gives

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

, yielding candidate roots

x = 5

and

x = -3

. Substituting

x = 5

into the original equation gives

\sqrt{25} = 5

, which is true. Substituting

x = -3

gives

\sqrt{9} = -3

, which is false because the principal square root must be non-negative. Therefore,

x = 5

is the only valid real solution, and its sum is 5.

Step-by-Step Solution

1

Square both sides of the equation to eliminate the radical.

2x + 15 = x^2

Squaring both sides removes the radical so the equation can be expressed in quadratic form.

2

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 2x - 15 = 0

Setting the quadratic expression equal to zero enables solution by factoring.

3

Factor the quadratic polynomial.

(x - 5)(x + 3) = 0 \implies x = 5 \text{ or } x = -3

Factoring determines the candidate roots of the quadratic equation.

4

Check candidate roots in the original equation

\sqrt{2x + 15} = x

for extraneous solutions.

For

x = 5

:

\sqrt{2(5) + 15} = \sqrt{25} = 5

(Valid). For

x = -3

:

\sqrt{2(-3) + 15} = \sqrt{9} = 3 \neq -3

(Extraneous).

Squaring both sides of an equation can introduce extraneous roots that violate the non-negativity constraint of principal square roots.

5

Sum all valid real solutions.

Sum = 5

Since

x = 5

is the only valid solution, the sum of all solutions is 5.

Key Concept

Quadratic Equations and Extraneous Solutions in Radical Equations

Question 156Question →

A logistics company operates two models of cargo trucks, Model X and Model Y, to transport freight between two distribution hubs. Model X consumes $10$ liters of fuel per $100$ kilometers driven in city traffic and $6$ liters of fuel per $100$ kilometers driven on open highway. Model Y consumes $14$ liters of fuel per $100$ kilometers in city traffic and $8$ liters per $100$ kilometers on open highway. On a trip between the two hubs along a route composed entirely of city traffic segments and open highway segments, Model X consumed a total of $42$ liters of fuel, and Model Y consumed a total of $58$ liters of fuel. What is the total length, in kilometers, of the open highway segments along this route?

Show answer & explanation

Answer: 200

Answer

The total length of the open highway segments along the route is 200 kilometers.

By converting the per-100-kilometer fuel consumption into per-kilometer rates, we obtain two linear equations in two variables:

0.10C + 0.06H = 42

and

0.14C + 0.08H = 58

. Solving this system by elimination yields

H = 200

kilometers for the open highway distance.

Step-by-Step Solution

1

Define variables for city traffic distance and open highway distance.

Let

C

equal the distance in km of city traffic segments, and

H

equal the distance in km of open highway segments.

Establishing explicit variables allows the translation of the word problem context into algebraic equations.

2

Convert consumption rates to liters per kilometer and formulate total fuel equations for both trucks.

Model X equation:

0.10C + 0.06H = 42

; Model Y equation:

0.14C + 0.08H = 58

.

Dividing the consumption per 100 km by 100 gives the unit rates per km for city and highway driving.

3

Simplify the system of linear equations by eliminating decimals and common factors.

First equation:

5C + 3H = 2100

; Second equation:

7C + 4H = 2900

.

Simplifying equations reduces computation errors during elimination.

4

Solve the system using elimination to isolate the highway distance variable

H

.

Multiplying the equations to match coefficients of

C

(

35C + 21H = 14700

and

35C + 20H = 14500

) and subtracting gives

H = 200

.

Eliminating

C

directly yields the target quantity

H

without needing additional substitution steps.

Key Concept

Formulating and Solving Systems of Linear Equations from Multi-Condition Word Problems

Estimated Time:2m 0s

Question 157Question →

A logistics company operates a fleet consisting of standard electric delivery vans and high-capacity electric delivery vans. A standard van consumes a fixed $2.5$ kilowatt-hours (kWh) of electricity per delivery route, while a high-capacity van consumes $50\%$ more electricity per route than a standard van. On a given day, the fleet completed a total of $40$ delivery routes and consumed a total of $130$ kWh of electricity. How many delivery routes were completed by high-capacity vans?

Show answer & explanation

Answer: 24

Answer

The total number of delivery routes completed by high-capacity vans is 24.

By modeling the relationship using the system

S + H = 40

and

2.5S + 3.75H = 130

, substituting

S = 40 - H

gives

100 + 1.25H = 130

, which resolves to

H = 24

high-capacity van routes.

Step-by-Step Solution

1

Calculate the power consumption per route for a high-capacity van.

High-capacity rate =

2.5 \times 1.5 = 3.75

kWh per route.

High-capacity vans consume

50\%

more energy than standard vans.

2

Formulate a system of linear equations representing total routes and total electricity used.

S + H = 40

and

2.5S + 3.75H = 130

, where

S

represents standard van routes and

H

represents high-capacity van routes.

The sum of routes equals 40 and the combined energy consumption equals 130 kWh.

3

Solve for the target variable

H

.

Substitute

S = 40 - H

into the energy equation:

2.5(40 - H) + 3.75H = 130 \implies 100 + 1.25H = 130 \implies 1.25H = 30 \implies H = 24

.

Substituting

S

yields a single-variable linear equation for

H

.

Key Concept

Algebraic Modeling of Systems of Linear Equations

Estimated Time:1m 30s

Question 158Question →

A software enterprise sells two annual licensing tiers, Tier 1 and Tier 2. In January, the company sold a combined total of $200$ licenses for the two tiers. The price of a Tier 1 license was $\$40$ and the price of a Tier 2 license was $\$60$ . In February, the price of a Tier 1 license increased by $25\%$ and the number of Tier 1 licenses sold increased by $50\%$ , while both the price and the number of Tier 2 licenses sold remained unchanged from January. If the total monthly revenue from these two tiers increased by $\$2,800$ in February compared to January, how many Tier 1 licenses were sold in January?

60

70

80

100

120

Show answer & explanation

Answer: 80

Answer

80 Tier 1 licenses were sold in January.

The correct answer is 80. Modeling the sales volume of Tier 1 in January as

x

, Tier 1 generated

40x

dollars in January. In February, with a

25\%

price increase (to

\$50

) and a

50\%

volume increase (to

1.5x

), Tier 1 generated

50(1.5x) = 75x

dollars. Because Tier 2 price and sales volume remained constant, the total revenue increase of

\$2,800

is equal to the change in Tier 1 revenue:

75x - 40x = 35x = 2,800

, which gives

x = 80

.

Step-by-Step Solution

1

Define variables for January sales volume and set up equations for each tier.

Let

x

be the number of Tier 1 licenses sold in January. The number of Tier 2 licenses sold in January is

200 - x

.

The total number of licenses sold in January was given as 200.

2

Calculate January revenue from Tier 1.

\text{Revenue}_{\text{Tier 1, Jan}} = 40x.

Tier 1 licenses cost

\$40

each in January.

3

Determine February price and quantity for Tier 1, and calculate February Tier 1 revenue.

\text{Price}_{\text{Tier 1, Feb}} = 40 \times 1.25 = \

50

; \text{Quantity}_{\text{Tier 1, Feb}} = 1.5x$. Thus, \text{Revenue}_{\text{Tier 1, Feb}} = 50(1.5x) = 75x.

Tier 1 price increased by

25\%

and volume increased by

50\%

.

4

Express the total revenue change between February and January.

Since Tier 2 price and volume did not change, net revenue change comes entirely from Tier 1:

75x - 40x = 35x

.

Unchanged terms cancel out when taking the difference (February Revenue minus January Revenue).

5

Solve for

x

using the given revenue increase of

\$2,800

.

35x = 2,800 \implies x = 80.

Dividing both sides by 35 gives the January quantity for Tier 1 licenses.

Key Concept

Algebraic Word Problems and Equation Modeling

Estimated Time:2m 0s

Question 159Question →

If the roots of the quadratic equation $x^2 - (k - 2)x + (k + 1) = 0$ have a sum that is equal to twice their product, what is the value of $k$ ?

Show answer & explanation

Answer: -4

Answer

The value of

k

is

-4

.

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

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

is

k-2

and the product of the roots is

k+1

. Equating the sum to twice the product gives

k - 2 = 2(k + 1)

, which simplifies to

k - 2 = 2k + 2

and yields

k = -4

. Evaluating the discriminant with

k = -4

gives

\Delta = (-6)^2 - 4(-3) = 48 > 0

, ensuring real roots exist.

Step-by-Step Solution

1

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

k

.

Sum of roots

= -\frac{b}{a} = k - 2

, and product of roots

= \frac{c}{a} = k + 1

.

For any 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

Set up an equation based on the condition that the sum of the roots equals twice their product.

k - 2 = 2(k + 1)

This directly translates the verbal condition given in the problem.

3

Solve the linear equation for

k

.

k - 2 = 2k + 2 \implies -4 = k

Subtracting

k

and 2 from both sides isolates

k

.

4

Check the discriminant to ensure real roots exist when

k = -4

.

\Delta = b^2 - 4ac = (-6)^2 - 4(1)(-3) = 36 + 12 = 48 > 0

A positive discriminant guarantees that the quadratic equation has two distinct real roots.

Key Concept

Vieta's Formulas for Quadratic Equations

Question 160Question →

If $x$ is a real number, what is the sum of all valid real solutions to the equation $\sqrt{3x - 2} - \sqrt{x + 3} = 1$ ?

1

3

4

6

7

Show answer & explanation

Answer: 6

Answer

The sum of all valid real solutions is 6.

To solve the radical equation, we isolate

\sqrt{3x - 2}

to get

\sqrt{3x - 2} = 1 + \sqrt{x + 3}

. Squaring both sides yields

3x - 2 = x + 4 + 2\sqrt{x + 3}

, which simplifies to

x - 3 = \sqrt{x + 3}

. Squaring both sides a second time produces the quadratic

x^2 - 7x + 6 = 0

, giving candidate solutions

x = 6

and

x = 1

. Checking in the original equation reveals that

x = 6

yields

\sqrt{16} - \sqrt{9} = 1

, which is true, whereas

x = 1

yields

\sqrt{1} - \sqrt{4} = -1 \neq 1

, making

x = 1

extraneous. The only valid solution is 6, so the sum of all valid solutions is 6.

Step-by-Step Solution

1

Isolate one radical expression on one side of the equation.

\sqrt{3x - 2} = 1 + \sqrt{x + 3}

Isolating a radical allows for clean squaring of both sides without creating cross-terms with multiple roots.

2

Square both sides of the equation.

3x - 2 = 1 + 2\sqrt{x + 3} + (x + 3) \implies 3x - 2 = x + 4 + 2\sqrt{x + 3}

Eliminates the radical on the left side.

3

Isolate the remaining radical term and simplify.

2x - 6 = 2\sqrt{x + 3} \implies x - 3 = \sqrt{x + 3}

Simplifies the algebraic expression prior to squaring a second time.

4

Square both sides again to obtain a quadratic equation.

(x - 3)^2 = x + 3 \implies x^2 - 6x + 9 = x + 3 \implies x^2 - 7x + 6 = 0

Eliminates all remaining radical terms.

5

Factor the quadratic equation to find algebraic candidate solutions.

(x - 6)(x - 1) = 0 \implies x = 6 \text{ or } x = 1

Identifies potential real roots.

6

Test candidate solutions in the original equation

\sqrt{3x - 2} - \sqrt{x + 3} = 1

to remove extraneous solutions.

For

x = 6

:

\sqrt{16} - \sqrt{9} = 4 - 3 = 1

(Valid). For

x = 1

:

\sqrt{1} - \sqrt{4} = 1 - 2 = -1 \neq 1

(Extraneous). Thus,

x = 6

is the sole valid solution.

Squaring equations during algebraic steps can introduce extraneous roots that do not satisfy the original radical equation.

Key Concept

Solving Radical Equations and Eliminating Extraneous Solutions

Estimated Time:2m 0s

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