Intermediate Algebra Practice Questions — ACT — Page 9

Question 161Question →

If $x$ and $y$ are real numbers such that $2^x \cdot 3^y = 24$ and $3^x \cdot 2^y = 54$ , what is the value of $x^2 + y^2$ ?

Show answer & explanation

Answer: 10

Answer

10

Multiplying the two equations gives

(2^x \cdot 3^y)(3^x \cdot 2^y) = 24 \cdot 54 \implies 6^{x+y} = 1296 = 6^4

, which yields

x + y = 4

. Dividing the first equation by the second equation gives \frac{2^x \cdot 3^y}{3^x \cdot 2^y} = \frac{24}{54} \implies (\frac{2}{3})^{x-y} = \frac{4}{9} = (\frac{2}{3})^2

, which yields

x - y = 2

. Solving the system of equations

x + y = 4

and

x - y = 2

gives

x = 3

and

y = 1

. The value of

x^2 + y^2

is therefore

3^2 + 1^2 = 10$.

Step-by-Step Solution

1

Multiply the two equations together.

6^{x+y} = 1296

, which simplifies to

x + y = 4

.

Multiplying the equations groups bases of 2 and 3 together to form base 6, allowing us to find the sum of the variables.

2

Divide the first equation by the second equation.

(\frac{2}{3})^{x-y} = \frac{4}{9}

, which simplifies to

x - y = 2

.

Dividing the equations groups bases of 2 and 3 to form base

2/3

, allowing us to find the difference of the variables.

3

Solve the system of equations for

x

and

y

.

x = 3

and

y = 1

.

Solving the linear system of equations

x + y = 4

and

x - y = 2

gives the individual values of

x

and

y

.

4

Calculate

x^2 + y^2

.

10

Substitute the values of

x

and

y

into the target expression.

Key Concept

Solving systems of exponential equations using properties of exponents and bases

Alternative Method

Take the logarithm of both sides of each equation to convert them into a system of linear equations in terms of

x

and

y

:

x \log 2 + y \log 3 = \log 24

and

x \log 3 + y \log 2 = \log 54

. Solving this system using elimination or substitution yields

x = 3

and

y = 1

, so

x^2 + y^2 = 10

.

Estimated Time:2m 30s

Question 162Question →

For a real number $a$ , the quadratic equation $x^2 + ax + (a - 3)^2 = 0$ has complex roots. One of these roots is $2 + i\sqrt{b}$ , where $b$ is a positive real number and $i = \sqrt{-1}$ . What is the value of $a + b$ ?

17

-21

41

-49

1

Show answer & explanation

Answer: 41

Answer

The value of

a + b

is

41

.

The correct answer is

41

. Since the quadratic equation has real coefficients, its complex roots must occur in conjugate pairs. Thus, the roots are

2 + i\sqrt{b}

and

2 - i\sqrt{b}

. Their sum is

4

, which by Vieta's formulas equals

-a

, giving

a = -4

. The product of the roots is

4 + b

, which equals the constant term

(a - 3)^2 = (-4 - 3)^2 = 49

. Solving for

b

gives

b = 45

. Therefore,

a + b = -4 + 45 = 41

.

Step-by-Step Solution

1

Identify the second root using the complex conjugate root theorem.

The conjugate root is

2 - i\sqrt{b}

.

Since the coefficients of the quadratic equation are real, complex roots must occur in conjugate pairs.

2

Find the value of

a

using the sum of the roots.

a = -4

By Vieta's formulas, the sum of the roots is equal to

-a

. The sum of the conjugate roots is

(2 + i\sqrt{b}) + (2 - i\sqrt{b}) = 4

, so

-a = 4

.

3

Calculate the constant term of the quadratic equation.

The constant term is

49

.

Substitute

a = -4

into the constant term

(a - 3)^2

to get

(-4 - 3)^2 = (-7)^2 = 49

.

4

Find the value of

b

using the product of the roots.

b = 45

By Vieta's formulas, the product of the roots is the constant term. The product is

(2 + i\sqrt{b})(2 - i\sqrt{b}) = 4 - i^2 b = 4 + b

(since

i^2 = -1

). Equating this to the constant term gives

4 + b = 49

, so

b = 45

.

5

Calculate the final value of

a + b

.

41

Substitute the values of

a

and

b

:

a + b = -4 + 45 = 41

.

Key Concept

Vieta's formulas and complex conjugate roots of quadratic equations with real coefficients.

Question 163Question →

If $x$ is a positive real number such that $\log_2(x) \cdot \log_3(x) - \log_2(x) - \log_3(x) = 1$ , what is the product of all possible real values of $x$ ?

\frac{1}{6}

5

6

\frac{1}{5}

12

Show answer & explanation

Answer: $6$

Answer

6

The correct answer is 6. By converting the equation using the change-of-base formula into natural logarithms, we obtain the quadratic equation

(\ln x)^2 - (\ln 2 + \ln 3)\ln x - \ln 2 \cdot \ln 3 = 0

. By setting

w = \ln x

, we have a quadratic in terms of

w

. The product of the two solutions

x_1

and

x_2

is

x_1 x_2 = e^{w_1} e^{w_2} = e^{w_1 + w_2}

. Using Vieta's formulas, the sum of the roots

w_1 + w_2 = \ln 2 + \ln 3 = \ln 6

. Therefore, the product of the solutions is

e^{\ln 6} = 6

.

Step-by-Step Solution

1

Rewrite the logarithmic equation using a common base.

Using the change-of-base formula

\log_b(a) = \frac{\ln a}{\ln b}

, the equation becomes:

\left(\frac{\ln x}{\ln 2}\right)\left(\frac{\ln x}{\ln 3}\right) - \frac{\ln x}{\ln 2} - \frac{\ln x}{\ln 3} = 1

This allows all logarithmic terms to be expressed in terms of the natural logarithm, making it easier to solve.

2

Eliminate denominators by multiplying the equation by

\ln 2 \cdot \ln 3

.

(\ln x)^2 - (\ln 3)\ln x - (\ln 2)\ln x = \ln 2 \cdot \ln 3

Multiplying through by the common denominator simplifies the rational equation into a standard polynomial form.

3

Rearrange the equation into a standard quadratic form.

(\ln x)^2 - (\ln 2 + \ln 3)\ln x - \ln 2 \cdot \ln 3 = 0

Grouping the coefficients of

\ln x

reveals a quadratic equation of the form

Aw^2 + Bw + C = 0

, where

w = \ln x

.

4

Define the relationship between the roots of the quadratic equation and the values of

x

.

Let the two real roots of the quadratic equation be

w_1

and

w_2

. These correspond to the solutions for

x

, namely

x_1 = e^{w_1}

and

x_2 = e^{w_2}

. The product of the solutions is

x_1 \cdot x_2 = e^{w_1} \cdot e^{w_2} = e^{w_1 + w_2}

.

To find the product of the solutions for

x

, we must compute the exponential of the sum of the roots of the quadratic equation.

5

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

w_1 + w_2

.

The sum of the roots is

w_1 + w_2 = -\frac{B}{A} = -\frac{-(\ln 2 + \ln 3)}{1} = \ln 2 + \ln 3

. Using the logarithmic product rule, this simplifies to

\ln(2 \cdot 3) = \ln 6

.

Vieta's formulas state that the sum of the roots of

Aw^2 + Bw + C = 0

is

-\frac{B}{A}

.

6

Calculate the product of the real values of

x

.

x_1 \cdot x_2 = e^{\ln 6} = 6

Substituting the sum of the roots back into the exponent gives the final product.

Key Concept

Solving equations involving logarithmic properties, change of base, and relating quadratic roots to exponential functions

Question 164Question →

A cleaning service company offers two types of packages: Standard Clean and Deep Clean. The number of packages sold during the first two weeks of the month is represented by matrix $Q$ :

Q = \begin{bmatrix} 20 & 15 \\ 25 & 18 \end{bmatrix}

where row 1 represents Week 1, row 2 represents Week 2, column 1 represents Standard Clean packages, and column 2 represents Deep Clean packages.

The price charged and the cost to perform each package (in dollars) are represented by matrix $P$ :

P = \begin{bmatrix} 80 & 30 \\ 150 & 60 \end{bmatrix}

where row 1 represents Standard Clean packages, row 2 represents Deep Clean packages, column 1 represents the price charged, and column 2 represents the cost.

Which of the following matrices represents the total revenue and total cost for Week 1 and Week 2, respectively?

\begin{bmatrix} 1,600 & 450 \\ 3,750 & 1,080 \end{bmatrix}

\begin{bmatrix} 2,350 & 1,740 \\ 4,500 & 3,330 \end{bmatrix}

\begin{bmatrix} 3,850 & 1,500 \\ 4,700 & 1,830 \end{bmatrix}

\begin{bmatrix} 3,800 & 1,500 \\ 4,000 & 1,830 \end{bmatrix}

\begin{bmatrix} 650 & 300 \\ 700 & 330 \end{bmatrix}

Show answer & explanation

Answer: $\begin{bmatrix} 3,850 & 1,500 \\ 4,700 & 1,830 \end{bmatrix}$

Answer

The matrix with elements 3,850 and 1,500 in the first row, and 4,700 and 1,830 in the second row.

The matrix representing the total revenue and total cost for Week 1 and Week 2 is found by multiplying matrix

Q

by matrix

P

. Since

Q

is a

2 \times 2

matrix (weeks by package types) and

P

is a

2 \times 2

matrix (package types by financial categories), the product

Q \times P

yields a

2 \times 2

matrix representing weeks by financial categories (revenue and cost).

Calculating each element:
- Row 1, Column 1 (Week 1 Revenue):

(20 \times 80) + (15 \times 150) = 1,600 + 2,250 = 3,850

- Row 1, Column 2 (Week 1 Cost):

(20 \times 30) + (15 \times 60) = 600 + 900 = 1,500

- Row 2, Column 1 (Week 2 Revenue):

(25 \times 80) + (18 \times 150) = 2,000 + 2,700 = 4,700

- Row 2, Column 2 (Week 2 Cost):

(25 \times 30) + (18 \times 60) = 750 + 1,080 = 1,830

This results in the matrix:

\begin{bmatrix} 3,850 & 1,500 \\ 4,700 & 1,830 \end{bmatrix}

Step-by-Step Solution

1

Identify the appropriate matrix operation to combine the weekly package sales with their respective price and cost.

We need to compute the product of matrix

Q

and matrix

P

(

Q \times P

), since multiplying a

2 \times 2

matrix of (weeks

\times

packages) by a

2 \times 2

matrix of (packages

\times

financial category) results in a

2 \times 2

matrix of (weeks

\times

financial category).

This alignment ensures that the package types cancel out, leaving the desired weekly revenue and cost breakdown.

2

Calculate the elements of the first row of the product matrix.

Row 1, Column 1 (Week 1 Revenue):

20(80) + 15(150) = 1,600 + 2,250 = 3,850

. Row 1, Column 2 (Week 1 Cost):

20(30) + 15(60) = 600 + 900 = 1,500

.

The first row represents Week 1 revenue and cost.

3

Calculate the elements of the second row of the product matrix.

Row 2, Column 1 (Week 2 Revenue):

25(80) + 18(150) = 2,000 + 2,700 = 4,700

. Row 2, Column 2 (Week 2 Cost):

25(30) + 18(60) = 750 + 1,080 = 1,830

.

The second row represents Week 2 revenue and cost.

Key Concept

Matrix multiplication represents linear combinations of variables, useful for calculating total cost and revenue in multi-variable business scenarios.

Question 165Question →

Let matrix $A = \begin{bmatrix} 3 & -2 \\ 5 & 1 \end{bmatrix}$ and matrix $B = \begin{bmatrix} -2 & 3 \\ 2 & -3 \end{bmatrix}$ . If matrix $C$ is defined by the equation $C = 3A - 2B$ , what is the value of the element in the second row and first column of $C$ ?

Show answer & explanation

Answer: 11

Answer

The element in the second row and first column of matrix

C

is 11.

To find the element in the second row and first column of matrix

C

, we apply the operations defined by

C = 3A - 2B

directly to the corresponding elements of

A

and

B

. The element in row 2, column 1 of matrix

A

is

5

, and of matrix

B

is

2

. Computing

3(5) - 2(2)

yields

15 - 4 = 11

.

Step-by-Step Solution

1

Identify the elements in the second row and first column of matrices

A

and

B

.

A_{21} = 5

and

B_{21} = 2

.

To find a specific element of the resulting matrix

C = 3A - 2B

, we only need to perform the operations on the elements in the corresponding position.

2

Set up the equation for the element in the second row and first column of

C

.

C_{21} = 3A_{21} - 2B_{21}

Matrix addition, subtraction, and scalar multiplication are performed element-wise.

3

Substitute the identified values into the equation and compute the result.

C_{21} = 3(5) - 2(2) = 15 - 4 = 11

Evaluating the expression gives the value of the target element.

Key Concept

Matrix scalar multiplication and element-wise subtraction

Question 166Question →

Let the functions $f$ and $g$ be defined by $f(x) = (x - 3)^2$ and $g(x) = 2x + 1$ . What is the value of the composite function $f(g(-2))$ ?

Show answer & explanation

Answer: 36

Answer

36

To evaluate the composite function

f(g(-2))

, evaluate from the inside out. First, evaluate the inner function

g(-2) = 2(-2) + 1 = -3

. Next, substitute this output value into the outer function

f(x)

to get

f(-3) = (-3 - 3)^2 = (-6)^2 = 36

.

Step-by-Step Solution

1

Evaluate the inner function

g(x)

at

x = -2

g(-2) = -3

Before evaluating the outer function

f

, we must determine the output of the inner function

g

at the given input value.

2

Evaluate the outer function

f(x)

at the result of the inner function

f(-3) = 36

Substitute the inner output

-3

as the input for

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

, giving

(-3 - 3)^2 = (-6)^2 = 36

.

Key Concept

Function composition and evaluation

Estimated Time:1m 0s

Question 167Question →

If the quadratic equation $x^2 + kx + 25 = 0$ has exactly one real solution, and $k > 0$ , what is the value of $k$ ?

Show answer & explanation

Answer: 10

Answer

The value of

k

is

10

.

For any quadratic equation in standard form,

ax^2 + bx + c = 0

, the equation has exactly one real solution when the discriminant

b^2 - 4ac

is equal to

0

. Here, the equation is

x^2 + kx + 25 = 0

, so

a = 1

,

b = k

, and

c = 25

. Setting the discriminant to

0

gives

k^2 - 4(1)(25) = 0

, which simplifies to

k^2 - 100 = 0

. Solving for

k

yields

k = \pm 10

. Since we are given that

k > 0

, the only valid solution is

10

.

Step-by-Step Solution

1

Set the discriminant of the quadratic equation to zero.

b^2 - 4ac = 0

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

2

Identify coefficients

a

,

b

, and

c

from the equation

x^2 + kx + 25 = 0

and substitute them into the discriminant equation.

k^2 - 4(1)(25) = 0

For

x^2 + kx + 25 = 0

, the coefficients are

a=1

,

b=k

, and

c=25

.

3

Simplify the equation and solve for

k

.

k^2 = 100

, so

k = 10

or

k = -10

Simplifying

k^2 - 100 = 0

gives

k^2 = 100

.

4

Apply the constraint

k > 0

to find the final value.

k = 10

The problem specifies that

k

must be greater than zero.

Key Concept

Using the discriminant (

b^2 - 4ac

) of a quadratic equation to determine when there is exactly one real solution.

Estimated Time:45s

Question 168Question →

A line with a positive slope passes through the point $(0, -5)$ and is tangent to the parabola $y = x^2 - 6x + 11$ . What is the slope of this line?

Show answer & explanation

Answer: 2

Answer

The slope of the line is 2.

The correct slope is 2. Representing the line as

y = mx - 5

and setting it equal to the parabola

y = x^2 - 6x + 11

results in the quadratic equation

x^2 - (6 + m)x + 16 = 0

. For the line to be tangent, this equation must have exactly one real solution, meaning its discriminant must equal zero:

(6 + m)^2 - 4(1)(16) = 0

. Solving this gives

6 + m = 8

or

6 + m = -8

, which results in

m = 2

or

m = -14

. Since the problem specifies that the slope is positive, the value of

m

must be 2.

Step-by-Step Solution

1

Write the equation of the line in slope-intercept form.

y = mx - 5

, where

m > 0

.

The line passes through

(0, -5)

, which represents the y-intercept of the line.

2

Equate the line and the parabola to set up the system of equations.

x^2 - 6x + 11 = mx - 5

Setting the two expressions equal allows us to find the x-coordinates of any intersection points.

3

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - (6 + m)x + 16 = 0

Writing the equation in standard form identifies the coefficients

a = 1

,

b = -(6 + m)

, and

c = 16

needed for the discriminant.

4

Apply the condition for tangency by setting the discriminant to zero.

b^2 - 4ac = (-(6 + m))^2 - 4(1)(16) = 0 \implies (6 + m)^2 - 64 = 0

A line is tangent to a parabola if they touch at exactly one point, meaning the quadratic equation has exactly one real root (discriminant equals zero).

5

Solve the quadratic equation for the slope

m

and filter for the positive value.

(6+m)^2 = 64 \implies 6+m = \pm 8

, yielding

m = 2

or

m = -14

. Since the slope is positive,

m = 2

.

Solving the equation gives two possible slope values for tangent lines, and we select the positive slope as specified in the problem statement.

Key Concept

Determining tangency between a linear and quadratic equation by setting the discriminant of the intersection equation to zero.

Question 169Question →

For all positive real numbers $x$ , which of the following expressions is equivalent to $\log_3(27x^4) - \log_3(3x^2)$ ?

2 + 6\log_3(x)

\log_3(24x^2)

2 + 2\log_3(x)

4\log_3(x)

3 + 2\log_3(x)

Show answer & explanation

Answer: $2 + 2\log_3(x)$

Answer

2 + 2\log_3(x)

The correct answer is found by applying the quotient property of logarithms to combine the terms, yielding

\log_3(9x^2)

. Then, applying the product property splits this into

\log_3(9) + \log_3(x^2)

. Finally, evaluating

\log_3(9) = 2

and using the power property to rewrite

\log_3(x^2)

as

2\log_3(x)

yields the simplified expression

2 + 2\log_3(x)

.

Step-by-Step Solution

1

Apply the quotient property of logarithms:

\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)

.

\log_3\left(\frac{27x^4}{3x^2}\right)

To combine the two logarithmic terms into a single logarithm.

2

Simplify the algebraic expression inside the logarithm.

\log_3(9x^2)

Dividing the coefficients (

27 \div 3 = 9

) and subtracting the exponents of the variable

x

(

4 - 2 = 2

).

3

Apply the product property of logarithms:

\log_b(CD) = \log_b(C) + \log_b(D)

.

\log_3(9) + \log_3(x^2)

To separate the constant and variable parts of the logarithmic argument.

4

Evaluate the numerical logarithm and apply the power property of logarithms:

\log_b(y^k) = k\log_b(y)

.

2 + 2\log_3(x)

Since

3^2 = 9

,

\log_3(9) = 2

, and the exponent of

x

can be moved in front of the logarithm as a multiplier.

Key Concept

Logarithmic and Exponential Expressions and Equations

Alternative Method

Alternatively, expand each logarithm first using the product and power properties of logarithms:
1.

\log_3(27x^4) = \log_3(27) + \log_3(x^4) = 3 + 4\log_3(x)

2.

\log_3(3x^2) = \log_3(3) + \log_3(x^2) = 1 + 2\log_3(x)

Subtracting the second expanded expression from the first yields:

(3 + 4\log_3(x)) - (1 + 2\log_3(x)) = 3 - 1 + 4\log_3(x) - 2\log_3(x) = 2 + 2\log_3(x)

.

Estimated Time:1m 30s

Question 170Question →

If $y$ is a real number such that $|3 - \frac{1}{2}y| = 5$ , what is the sum of all possible values of $y$ ?

-4

12

16

-12

20

Show answer & explanation

Answer: 12

Answer

12

Solving the equation

|3 - \frac{1}{2}y| = 5

requires setting up two cases:

3 - \frac{1}{2}y = 5

and

3 - \frac{1}{2}y = -5

. Solving the first case gives

y = -4

, and solving the second case gives

y = 16

. The sum of these two solutions is

-4 + 16 = 12

.

Step-by-Step Solution

1

Set up the two equations represented by the absolute value expression.

3 - \frac{1}{2}y = 5

and

3 - \frac{1}{2}y = -5

An absolute value equation

|u| = C

(where

C \ge 0

) splits into two cases:

u = C

and

u = -C

.

2

Solve the first equation

3 - \frac{1}{2}y = 5

.

y = -4

Subtracting 3 from both sides yields

-\frac{1}{2}y = 2

. Multiplying both sides by

-2

isolates

y

.

3

Solve the second equation

3 - \frac{1}{2}y = -5

.

y = 16

Subtracting 3 from both sides yields

-\frac{1}{2}y = -8

. Multiplying both sides by

-2

isolates

y

.

4

Calculate the sum of all possible values of

y

.

12

Add the two solutions together:

-4 + 16 = 12

.

Key Concept

Absolute Value Equations

Estimated Time:1m 30s

Question 171Question →

Matrices $C$ and $D$ are defined as follows:

C = \begin{bmatrix} 4 & 1 \\ -2 & k \end{bmatrix}, \quad D = \begin{bmatrix} 3 & -1 \\ 5 & 2 \end{bmatrix}

If the element in the second row and first column of the product matrix $CD$ is $-11$ , what is the value of $k$ ?

Show answer & explanation

Answer: -1

Answer

The value of

k

is

-1

.

To find the element in the second row and first column of the product matrix

CD

, we take the dot product of the second row of

C

, which is

\begin{bmatrix} -2 & k \end{bmatrix}

, and the first column of

D

, which is

\begin{bmatrix} 3 \\ 5 \end{bmatrix}

. This gives

(-2)(3) + (k)(5) = -6 + 5k

. Setting this equal to the given value of

-11

yields the equation

-6 + 5k = -11

. Adding

6

to both sides gives

5k = -5

, and dividing by

5

results in

k = -1

.

Step-by-Step Solution

1

Identify the second row of matrix

C

and the first column of matrix

D

.

Row 2 of

C

is

\begin{bmatrix} -2 & k \end{bmatrix}

and Column 1 of

D

is

\begin{bmatrix} 3 \\ 5 \end{bmatrix}

.

To find the element in the second row and first column of the product matrix

CD

, we must compute the dot product of the second row of the first matrix (

C

) and the first column of the second matrix (

D

).

2

Multiply the corresponding elements of the row and column and add the products.

The element at row 2, column 1 of

CD

is

(-2)(3) + (k)(5) = -6 + 5k

.

This defines the matrix multiplication rule for that specific position in the resulting matrix.

3

Set the expression

-6 + 5k

equal to the given value of

-11

and solve the linear equation for

k

.

-6 + 5k = -11 \implies 5k = -5 \implies k = -1

.

Solving the equation yields the value of the unknown variable

k

.

Key Concept

2x2 Matrix Multiplication

Question 172Question →

If $\log_3(x + 4) = 4$ , what is the value of $x$ ?

Show answer & explanation

Answer: 77

Answer

The value of

x

is 77.

The correct answer is 77. To solve the equation

\log_3(x + 4) = 4

, convert the equation from its logarithmic form to its exponential form. Since

\log_b(a) = c

means

b^c = a

, the equation becomes

3^4 = x + 4

. Calculating

3^4

yields 81. Thus,

81 = x + 4

. Subtracting 4 from both sides gives

x = 77

.

Step-by-Step Solution

1

Rewrite the logarithmic equation in exponential form.

x + 4 = 3^4

By definition,

\log_b(a) = c

is equivalent to

b^c = a

.

2

Calculate the value of the exponential expression.

3^4 = 81

3^4 = 3 \times 3 \times 3 \times 3 = 81

.

3

Solve for

x

by isolating the variable.

x = 77

Subtract 4 from both sides of the equation:

81 - 4 = 77

.

Key Concept

Converting logarithmic equations to exponential form

Question 173Question →

For a real constant $k$ , the quadratic equation $x^2 - 2(k - 2)x + (k^2 - 3k + 2) = 0$ has two real roots, $r_1$ and $r_2$ . What is the value of $k$ that minimizes the sum of the squares of these roots, $r_1^2 + r_2^2$ ?

Show answer & explanation

Answer: 2

Answer

The value of

k

that minimizes the sum of the squares of the roots is 2.

Applying Vieta's formulas and algebraic identities, the sum of the squares of the roots is expressed as

2k^2 - 10k + 12

. The discriminant condition for the roots to be real requires

k \leq 2

. Since the vertex of the upward-opening parabola

2k^2 - 10k + 12

is at

k = 2.5

, the function is strictly decreasing for all

k \leq 2

. Thus, the minimum value on the interval

k \leq 2

occurs at the boundary

k = 2

.

Step-by-Step Solution

1

Use Vieta's formulas to find the sum and product of the roots.

r_1 + r_2 = 2(k - 2)

and

r_1 r_2 = k^2 - 3k + 2

Vieta's formulas relate the roots of a quadratic equation

Ax^2 + Bx + C = 0

to its coefficients by

r_1 + r_2 = -B/A

and

r_1 r_2 = C/A

.

2

Express the sum of the squares of the roots,

r_1^2 + r_2^2

, in terms of

k

.

r_1^2 + r_2^2 = 2k^2 - 10k + 12

Since

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2

, substituting the Vieta relations gives

[2(k-2)]^2 - 2(k^2 - 3k + 2) = 4(k^2 - 4k + 4) - 2k^2 + 6k - 4 = 2k^2 - 10k + 12

.

3

Determine the condition for the quadratic equation to have real roots using the discriminant.

\Delta = -4k + 8 \geq 0 \implies k \leq 2

For the roots

r_1

and

r_2

to be real, the discriminant

\Delta = B^2 - 4AC

must be greater than or equal to 0.

4

Minimize the quadratic function

f(k) = 2k^2 - 10k + 12

subject to the constraint

k \leq 2

.

k = 2

The parabola

f(k)

opens upwards with its vertex at

k = 2.5

. For

k \leq 2

, the function is strictly decreasing, meaning its minimum value on this interval occurs at the upper boundary,

k = 2

.

Key Concept

Quadratic Equations and the Quadratic Formula

Question 174Question →

For all real values of $q$ that satisfy the inequality $8 - 2|3q - 1| \ge -6$ , which of the following inequality expressions represents the complete solution set for $q$ ?

$

-2 \le q \le \frac{8}{3}

q \le -2\text{ or }q \ge \frac{8}{3}

q \le \frac{8}{3}

q \ge \frac{8}{3}

q \ge 0

Show answer & explanation

Answer: $ $-2 \le q \le \frac{8}{3}$

Answer

-2 \le q \le \frac{8}{3}

The correct answer is the interval containing all real numbers between

-2

and

\frac{8}{3}

inclusive. This is determined by subtracting 8 from both sides of the inequality, dividing by

-2

(which reverses the inequality sign to yield

|3q - 1| \le 7

), expressing this as the compound inequality

-7 \le 3q - 1 \le 7

, and isolating

q

.

Step-by-Step Solution

1

Subtract 8 from both sides of the inequality to isolate the absolute value term.

-2|3q - 1| \ge -14

To solve for the variable, we must first isolate the term containing the absolute value by performing inverse operations.

2

Divide both sides by

-2

and reverse the inequality sign.

|3q - 1| \le 7

Dividing both sides of an inequality by a negative number requires reversing the direction of the inequality sign.

3

Rewrite the absolute value inequality as a compound inequality.

-7 \le 3q - 1 \le 7

An inequality of the form

|x| \le d

(where

d \ge 0

) is equivalent to the compound inequality

-d \le x \le d

.

4

Add 1 to all three parts of the compound inequality.

-6 \le 3q \le 8

Adding 1 eliminates the constant term from the middle section of the inequality.

5

Divide all three parts by 3 to solve for

q

.

-2 \le q \le \frac{8}{3}

Dividing by 3 isolates the variable

q

, yielding the complete solution set.

Key Concept

Solving multi-step absolute value inequalities, including reversing the inequality sign when dividing by a negative number and setting up a compound inequality to represent both positive and negative cases.

Question 175Question →

A system of equations consists of the line $2x - y = 5$ and the parabola $y = x^2 - 4x + c$ , where $c$ is a constant. If the line and the parabola intersect at two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ such that the positive difference between their $x$ -coordinates is 2, what is the value of $c$ ?

0

3

5

8

11

Show answer & explanation

Answer: 3

Answer

3

To find the constant

c

, we equate the line and the parabola equations:

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

. Bringing all terms to one side gives the quadratic equation

x^2 - 6x + (c + 5) = 0

. Applying the quadratic formula, the

x

-coordinates of the intersection points are

x = 3 \pm \sqrt{4 - c}

. The positive difference between these coordinates is

(3 + \sqrt{4 - c}) - (3 - \sqrt{4 - c}) = 2\sqrt{4 - c}

. Setting this difference equal to the given value of 2 gives

2\sqrt{4 - c} = 2

, which simplifies to

\sqrt{4 - c} = 1

. Squaring both sides yields

4 - c = 1

, which gives

c = 3

. This corresponds to the correct option.

Step-by-Step Solution

1

Equate the equations for the line and the parabola to set up the equation for their intersection points.

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

The intersection points of the system of equations occur where the

y

-values are equal.

2

Rearrange the equation into standard quadratic form:

ax^2 + bx + c = 0

.

x^2 - 6x + (c + 5) = 0

Grouping the terms allows us to identify the coefficients:

a = 1

,

b = -6

, and the constant term is

c + 5

.

3

Apply the quadratic formula to find the

x

-coordinates of the intersection points in terms of

c

.

x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(c + 5)}}{2} = 3 \pm \sqrt{4 - c}

The quadratic formula yields the roots

x_1 = 3 + \sqrt{4 - c}

and

x_2 = 3 - \sqrt{4 - c}

.

4

Set up an equation representing the positive difference between the

x

-coordinates and solve for

c

.

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

The difference between the two coordinates is given as 2, which allows us to isolate and solve for

c

.

Key Concept

Solving systems of linear and non-linear equations by finding the intersection of a line and a parabola and using root properties to determine unknown constants

Alternative Method

Instead of using the quadratic formula, you can apply Vieta's formulas. Let the roots of

x^2 - 6x + (c + 5) = 0

be

x_1

and

x_2

. Vieta's formulas state that

x_1 + x_2 = 6

and

x_1 x_2 = c + 5

. We are given that

|x_1 - x_2| = 2

. Squaring this equation gives

(x_1 - x_2)^2 = 4

. Since

(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1 x_2

, we substitute the known values:

6^2 - 4(c + 5) = 4 \implies 36 - 4c - 20 = 4 \implies 16 - 4c = 4 \implies 4c = 12 \implies c = 3

.

Estimated Time:2m 0s

Question 176Question →

An arithmetic sequence $a_1, a_2, a_3, \dots$ has a first term $a_1 = a$ and a non-zero common difference $d$ . A geometric sequence $g_1, g_2, g_3, \dots$ has a first term $g_1 = a$ and a common ratio $r > 1$ . The third term of the arithmetic sequence is equal to the second term of the geometric sequence ( $a_3 = g_2$ ), and the seventh term of the arithmetic sequence is equal to the third term of the geometric sequence ( $a_7 = g_3$ ). If the sum of the first five terms of the arithmetic sequence is 150, what is the value of the fifth term of the geometric sequence, $g_5$ ?

240

480

800

96

810,000

Show answer & explanation

Answer: 240

Answer

The fifth term of the geometric sequence is 240.

The correct answer is 240. The term relationships

a_3 = g_2

and

a_7 = g_3

translate to

a + 2d = ar

and

a + 6d = ar^2

. Expressing the first equation as

2d = a(r - 1)

and substituting it into the second equation yields

a + 3a(r - 1) = ar^2

. Dividing by

a

gives the quadratic equation

r^2 - 3r + 2 = 0

. Since

r > 1

, we find

r = 2

, which implies

d = a/2

. The sum of the first five terms of the arithmetic sequence is

5(a + 2d) = 150

, which simplifies to

a + 2d = 30

. Substituting

d = a/2

gives

2a = 30

, so

a = 15

. The fifth term of the geometric sequence is then

g_5 = a \cdot r^4 = 15 \cdot 2^4 = 240

.

Step-by-Step Solution

1

Express the given sequence term relationships in terms of

a

,

d

, and

r

.

a_3 = a + 2d

,

a_7 = a + 6d

,

g_2 = ar

, and

g_3 = ar^2

.

To represent the equality of terms algebraically.

2

Solve the system of equations

a + 2d = ar

and

a + 6d = ar^2

for the common ratio

r

.

From the first equation,

2d = a(r - 1) \Rightarrow d = \frac{a(r - 1)}{2}

. Substituting this into the second equation gives

a + 3a(r - 1) = ar^2

. Dividing by

a

(since

a \neq 0

) yields

r^2 - 3r + 2 = 0

, which factors as

(r-1)(r-2) = 0

. Since

r > 1

, we must have

r = 2

. Thus,

d = a/2

.

To find the relationship between the arithmetic common difference, the geometric common ratio, and the first term.

3

Use the sum of the first five terms of the arithmetic sequence to solve for

a

.

S_5 = 5(a + 2d) = 150 \Rightarrow a + 2d = 30

. Substituting

d = a/2

gives

a + a = 30 \Rightarrow 2a = 30 \Rightarrow a = 15

.

To determine the numerical value of the first term.

4

Calculate the fifth term of the geometric sequence.

g_5 = a \cdot r^4 = 15 \cdot 2^4 = 15 \cdot 16 = 240

.

To find the requested term of the geometric sequence.

Key Concept

Solving systems of linear and exponential relationships using arithmetic and geometric sequence properties.

Question 177Question →

For the functions $f(x) = (x - 3)^2 - 2$ and $g(x) = |3x - 10|$ , what is the value of the composite function $g(f(1))$ ?

-4

4

24

28

40

Show answer & explanation

Answer: 4

Answer

The correct answer is 4

Evaluating

f(1)

gives

(1 - 3)^2 - 2 = (-2)^2 - 2 = 4 - 2 = 2

. Substituting this result into

g(x)

gives

g(2) = |3(2) - 10| = |6 - 10| = |-4| = 4

. Therefore, the value of the composite function is 4.

Step-by-Step Solution

1

Evaluate the inner function

f(x)

at

x = 1

.

f(1) = (1 - 3)^2 - 2 = (-2)^2 - 2 = 4 - 2 = 2

To evaluate the composite function

g(f(1))

, the output of the inner function

f(1)

must be calculated first.

2

Substitute the output

f(1) = 2

into the outer function

g(x)

.

g(2) = |3(2) - 10| = |6 - 10| = |-4| = 4

The output of the inner function becomes the input for the outer function

g(x)

.

Key Concept

Evaluating a composite function by first calculating the value of the inner function and then using that result as the input for the outer function.

Estimated Time:1m 0s

Question 178Question →

If $x$ and $y$ are positive real numbers such that $\log_x(y) = 2$ and $\log_4(x) + \log_2(y) = 5$ , what is the value of $y$ ?

Show answer & explanation

Answer: 16

Answer

16

Applying the definition of logarithms to

\log_x(y) = 2

yields

y = x^2

. Substituting this relationship into the second equation gives

\log_4(x) + \log_2(x^2) = 5

. Using the change of base formula, we rewrite

\log_4(x)

as

\frac{\log_2(x)}{\log_2(4)} = \frac{1}{2}\log_2(x)

, and using the power property, we rewrite

\log_2(x^2)

as

2\log_2(x)

. Combining the terms results in

\frac{5}{2}\log_2(x) = 5

, which simplifies to

\log_2(x) = 2

. Converting this back to exponential form gives

x = 2^2 = 4

. Substituting this value back into the relation

y = x^2

yields

y = 4^2 = 16

.

Step-by-Step Solution

1

Apply the definition of a logarithm to the first equation.

y = x^2

By definition,

\log_b(a) = c

is equivalent to

b^c = a

.

2

Substitute

y = x^2

into the second equation.

\log_4(x) + \log_2(x^2) = 5

Substitution reduces the system to a single equation in terms of

x

.

3

Express both logarithmic terms using base 2 properties.

\frac{1}{2}\log_2(x) + 2\log_2(x) = 5

The change of base formula gives

\log_4(x) = \frac{\log_2(x)}{\log_2(4)} = \frac{1}{2}\log_2(x)

, and the power property gives

\log_2(x^2) = 2\log_2(x)

.

4

Combine the coefficients and solve for

x

.

\log_2(x) = 2 \Rightarrow x = 2^2 = 4

Adding the coefficients gives

\frac{5}{2}\log_2(x) = 5

. Multiplying by

\frac{2}{5}

isolates

\log_2(x)

to solve for

x

.

5

Calculate the value of

y

using the relation from Step 1.

y = 4^2 = 16

Since

y = x^2

and

x = 4

, evaluating the square gives the final answer.

Key Concept

Solving systems of logarithmic equations using base conversion and definition of logarithms

Estimated Time:2m 0s

Question 179Question →

What is the complete solution set for the inequality $|4 - 3v| \ge 7$ ?

v \ge -1

or

v \le \frac{11}{3}

-1 \le v \le \frac{11}{3}

v \le -1

or

v \ge \frac{11}{3}

v \le -1

v \le -7

or

v \ge 7

Show answer & explanation

Answer: $v \le -1$ or $v \ge \frac{11}{3}$

Answer

The solution set is the union of two open-ended intervals, representing all values of

v

such that

v

is less than or equal to

-1

or

v

is greater than or equal to

\frac{11}{3}

.

To solve the absolute value inequality

|4 - 3v| \ge 7

, we rewrite it as two separate inequalities:

4 - 3v \ge 7

or

4 - 3v \le -7

. For the first case, subtracting 4 from both sides of

4 - 3v \ge 7

results in

-3v \ge 3

. Dividing by

-3

and reversing the inequality sign gives

v \le -1

. For the second case, subtracting 4 from both sides of

4 - 3v \le -7

results in

-3v \le -11

. Dividing by

-3

and reversing the inequality sign gives

v \ge \frac{11}{3}

. Combining these gives the correct solution set:

v \le -1

or

v \ge \frac{11}{3}

.

Step-by-Step Solution

1

Set up two separate linear inequalities based on the absolute value inequality template.

4 - 3v \ge 7

or

4 - 3v \le -7

An absolute value inequality of the form

|A| \ge B

where

B > 0

is equivalent to the compound statement

A \ge B

or

A \le -B

.

2

Solve the first inequality:

4 - 3v \ge 7

. Subtract 4 from both sides, then divide by

-3

and reverse the inequality sign.

v \le -1

Subtracting 4 yields

-3v \ge 3

. Dividing by the negative number

-3

requires reversing the direction of the inequality sign.

3

Solve the second inequality:

4 - 3v \le -7

. Subtract 4 from both sides, then divide by

-3

and reverse the inequality sign.

v \ge \frac{11}{3}

Subtracting 4 yields

-3v \le -11

. Dividing by the negative number

-3

requires reversing the direction of the inequality sign.

4

Combine the individual solutions into a single compound statement.

v \le -1

or

v \ge \frac{11}{3}

The solution to a 'greater than or equal to' absolute value inequality is the union of the individual solutions.

Key Concept

Solving absolute value inequalities by splitting them into two linear cases and correctly reversing the inequality sign when dividing by a negative number.

Alternative Method

Alternatively, test test-values from each interval. For instance, choosing

v = 0

(which is in the middle interval) yields

|4 - 0| = 4 \ge 7

, which is false. Choosing

v = -2

(which is in the left interval) yields

|4 - 3(-2)| = |10| = 10 \ge 7

, which is true. Choosing

v = 4

(which is in the right interval since

4 > \frac{11}{3}

) yields

|4 - 12| = |-8| = 8 \ge 7

, which is true. This confirms the solution set must cover

v \le -1

and

v \ge \frac{11}{3}

.

Estimated Time:1m 30s

Question 180Question →

If $a$ , $b$ , and $c$ are positive real numbers greater than $1$ such that $\log_b(a) = \frac{3}{2}$ and $\log_c(b) = \frac{4}{3}$ , what is the value of $\log_a(abc)$ ?

\frac{1}{3}

\frac{13}{6}

\frac{23}{9}

3

\frac{9}{2}

Show answer & explanation

Answer: $\frac{13}{6}$

Answer

\frac{13}{6}

The correct answer is obtained by expanding

\log_a(abc)

into

\log_a(a) + \log_a(b) + \log_a(c)

. Since

\log_a(a) = 1

and

\log_a(b)

is the reciprocal of

\log_b(a)

, we find

\log_a(b) = \frac{2}{3}

. Using the chain rule of change of base, we calculate

\log_a(c) = \log_a(b) \cdot \log_b(c) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}

. Adding these three parts together yields

1 + \frac{2}{3} + \frac{1}{2} = \frac{13}{6}

.

Step-by-Step Solution

1

Apply the logarithmic product rule to expand the target expression.

\log_a(abc) = \log_a(a) + \log_a(b) + \log_a(c)

The logarithm of a product is equal to the sum of the logarithms of its individual factors.

2

Evaluate the first two terms of the expanded expression.

\log_a(a) = 1

, and since

\log_b(a) = \frac{3}{2}

, we have

\log_a(b) = \frac{1}{\log_b(a)} = \frac{2}{3}

.

The logarithm of a base to itself is always

1

, and the base reciprocal property states that

\log_x(y) = \frac{1}{\log_y(x)}

.

3

Evaluate the third term

\log_a(c)

using base properties and the change of base formula.

Since

\log_c(b) = \frac{4}{3}

, we have

\log_b(c) = \frac{3}{4}

. Then,

\log_a(c) = \log_a(b) \cdot \log_b(c) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}

.

By applying the change of base formula, we can rewrite

\log_a(c)

in terms of base

b

as

\frac{\log_b(c)}{\log_b(a)} = \log_a(b) \cdot \log_b(c)

.

4

Sum the three evaluated logarithmic terms together.

1 + \frac{2}{3} + \frac{1}{2} = \frac{6}{6} + \frac{4}{6} + \frac{3}{6} = \frac{13}{6}

Combining the values of the individual terms gives the final value of the expanded expression.

Key Concept

Properties of Logarithms and Change of Base Formula

Estimated Time:2m 0s

Intermediate Algebra

Answer

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