Advanced Math Practice Questions — SAT — Page 9

Question 161Question →

\begin{aligned} y &= 3x^2 - 4x - 5 \\ y - 2x &= 4 \end{aligned}

If $(x_1, y_1)$ and $(x_2, y_2)$ are the solutions to the system of equations above, and $y_1 > y_2$ , what is the value of $x_1 + y_2$ ?

-3

-1

5

1

Show answer & explanation

Answer: 5

Answer

5

To find the solutions to the system of equations, substitute the expression for

y

from the second equation into the first equation. First, rewrite the second equation as

y = 2x + 4

. Substituting this into the first equation yields

2x + 4 = 3x^2 - 4x - 5

. Rearranging terms to set the equation to zero gives

3x^2 - 6x - 9 = 0

. Dividing the entire equation by 3 simplifies it to

x^2 - 2x - 3 = 0

. Factoring this quadratic equation gives

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

, which yields the solutions

x_1 = 3

and

x_2 = -1

. Next, find the corresponding

y

-coordinates by substituting these

x

-values back into the linear equation

y = 2x + 4

. For

x = 3

,

y_1 = 2(3) + 4 = 10

. For

x = -1

,

y_2 = 2(-1) + 4 = 2

. We are given that

y_1 > y_2

, which confirms that

(x_1, y_1) = (3, 10)

and

(x_2, y_2) = (-1, 2)

. Finally, calculate

x_1 + y_2

, which is

3 + 2 = 5

.

Step-by-Step Solution

1

Rewrite the linear equation to express

y

in terms of

x

.

y = 2x + 4

This allows for substitution into the quadratic equation.

2

Substitute

y = 2x + 4

into the quadratic equation and set to zero.

3x^2 - 6x - 9 = 0

To find the

x

-coordinates of the intersection points.

3

Simplify and factor the quadratic equation.

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

, giving

x = 3

and

x = -1

.

To solve for the

x

-values of the intersection points.

4

Substitute the

x

-values back into the linear equation to find the corresponding

y

-values.

For

x = 3

,

y = 10

. For

x = -1

,

y = 2

. The intersection points are

(3, 10)

and

(-1, 2)

.

To find the complete coordinates of the intersection points.

5

Identify

(x_1, y_1)

and

(x_2, y_2)

using the condition

y_1 > y_2

, and compute

x_1 + y_2

.

Since

10 > 2

,

y_1 = 10

(with

x_1 = 3

) and

y_2 = 2

(with

x_2 = -1

). Then,

x_1 + y_2 = 3 + 2 = 5

.

To compute the required target expression.

Key Concept

Solving a nonlinear system of equations by substituting a linear expression into a quadratic equation and solving the resulting quadratic equation.

Question 162Question →

If $3^{2x - 1} \cdot 9^{y + 1} = 27^{x + y}$ , which of the following equations expresses $x$ in terms of $y$ ?

x = 1 - y

x = y + 1

x = -y

x = -y - 1

Show answer & explanation

Answer: $x = 1 - y$

Answer

The equation expressing

x

in terms of

y

is

x = 1 - y

.

The correct option is

x = 1 - y

. To solve the equation, each term is rewritten with base 3:

3^{2x-1} \cdot (3^2)^{y+1} = (3^3)^{x+y}

. Using the exponent rule

(b^m)^n = b^{mn}

, this becomes

3^{2x-1} \cdot 3^{2y+2} = 3^{3x+3y}

. Using the rule

b^m \cdot b^n = b^{m+n}

, the left side simplifies to

3^{2x+2y+1}

. Equating the exponents gives

2x+2y+1 = 3x+3y

. Subtracting

2x

and

2y

from both sides results in

1 = x+y

, which isolated for

x

gives

x = 1-y

.

Step-by-Step Solution

1

Rewrite all bases in the equation as powers of 3.

Since

9 = 3^2

and

27 = 3^3

, the equation becomes

3^{2x - 1} \cdot (3^2)^{y + 1} = (3^3)^{x + y}

.

To apply exponent rules and combine terms, all bases must be identical.

2

Apply the power of a power rule

(b^m)^n = b^{mn}

to simplify the exponents.

3^{2x - 1} \cdot 3^{2y + 2} = 3^{3x + 3y}

Multiplying the outer exponent by the inner exponent simplifies the terms on both sides.

3

Apply the product of powers rule

b^m \cdot b^n = b^{m+n}

to combine the terms on the left side of the equation.

3^{(2x - 1) + (2y + 2)} = 3^{3x + 3y}

, which simplifies to

3^{2x + 2y + 1} = 3^{3x + 3y}

.

Adding exponents when multiplying expressions with the same base allows the left side to be represented as a single exponential term.

4

Set the exponents equal to each other and solve for

x

in terms of

y

.

2x + 2y + 1 = 3x + 3y \implies 1 = x + y \implies x = 1 - y

If two exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Key Concept

Solving exponential equations by expressing terms with a common base and applying exponent laws.

Estimated Time:1m 30s

Question 163Question →

If $(x - 4)^2 = 81$ and $x > 0$ , what is the value of $x$ ?

Show answer & explanation

Answer: 13

Answer

The value of

x

is 13.

Taking the square root of both sides of

(x - 4)^2 = 81

yields

x - 4 = 9

or

x - 4 = -9

. Solving these linear equations gives

x = 13

or

x = -5

. Since

x > 0

is specified, the correct value of

x

is 13.

Step-by-Step Solution

1

Take the square root of both sides of the equation.

x - 4 = 9

or

x - 4 = -9

Applying the square root property to solve the quadratic equation.

2

Solve each linear equation for

x

.

x = 13

or

x = -5

Adding 4 to both sides of each equation.

3

Apply the given constraint that

x > 0

.

x = 13

Since

-5

is not greater than 0, the only positive solution is 13.

Key Concept

Solving quadratic equations by taking square roots

Question 1640% of students answered this correctlyQuestion →

For all $x > 3$ , which of the following is equivalent to the expression $\frac{2x^2 - 5x - 3}{x^2 - 9} - \frac{x}{x+3}$ ?

\frac{x + 1}{x + 3}

\frac{x - 1}{x + 3}

\frac{3x + 1}{x + 3}

\frac{2x^2 - 6x - 3}{x^2 - 9}

Show answer & explanation

Answer: $\frac{x + 1}{x + 3}$

Answer

\frac{x + 1}{x + 3}

To simplify the expression, we factor the first term:

\frac{2x^2 - 5x - 3}{x^2 - 9} = \frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)}

. Canceling the common factor of

x - 3

leaves

\frac{2x + 1}{x + 3}

. Subtracting the second term gives

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

. This matches the correct expression.

Step-by-Step Solution

1

Factor the numerator and denominator of the first rational expression.

The numerator

2x^2 - 5x - 3

factors into

(2x + 1)(x - 3)

. The denominator

x^2 - 9

is a difference of squares and factors into

(x - 3)(x + 3)

.

Factoring allows for the simplification of the rational expression by identifying common factors in the numerator and denominator.

2

Simplify the first rational expression by canceling the common factor.

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

for all

x > 3

.

Since

x > 3

, the term

x - 3

is non-zero, allowing us to divide both the numerator and denominator by

x - 3

.

3

Subtract the second expression from the simplified first expression.

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

Since both fractions now have the same denominator,

x + 3

, we subtract their numerators.

Key Concept

Simplifying rational expressions by factoring polynomials and performing operations on fractions with common denominators.

Estimated Time:1m 30s

Question 165Question →

A scientist models the population of a bacteria culture using the function $P(t) = P_0 \cdot b^t$ , where $P(t)$ is the population $t$ hours after the start of the experiment, $P_0$ is the initial population, and $b$ is a constant. The table below shows the population at two different times:

$t$ (hours)	$P(t)$
$2$	$1,800$
$5$	$48,600$

If the population of the bacteria culture is $1,312,200$ after $k$ hours, what is the value of $k$ ?

4

8

9

14

Show answer & explanation

Answer: 8

Answer

8

The correct value is 8. By setting up the ratio of the population at

t = 5

to

t = 2

, we find that

\frac{P(5)}{P(2)} = b^3 = \frac{48,600}{1,800} = 27

, which yields an hourly growth factor of

b = 3

. Using

P(2) = P_0 \cdot 3^2 = 1,800

, we determine the initial population

P_0

is 200. To find the hour

k

when the population is

1,312,200

, we solve

200 \cdot 3^k = 1,312,200

, which simplifies to

3^k = 6,561

. Since

3^8 = 6,561

, we find

k = 8

. Alternatively, we can calculate

\frac{P(k)}{P(5)} = \frac{1,312,200}{48,600} = 27 = 3^3

, meaning the population triples 3 more times after

t = 5

, giving

k = 5 + 3 = 8

.

Step-by-Step Solution

1

Set up the ratio of the population at

t = 5

to the population at

t = 2

to find the growth factor

b

.

\frac{P(5)}{P(2)} = \frac{P_0 \cdot b^5}{P_0 \cdot b^2} = b^3 = \frac{48,600}{1,800} = 27

Dividing the function values eliminates the initial population

P_0

and isolates the base

b

.

2

Solve for the growth factor

b

.

b = 3

Since

b^3 = 27

, taking the cube root of both sides gives

b = 3

.

3

Find the initial population

P_0

using

P(2) = 1,800

and

b = 3

.

1,800 = P_0 \cdot 3^2 \Rightarrow 1,800 = 9P_0 \Rightarrow P_0 = 200

Substituting the known values into the function formula allows us to solve for

P_0

.

4

Set up the equation for the population after

k

hours and solve for

k

.

1,312,200 = 200 \cdot 3^k \Rightarrow 3^k = \frac{1,312,200}{200} = 6,561

Dividing both sides by 200 isolates the exponential term

3^k

.

5

Find the exponent

k

by expressing 6,561 as a power of 3.

3^k = 3^8 \Rightarrow k = 8

Since

3^8 = 6,561

, the exponents must be equal, so

k = 8

.

Key Concept

Determining parameters of an exponential growth function from given data points and using the function to solve for an unknown time variable.

Alternative Method

Instead of solving for the initial population

P_0

, we can compare the target population of

1,312,200

to the population at

t = 5

. Since the population triples every hour (

b = 3

), the ratio of the population at

t = k

to the population at

t = 5

is

3^{k-5}

. We have

\frac{1,312,200}{48,600} = 27

, which is

3^3

. Therefore,

3^{k-5} = 3^3

, which directly gives

k - 5 = 3

, or

k = 8

.

Estimated Time:1m 15s

Question 166Question →

If $x$ and $y$ are positive numbers, which of the following is equivalent to the expression $\frac{\sqrt{36x^5 y^3}}{3x(xy^3)^{1/2}}$ ?

2x\sqrt{x}

2x

12x

2xy

Show answer & explanation

Answer: $2x$

Answer

The expression is equivalent to

2x

.

To find the equivalent expression, simplify the numerator and the denominator separately. The numerator

\sqrt{36x^5 y^3}

simplifies to

6x^{5/2}y^{3/2}

. The denominator

3x(xy^3)^{1/2}

simplifies to

3x^{3/2}y^{3/2}

. Dividing the simplified numerator by the simplified denominator yields

\frac{6x^{5/2}y^{3/2}}{3x^{3/2}y^{3/2}} = 2x

.

Step-by-Step Solution

1

Convert the radical expression in the numerator to an expression with fractional exponents.

\sqrt{36x^5 y^3} = (36x^5 y^3)^{1/2} = 6x^{5/2}y^{3/2}

The square root of a product is the product of the square roots of each factor, where

\sqrt{a} = a^{1/2}

.

2

Apply exponent rules to simplify the denominator.

3x(xy^3)^{1/2} = 3x^1 \cdot x^{1/2} y^{3/2} = 3x^{3/2}y^{3/2}

When multiplying terms with the same base, add their exponents (

x^a \cdot x^b = x^{a+b}

).

3

Divide the simplified numerator by the simplified denominator.

\frac{6x^{5/2}y^{3/2}}{3x^{3/2}y^{3/2}} = 2x^{5/2 - 3/2}y^{3/2 - 3/2} = 2x^1 y^0 = 2x

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

\frac{x^a}{x^b} = x^{a-b}

).

Key Concept

Equivalent Algebraic Expressions

Estimated Time:1m 15s

Question 167Question →

Which of the following expressions is equivalent to $\frac{x^2 - 4x - 12}{2x - 12} - \frac{x}{4}$ for any real number $x > 6$ ?

\frac{x - 4}{4}

\frac{x + 2}{4}

\frac{x + 4}{4}

\frac{3x + 4}{4}

Show answer & explanation

Answer: $\frac{x + 4}{4}$

Answer

The expression stating that the equivalent form is the fraction with

x+4

in the numerator and

4

in the denominator

The correct equivalent expression is found by first factoring the rational expression

\frac{x^2 - 4x - 12}{2x - 12}

into

\frac{(x - 6)(x + 2)}{2(x - 6)}

. Since

x > 6

, the factor

x - 6

is non-zero and can be canceled, leaving

\frac{x + 2}{2}

. Converting

\frac{x + 2}{2}

to have a common denominator of

4

yields

\frac{2x + 4}{4}

. Finally, subtracting

\frac{x}{4}

gives the simplified result of

\frac{x + 4}{4}

.

Step-by-Step Solution

1

Factor the quadratic numerator and the linear denominator of the first rational term.

\frac{x^2 - 4x - 12}{2x - 12} = \frac{(x - 6)(x + 2)}{2(x - 6)}

To identify and cancel common algebraic factors in the fraction.

2

Cancel the common factor

x - 6

from the numerator and denominator.

\frac{x + 2}{2}

Since

x > 6

, the term

x - 6

is non-zero and can be canceled.

3

Find a common denominator of

4

to perform subtraction with the second term,

-\frac{x}{4}

.

\frac{2(x + 2)}{4} - \frac{x}{4} = \frac{2x + 4}{4} - \frac{x}{4}

To subtract fractions, their denominators must be identical.

4

Subtract the numerators and combine terms over the common denominator.

\frac{2x + 4 - x}{4} = \frac{x + 4}{4}

Subtracting like terms in the numerator yields the final simplified expression.

Key Concept

Equivalent Algebraic Expressions

Alternative Method

Instead of simplifying algebraically, you can substitute a value for

x

that is greater than

6

. For example, if

x = 8

, the original expression evaluates to

\frac{8^2 - 4(8) - 12}{2(8) - 12} - \frac{8}{4} = \frac{20}{4} - 2 = 3

. Evaluating the correct option with

x = 8

yields

\frac{8+4}{4} = 3

, confirming equivalence.

Estimated Time:1m 30s

Question 168Question →

The value of a certain car model decreases by $12\%$ each year. If the initial value of the car is $\$25,000$ , which of the following functions models the value of the car, $V(t)$ , in dollars, $t$ years after it was purchased?

V(t) = 25,000 - 0.12t

V(t) = 25,000(0.12)^t

V(t) = 25,000(0.88)^t

V(t) = 25,000(1.12)^t

Show answer & explanation

Answer: $V(t) = 25,000(0.88)^t$

Answer

The function

V(t) = 25,000(0.88)^t

The initial value of the car is

25,000

. An annual decrease of

12\%

means that the car retains

100\% - 12\% = 88\%

of its value from the previous year. This results in a decay factor of

1 - 0.12 = 0.88

. Substituting these values into the exponential model

V(t) = a(b)^t

yields

V(t) = 25,000(0.88)^t

.

Step-by-Step Solution

1

Identify the initial value of the car.

The initial value is

25,000

.

This is the value of the car at

t = 0

, which corresponds to the coefficient

a

in the general exponential model

V(t) = a(b)^t

.

2

Determine the growth or decay factor base,

b

.

The base is

1 - 0.12 = 0.88

.

Since the car's value decreases by

12\%

each year, it retains

100\% - 12\% = 88\%

of its value. This is represented as a decay factor of

0.88

.

3

Formulate the exponential decay equation.

The equation is

V(t) = 25,000(0.88)^t

.

By substituting the initial value

a = 25,000

and the decay base

b = 0.88

into

V(t) = a(b)^t

.

Key Concept

Exponential decay models and interpretation of growth/decay factors

Alternative Method

Calculate the value of the car after one year (

t = 1

). A

12\%

decrease on a

\$25,000

car means its value drops by

\$3,000

to

\$22,000

. Substituting

t=1

into the correct function should yield

22,000

. Calculating for

V(t) = 25,000(0.88)^t

at

t = 1

gives

25,000(0.88)^1 = 22,000

, confirming the choice.

Estimated Time:45s

Question 169Question →

A quadratic function $f$ is defined by $f(x) = 2(x + 3)^2 - 8$ . What are the coordinates of the vertex of the graph of $y = f(x)$ in the $xy$ -plane?

(-3, -8)

(3, -8)

(-3, 8)

(3, 8)

Show answer & explanation

Answer: $(-3, -8)$

Answer

The correct answer is the coordinate pair

(-3, -8)

.

The vertex form of a quadratic function is

f(x) = a(x - h)^2 + k

, where the vertex of the graph of the function in the

xy

-plane is

(h, k)

. For the function

f(x) = 2(x + 3)^2 - 8

, the expression inside the parentheses can be rewritten as

(x - (-3))

, showing that

h = -3

. The constant term outside the parentheses is

-8

, showing that

k = -8

. Therefore, the vertex of the graph of the function is

(-3, -8)

.

Step-by-Step Solution

1

Identify the standard vertex form of a quadratic function.

The vertex form is

y = a(x - h)^2 + k

, where

(h, k)

is the vertex of the parabola.

Comparing the given equation to this template allows us to read the coordinates directly.

2

Rewrite the given equation to align with the negative sign inside the parentheses of the vertex form.

The term

(x + 3)^2

can be rewritten as

(x - (-3))^2

, and the subtraction of

8

can be written as adding

-8

. This yields

f(x) = 2(x - (-3))^2 + (-8)

.

Matching the template exactly prevents sign errors when identifying the values of

h

and

k

.

3

Extract the values of

h

and

k

.

h = -3

and

k = -8

, which means the vertex is located at

(-3, -8)

.

The vertex

(h, k)

coordinates represent the minimum point of this upward-opening parabola.

Key Concept

Identifying the vertex coordinates directly from the vertex form of a quadratic function.

Alternative Method

An alternative method is to expand the function into the standard form

f(x) = ax^2 + bx + c

. Expanding

f(x) = 2(x + 3)^2 - 8

gives

f(x) = 2(x^2 + 6x + 9) - 8

, which simplifies to

f(x) = 2x^2 + 12x + 10

. The

x

-coordinate of the vertex can be found using the formula

x = -\frac{b}{2a} = -\frac{12}{2(2)} = -3

. Substituting

x = -3

back into the function gives

f(-3) = 2(-3 + 3)^2 - 8 = -8

. This confirms the vertex is at

(-3, -8)

.

Estimated Time:45s

Question 170Question →

In the quadratic equation $x^2 - 15x + k = 0$ , where $k$ is a constant, one of the real solutions is $4$ times the other solution. What is the value of $k$ ?

Show answer & explanation

Answer: 36

Answer

36

The correct answer is 36. By Vieta's formulas, the sum of the roots of the quadratic equation

x^2 - 15x + k = 0

is

15

, and the product of the roots is

k

. Let the roots be

r

and

4r

. Their sum is

r + 4r = 5r = 15

, which gives

r = 3

. The roots are

3

and

4(3) = 12

. The product of the roots is

3 \times 12 = 36

, which represents the constant term

k

.

Step-by-Step Solution

1

Express the relationship between the two roots.

Let the two roots be

r

and

4r

.

We are given that one of the real solutions is 4 times the other solution.

2

Use Vieta's formulas to set up an equation for the sum of the roots.

r + 4r = 15 \implies 5r = 15

For a quadratic equation in the form

x^2 - sx + p = 0

, the sum of the roots is equal to

s

.

3

Solve for the variable

r

and find both roots.

r = 3

, so the roots are

3

and

4(3) = 12

.

Dividing both sides of

5r = 15

by 5 gives

r = 3

. Substituting this back gives the two roots.

4

Calculate the product of the roots to find the value of

k

.

k = 3 \times 12 = 36

For a quadratic equation in the form

x^2 - sx + p = 0

, the product of the roots is equal to

p

, which in this case is

k

.

Key Concept

Using Vieta's formulas to find relationships between the coefficients and roots of a quadratic equation.

Question 171Question →

\begin{aligned} y &= 2x^2 - kx + 8 \\ y &= 4x - 10 \end{aligned}

In the system of equations above, $k$ is a constant. If the system has exactly one real solution, what is the product of all possible values of $k$ ?

-48

-128

-20

128

Show answer & explanation

Answer: -128

Answer

-128

To find the number of solutions to the system, set the two equations equal to each other:

2x^2 - kx + 8 = 4x - 10

. Moving all terms to one side gives

2x^2 - (k + 4)x + 18 = 0

. For this quadratic equation to have exactly one real solution, its discriminant,

\Delta = B^2 - 4AC

, must equal zero. Substituting the coefficients

A = 2

,

B = -(k + 4)

, and

C = 18

into the discriminant formula gives

(-(k + 4))^2 - 4(2)(18) = 0

, which simplifies to

(k + 4)^2 - 144 = 0

. Solving this equation gives

k + 4 = 12 \implies k = 8

, or

k + 4 = -12 \implies k = -16

. The product of these possible values is

8 \times (-16) = -128

.

Step-by-Step Solution

1

Equate the quadratic and linear equations to find the condition for intersection.

2x^2 - kx + 8 = 4x - 10

To find the points where the two graphs intersect, their

y

-values must be set equal to each other.

2

Rearrange the terms to form a standard quadratic equation of the form

Ax^2 + Bx + C = 0

.

2x^2 - (k + 4)x + 18 = 0

Grouping terms allows us to determine the coefficients

A = 2

,

B = -(k + 4)

, and

C = 18

to use in the discriminant.

3

Set the discriminant

\Delta = B^2 - 4AC

of the quadratic equation to

0

.

[-(k + 4)]^2 - 4(2)(18) = 0 \implies (k + 4)^2 - 144 = 0

A system of a line and a parabola has exactly one real solution (meaning the line is tangent to the parabola) when the resulting quadratic equation has exactly one real root, which corresponds to a discriminant of zero.

4

Solve for the possible values of

k

and calculate their product.

k + 4 = \pm 12 \implies k = 8

or

k = -16

. The product is

8 \times (-16) = -128

.

Extracting the square root yields two possible values for

k

, and multiplying them gives the requested product.

Key Concept

Using the discriminant of a quadratic equation to find parameter values that make a linear-quadratic system have exactly one real solution.

Alternative Method

Once the quadratic equation

(k + 4)^2 - 144 = 0

is simplified to

k^2 + 8k - 128 = 0

, Vieta's formulas can be applied directly to find the product of the possible values of

k

. According to Vieta's formulas, the product of the roots of a quadratic equation

ak^2 + bk + c = 0

is given by

c/a

. Here,

a = 1

and

c = -128

, so the product of the roots is

-128 / 1 = -128

.

Estimated Time:2m 0s

Question 172Question →

The population of a colony of bacteria doubles every $15$ minutes. The population of the colony, $P(t)$ , after $t$ hours can be modeled by the function $P(t) = P_0 \cdot 8^{kt}$ , where $P_0$ is the initial population of the colony and $k$ is a constant. What is the value of $k$ ?

\frac{1}{12}

\frac{3}{4}

\frac{4}{3}

12

Show answer & explanation

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

Answer

The correct value of

k

is

\frac{4}{3}

.

Since the population doubles every

15

minutes, it doubles

4

times per hour. Therefore, the population after

t

hours is

P_0 \cdot 2^{4t}

. To express this model in the form

P_0 \cdot 8^{kt}

, we equate the growth factors:

2^{4t} = 8^{kt}

. Since

8 = 2^3

, this can be rewritten as

2^{4t} = (2^3)^{kt} = 2^{3kt}

. Setting the exponents equal gives

4t = 3kt

, which simplifies to

4 = 3k

and yields

k = \frac{4}{3}

.

Step-by-Step Solution

1

Determine the number of doubling periods per hour.

The population doubles

4

times per hour.

Since there are

60

minutes in an hour and the population doubles every

15

minutes, the number of doubling periods in one hour is

\frac{60}{15} = 4

.

2

Formulate the population growth equation in terms of base 2.

P(t) = P_0 \cdot 2^{4t}

If the population doubles

4

times per hour, then after

t

hours, it has doubled

4t

times, representing a growth factor of

2^{4t}

.

3

Equate the growth model to the given function and express using a common base.

2^{4t} = (2^3)^{kt} \implies 2^{4t} = 2^{3kt}

Set the two expressions for the population equal:

P_0 \cdot 2^{4t} = P_0 \cdot 8^{kt}

. Dividing by

P_0

gives

2^{4t} = 8^{kt}

. Since

8 = 2^3

, rewrite the right side to have a base of

2

.

4

Solve for the constant

k

by equating the exponents.

4t = 3kt \implies 4 = 3k \implies k = \frac{4}{3}

Since the bases on both sides of the equation are equal, their exponents must also be equal. Divide both sides by

t

(since

t > 0

) and isolate

k

.

Key Concept

Converting bases in exponential functions and adjusting exponents for different units of time.

Question 173Question →

If $5^{a + 2} = 25^{b}$ and $a + b = 10$ , what is the value of $b$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

The correct answer is 4. By rewriting

25^b

as

(5^2)^b = 5^{2b}

, we can set the exponents equal to get

a + 2 = 2b

. Substituting

a = 10 - b

from the second equation into this relation yields

(10 - b) + 2 = 2b

. Simplifying this gives

12 = 3b

, which results in

b = 4

.

Step-by-Step Solution

1

Rewrite the exponential equation using a common base.

5^{a + 2} = 5^{2b}

Since

25 = 5^2

, we write

25^b

as

(5^2)^b = 5^{2b}

to match the base on the left side.

2

Equate the exponents from the base 5 terms.

a + 2 = 2b

When bases are equal, their exponents must be equal for the equation to hold.

3

Express

a

in terms of

b

using the linear equation

a + b = 10

.

a = 10 - b

This allows for substitution into the equation from Step 2 to solve for a single variable.

4

Substitute

a = 10 - b

into

a + 2 = 2b

and solve for

b

.

b = 4

Substituting gives

(10 - b) + 2 = 2b \implies 12 - b = 2b \implies 12 = 3b \implies b = 4

.

Key Concept

Solving systems of equations involving exponential equations by finding a common base.

Question 174Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = a(x-3)(x-11)$ , where $a$ is a constant, has a vertex at $(h, k)$ . The graph of the quadratic function $g(x) = -(x-h)^2 + q$ has its vertex at $(h, k+8)$ . If the graphs of $f$ and $g$ intersect at a point with an $x$ -coordinate of $5$ , what is the value of $a$ ?

Show answer & explanation

Answer: 1

Answer

1

The value of

a

is 1. By finding the vertex of

f(x)

at

(7, -16a)

and using the vertex of

g(x)

at

(7, -16a+8)

, we evaluate both functions at

x=5

. Setting

f(5) = g(5)

yields the linear equation

-12a = -16a + 4

, which simplifies to

a = 1

.

Step-by-Step Solution

1

Find the x-coordinate of the vertex of

f(x)

h = 7

The x-coordinate of the vertex of a quadratic function of the form

f(x) = a(x-r_1)(x-r_2)

is the midpoint of its x-intercepts,

r_1 = 3

and

r_2 = 11

.

2

Find the y-coordinate of the vertex of

f(x)

in terms of

a

k = -16a

Evaluate the function at the vertex x-coordinate:

f(7) = a(7-3)(7-11) = -16a

.

3

Write the equation of

g(x)

using the given vertex information

g(x) = -(x-7)^2 + k+8

The vertex of

g(x)

is given as

(h, k+8)

. Since

h = 7

, the vertex form of

g(x)

is

g(x) = -(x-7)^2 + k+8

.

4

Set up the equation for the intersection point at

x = 5

f(5) = g(5)

Since the graphs of

f

and

g

intersect at

x=5

, their y-values must be equal at this point.

5

Substitute expressions and solve for

a

a = 1

Evaluate both sides:

f(5) = a(5-3)(5-11) = -12a

, and

g(5) = -(5-7)^2 + k+8 = -4 + k+8 = k+4

. Substituting

k = -16a

gives

-12a = -16a + 4 \implies 4a = 4 \implies a = 1

.

Key Concept

Quadratic Functions and Graphs

Question 175Question →

The population of a certain species of wildflower in a national park increases by $300\%$ every $8$ years. The population can be modeled by the function $P(t) = P_0(1 + r)^t$ , where $P_0$ is the initial population, $r$ is the constant annual growth rate, and $t$ is the time in years. Which of the following is closest to the value of $r$ ?

0.15

0.19

0.38

1.19

Show answer & explanation

Answer: 0.19

Answer

The correct value of the annual growth rate is

0.19

.

An increase of

300\%

means the population becomes

100\% + 300\% = 400\%

of its initial value, which corresponds to a growth factor of

4

over an

8

-year period. The annual growth rate

r

satisfies the equation

(1 + r)^8 = 4

. Solving for

1 + r

gives

1 + r = 4^{1/8} = (2^2)^{1/8} = 2^{1/4} \approx 1.1892

. Subtracting

1

from both sides yields the annual growth rate

r \approx 0.1892

, which is closest to

0.19

.

Step-by-Step Solution

1

Determine the total growth factor over the 8-year period from the percentage increase.

The population increases by

300\%

, meaning the new population is

100\% + 300\% = 400\%

of the initial population. This corresponds to a growth factor of

4

every

8

years.

An increase of

300\%

adds

3

times the original value to the original value, resulting in a total of

4

times the original value.

2

Set up the equation relating the annual growth rate

r

to the 8-year growth factor.

(1+r)^8 = 4

Since the population is multiplied by

1+r

each year, after

8

years it is multiplied by

(1+r)^8

, which must equal the total growth factor of

4

.

3

Solve the equation for

r

and round to the nearest hundredth.

1+r = 4^{1/8} \approx 1.1892 \implies r \approx 0.1892 \approx 0.19

Taking the eighth root of both sides gives the annual growth factor

1+r \approx 1.19

. Subtracting

1

isolates the annual growth rate

r

.

Key Concept

Relating exponential growth rates and growth factors across different time periods

Alternative Method

Instead of solving

(1+r)^8 = 4

, we can write the function directly using the 8-year growth factor as

P(t) = P_0 \cdot 4^{t/8}

. Rewriting this in the form

P(t) = P_0(4^{1/8})^t

shows that the annual growth factor is

4^{1/8}

. We can approximate

4^{1/8} \approx 1.1892

, which corresponds to an annual growth rate

r = 1.1892 - 1 = 0.1892

, or approximately

0.19

.

Estimated Time:1m 30s

Question 176Question →

If $4^{2x + 1} = 8^{x - 1}$ , what is the value of $x$ ?

-5

-2

-1

5

Show answer & explanation

Answer: -5

Answer

-5

The correct answer is

-5

. By converting the bases of both sides to

2

, the equation becomes

2^{2(2x+1)} = 2^{3(x-1)}

. Equating the exponents gives

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

, which expands to

4x+2 = 3x-3

. Solving for

x

yields

x = -5

.

Step-by-Step Solution

1

Express both bases as powers of 2

4 = 2^2

and

8 = 2^3

To solve an exponential equation, we need to rewrite the terms with a common base.

2

Substitute the bases back into the equation and apply exponent power rules

(2^2)^{2x + 1} = (2^3)^{x - 1} \Rightarrow 2^{2(2x + 1)} = 2^{3(x - 1)}

Applying the power of a power property,

(b^m)^n = b^{mn}

, allows us to simplify the exponents.

3

Set the exponents equal to each other

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

Since the bases are identical and positive, their exponents must be equal.

4

Distribute and solve the linear equation for

x

4x + 2 = 3x - 3 \Rightarrow x = -5

Distributing the constants gives

4x + 2 = 3x - 3

. Subtracting

3x

from both sides gives

x + 2 = -3

, and subtracting

2

from both sides isolates the variable to find

x = -5

.

Key Concept

Solving exponential equations by expressing both sides with a common base

Question 177Question →

The quadratic equation $x^2 - (k + 3)x + (2k + 3) = 0$ has exactly one real solution, where $k$ is a constant. If $k > 1$ , what is the value of $k$ ?

-1

1

3

5

Show answer & explanation

Answer: 3

Answer

The value of

k

is

3

.

For the quadratic equation to have exactly one real solution, its discriminant must be equal to

0

. Setting the discriminant of

x^2 - (k + 3)x + (2k + 3) = 0

to

0

yields

[-(k + 3)]^2 - 4(1)(2k + 3) = 0

. Simplifying this yields

k^2 - 2k - 3 = 0

, which factors as

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

. This gives two potential values for

k

:

3

and

-1

. Because the problem specifies that

k > 1

, the value of

k

must be

3

.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation and state the condition for having exactly one real solution.

The coefficients of the quadratic equation

x^2 - (k + 3)x + (2k + 3) = 0

are

a = 1

,

b = -(k + 3)

, and

c = 2k + 3

. For a quadratic equation to have exactly one real solution, its discriminant

b^2 - 4ac

must equal

0

.

A discriminant of zero indicates that the quadratic equation has two equal real roots, which is equivalent to exactly one distinct real solution.

2

Set up the discriminant equation in terms of

k

and simplify it.

[-(k + 3)]^2 - 4(1)(2k + 3) = 0

(k^2 + 6k + 9) - (8k + 12) = 0

k^2 - 2k - 3 = 0

This algebraic expansion and simplification transforms the discriminant condition into a solvable quadratic equation for

k

.

3

Solve the quadratic equation

k^2 - 2k - 3 = 0

by factoring.

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

, which gives

k = 3

or

k = -1

.

Factoring is the most direct method to find the roots of this quadratic equation.

4

Apply the constraint

k > 1

to select the correct value of

k

.

Since

k

must be greater than

1

, the value

k = -1

is discarded, leaving

k = 3

.

The solution must satisfy all given constraints in the problem statement.

Key Concept

Discriminant of a Quadratic Equation

Alternative Method

Instead of solving the quadratic equation algebraically, you can test the given choices for

k

. For the correct option of

3

, substituting

k = 3

into the original equation yields

x^2 - 6x + 9 = 0

. Factoring this expression gives

(x - 3)^2 = 0

, which clearly has exactly one real solution (

x = 3

). Testing the other options would not produce a perfect square trinomial.

Estimated Time:2m 0s

Question 178Question →

Consider the system of equations below:

\begin{aligned} y &= 2x^2 + kx + 9 \\ y &= -2x + 1 \end{aligned}

In the system of equations above, $k$ is a positive constant. If the system has exactly one real solution, what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the positive constant

k

is 6.

By substituting the linear equation into the quadratic equation, we obtain the equation

2x^2 + (k+2)x + 8 = 0

. For the system to have exactly one real solution, the discriminant of this quadratic equation,

\Delta = b^2 - 4ac

, must be equal to zero. Substituting

a = 2

,

b = k+2

, and

c = 8

into the discriminant formula gives

(k+2)^2 - 4(2)(8) = 0

, which simplifies to

(k+2)^2 = 64

. Solving for

k

gives

k+2 = 8

or

k+2 = -8

, resulting in

k = 6

or

k = -10

. Since

k

is specified to be a positive constant, the value of

k

must be 6.

Step-by-Step Solution

1

Substitute the expression for

y

from the linear equation into the quadratic equation.

2x^2 + kx + 9 = -2x + 1

Setting the two expressions for

y

equal to each other allows us to solve for the

x

-coordinates of the intersection points.

2

Rearrange the equation into the standard quadratic form

ax^2 + bx + c = 0

.

2x^2 + (k+2)x + 8 = 0

Grouping the terms by power of

x

lets us identify the coefficients

a = 2

,

b = k+2

, and

c = 8

.

3

Set the discriminant of the quadratic equation to zero.

(k+2)^2 - 4(2)(8) = 0

, which simplifies to

(k+2)^2 - 64 = 0

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

b^2 - 4ac

, is equal to zero.

4

Solve for the positive constant

k

.

k = 6

Solving

(k+2)^2 = 64

yields

k+2 = 8

or

k+2 = -8

, which means

k = 6

or

k = -10

. Since

k

must be a positive constant, we select 6.

Key Concept

Determining the number of solutions of a quadratic-linear system using the discriminant.

Alternative Method

Alternatively, the quadratic equation

2x^2 + (k+2)x + 8 = 0

has exactly one real solution when it is a perfect square trinomial. Factoring out the leading coefficient gives

2(x^2 + \frac{k+2}{2}x + 4) = 0

. For the term inside the parentheses to be a perfect square trinomial of the form

x^2 + 2dx + d^2

, we must have

d^2 = 4

, which means

d = \pm 2

. Thus, the middle coefficient

\frac{k+2}{2}

must equal

2d = \pm 4

. Since

k

is positive, we set

\frac{k+2}{2} = 4

, which solves to

k = 6

.

Estimated Time:1m 30s

Question 179Question →

What is the solution to the equation $\sqrt{3x + 13} = x + 3$ ?

1

-4

1 and -4

4

Show answer & explanation

Answer: 1

Answer

1

The value

1

is the correct answer because substituting it back into the original equation results in a true mathematical statement:

\sqrt{3(1) + 13} = 1 + 3

, which simplifies to

4 = 4

.

Step-by-Step Solution

1

Square both sides of the equation to eliminate the radical.

3x + 13 = (x + 3)^2

Squaring both sides is the standard method to solve equations containing square roots.

2

Expand the right side of the equation.

3x + 13 = x^2 + 6x + 9

Expanding the squared binomial gives a quadratic expression.

3

Move all terms to one side to set the quadratic equation to zero.

x^2 + 3x - 4 = 0

Setting the quadratic equation to zero allows us to solve for

x

by factoring.

4

Factor the quadratic equation.

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

Finding two numbers that multiply to

-4

and add to

3

gives the binomial factors.

5

Solve for the potential values of

x

.

x = -4

or

x = 1

Setting each factor equal to zero yields the potential solutions.

6

Substitute each potential solution back into the original equation to check for extraneous solutions.

For

x = 1

:

\sqrt{3(1) + 13} = 1 + 3 \Rightarrow 4 = 4

(valid). For

x = -4

:

\sqrt{3(-4) + 13} = -4 + 3 \Rightarrow 1 = -1

(invalid). Thus, the only solution is

1

.

Squaring both sides can introduce extraneous solutions, so checking all potential solutions in the original equation is necessary to verify their validity.

Key Concept

Solving radical equations by squaring both sides and checking for extraneous solutions.

Question 180Question →

The function $f$ is defined by $f(x) = -(x - 5)^2 + 4$ . What is the maximum value of the function $f$ ?

-4

4

5

-5

Show answer & explanation

Answer: $4$

Answer

The maximum value of the function is 4.

The quadratic function

f(x) = -(x - 5)^2 + 4

is in vertex form

f(x) = a(x - h)^2 + k

, where the vertex of the parabola is

(h, k) = (5, 4)

. Because the leading coefficient

a = -1

is negative, the parabola opens downward, which means the vertex is the highest point on the graph. The maximum value of the function is the y-coordinate of this vertex, which is 4.

Step-by-Step Solution

1

Identify the form of the quadratic function.

The function

f(x) = -(x - 5)^2 + 4

is written in vertex form,

f(x) = a(x - h)^2 + k

.

Recognizing the vertex form allows us to directly identify the vertex coordinates of the parabola.

2

Identify the vertex

(h, k)

and the direction the parabola opens.

The vertex is

(5, 4)

and the leading coefficient is

a = -1

.

Since the leading coefficient

a

is negative, the parabola opens downward, meaning the vertex is the highest point on the graph.

3

Determine the maximum value of the function.

The maximum value of the function is the y-coordinate of the vertex, which is 4.

The maximum value of a function refers to its largest output (y-value), which corresponds to the y-coordinate of its highest point.

Key Concept

Quadratic Functions and Graphs

Advanced Math

Answer

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