Advanced Math Practice Questions — SAT — Page 13

Question 241Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, has a vertex $(h, k)$ that lies on the line $y = 4x + 8$ . If the graph of $y = f(x)$ has $x$ -intercepts at $x = -2$ and $x = 6$ , what is the value of $c$ ?

Show answer & explanation

Answer: 12

Answer

The value of c is 12.

The correct answer is 12. The axis of symmetry of the quadratic function lies midway between its x-intercepts at

x = -2

and

x = 6

, giving an x-coordinate of

h = 2

for the vertex. Since the vertex lies on the line

y = 4x + 8

, its y-coordinate is

k = 4(2) + 8 = 16

. Substituting the vertex

(2, 16)

into the vertex form of a quadratic function gives

f(x) = a(x - 2)^2 + 16

. Using the x-intercept at

(6, 0)

to solve for

a

gives

0 = a(6 - 2)^2 + 16

, which yields

a = -1

. Evaluating the function at

x = 0

to find the constant term

c

gives

c = f(0) = -(0 - 2)^2 + 16 = 12

.

Step-by-Step Solution

1

Determine the x-coordinate of the vertex of the quadratic function.

h = 2

The axis of symmetry of a parabola lies midway between its x-intercepts. The midpoint of the intercepts

x = -2

and

x = 6

is calculated as

\frac{-2 + 6}{2} = 2

.

2

Determine the y-coordinate of the vertex using the given linear equation.

k = 16

Since the vertex

(h, k)

lies on the line

y = 4x + 8

, substituting

x = h = 2

into the line equation gives

k = 4(2) + 8 = 16

.

3

Write the quadratic function in vertex form and solve for the leading coefficient

a

.

a = -1

The vertex form is

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

. Substituting the vertex

(2, 16)

yields

f(x) = a(x - 2)^2 + 16

. Using the x-intercept

(6, 0)

, substitute

x = 6

and

f(x) = 0

into the equation:

0 = a(6 - 2)^2 + 16

, which simplifies to

16a = -16

, resulting in

a = -1

.

4

Expand the quadratic function or evaluate it at

x = 0

to find the value of

c

.

c = 12

In standard form

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

, the constant term

c

is equal to

f(0)

. Substituting

x = 0

into

f(x) = -(x - 2)^2 + 16

yields

f(0) = -(0 - 2)^2 + 16 = -4 + 16 = 12

.

Key Concept

Using symmetry and the vertex form of a quadratic function to determine its standard form coefficients.

Alternative Method

Alternatively, since the x-intercepts are

-2

and

6

, the quadratic function can be written in factored form as

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

. Expanding this gives

f(x) = a(x^2 - 4x - 12) = ax^2 - 4ax - 12a

. Comparing this to the standard form

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

, we see that the x-coordinate of the vertex is

h = -\frac{b}{2a} = -\frac{-4a}{2a} = 2

. Using the line equation

y = 4x + 8

at

x = 2

, we find the vertex y-coordinate is

k = 16

. Since the vertex is

(2, 16)

, we evaluate the factored form at

x = 2

:

16 = a(2 + 2)(2 - 6) \Rightarrow 16 = -16a \Rightarrow a = -1

. The constant term

c

is

-12a

, so

c = -12(-1) = 12

.

Estimated Time:2m 30s

Question 242Question →

If $(x, y)$ is a solution to the system of equations below and $x > 1$ , what is the value of $x$ ?

\begin{aligned} y - 2x &= 3 \\ y^2 - 5x^2 &= 20 \end{aligned}

Show answer & explanation

Answer: 11

Answer

The value of

x

is 11.

The system can be solved by substituting the first equation into the second. Rearranging the first equation gives

y = 2x + 3

. Substituting this expression for

y

in the second equation gives

(2x + 3)^2 - 5x^2 = 20

. Expanding the binomial yields

4x^2 + 12x + 9 - 5x^2 = 20

, which simplifies to

-x^2 + 12x + 9 = 20

. Rearranging into standard quadratic form gives

x^2 - 12x + 11 = 0

. Factoring the quadratic equation results in

(x - 11)(x - 1) = 0

. Thus,

x = 11

or

x = 1

. Since the problem specifies

x > 1

, the value of

x

must be 11.

Step-by-Step Solution

1

Rearrange the first equation to express

y

in terms of

x

.

y = 2x + 3

This allows for substitution into the second equation.

2

Substitute

y = 2x + 3

into the second equation.

(2x + 3)^2 - 5x^2 = 20

To eliminate

y

and obtain an equation in terms of

x

only.

3

Expand and simplify the equation.

x^2 - 12x + 11 = 0

Expanding

(2x + 3)^2

gives

4x^2 + 12x + 9

. Subtracting

5x^2

yields

-x^2 + 12x + 9 = 20

. Subtracting 20 from both sides gives

-x^2 + 12x - 11 = 0

. Multiplying the entire equation by

-1

results in standard form.

4

Factor the quadratic equation.

(x - 11)(x - 1) = 0

To find the possible values of

x

.

5

Apply the given constraint

x > 1

.

x = 11

The solutions are

x = 11

and

x = 1

. The constraint

x > 1

excludes

x = 1

, leaving

x = 11

as the only valid solution.

Key Concept

Solving a system of nonlinear equations by substitution and factoring a quadratic equation.

Question 243Question →

If $x$ is a real number that satisfies the equation $\frac{x}{x-3} - \frac{6}{x^2-3x} = \frac{2}{x}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

2

The correct answer is 2. Multiplying the entire equation by the lowest common denominator,

x(x-3)

, yields

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

. Distributing the right side gives

x^2 - 6 = 2x - 6

. Adding 6 to both sides and subtracting

2x

results in

x^2 - 2x = 0

. Factoring this equation gives

x(x-2) = 0

, which has solutions

x = 0

and

x = 2

. However,

x = 0

is extraneous because it results in division by zero in the original equation. Thus, the only valid real solution is 2.

Step-by-Step Solution

1

Identify the common denominator and multiply the equation by it.

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

To clear the fractions and simplify the rational equation.

2

Distribute and simplify the equation.

x^2 - 2x = 0

To set up the quadratic equation in standard form.

3

Factor and solve for

x

.

x = 0

or

x = 2

To find potential solutions to the equation.

4

Check for extraneous solutions by substituting the values back into the original denominators.

x = 2

x = 0

makes the denominators equal to zero and is therefore extraneous.

Key Concept

Solving rational equations and identifying extraneous solutions.

Question 244Question →

If $\frac{x - 3}{4} = \frac{9}{x - 3}$ and $x > 3$ , what is the value of $x$ ?

Show answer & explanation

Answer: 9

Answer

9

Cross-multiplying the equation

\frac{x - 3}{4} = \frac{9}{x - 3}

gives

(x - 3)^2 = 36

. Taking the square root of both sides gives

x - 3 = 6

or

x - 3 = -6

. Solving for

x

yields

x = 9

or

x = -3

. Since the question specifies the constraint

x > 3

, the only valid solution is

9

.

Step-by-Step Solution

1

Cross-multiply the equation to eliminate the denominators.

(x - 3)^2 = 36

Multiplying both sides of the equation by

4(x - 3)

simplifies the rational equation into a quadratic form.

2

Take the square root of both sides of the equation.

x - 3 = 6

or

x - 3 = -6

Taking the square root of a squared term yields both positive and negative root options.

3

Solve each linear equation for

x

and apply the constraint

x > 3

.

x = 9

Adding

3

to both sides gives

x = 9

or

x = -3

. The constraint

x > 3

excludes the negative solution, leaving

9

as the only valid value.

Key Concept

Solving rational equations by cross-multiplication and factoring the resulting quadratic equation while adhering to domain constraints.

Question 245Question →

A parabola defined by the equation $y = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, has its vertex in the second quadrant of the coordinate plane. If the parabola passes through the point $(0,0)$ , which of the following inequalities must be true?

ab > 0

ab < 0

a > 0

b > 0

Show answer & explanation

Answer: $ab > 0$

Answer

The inequality

ab > 0

must be true.

Since the parabola passes through the point

(0,0)

and its vertex

(h, k)

lies in the second quadrant where

k > 0

, the vertex must represent a maximum value. Therefore, the parabola opens downward, so the leading coefficient

a

is negative. The x-coordinate of the vertex,

h

, is also negative because it lies in the second quadrant. The vertex x-coordinate is defined by

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

, which means

b = -2ah

. Because both

a

and

h

are negative, their product

ah

is positive, which makes

b

negative when multiplied by

-2

. Finally, since both

a

and

b

are negative, their product

ab

must be positive.

Step-by-Step Solution

1

Determine the signs of the vertex coordinates from its quadrant.

The vertex

(h, k)

lies in the second quadrant, which means

h < 0

and

k > 0

.

Points in the second quadrant have negative x-coordinates and positive y-coordinates.

2

Determine the value of the constant

c

.

c = 0

Since the parabola passes through

(0,0)

, substituting

x = 0

must result in

y = 0

.

3

Determine the sign of the leading coefficient

a

.

a < 0

The vertex is at

(h, k)

with

k > 0

, and the graph passes through

(0, 0)

where the y-value is

0

. Since the maximum value must be at least the value at any other point, the vertex

(h,k)

is a maximum. A parabola with a maximum opens downward, so

a < 0

.

4

Determine the sign of the coefficient

b

using the vertex formula.

b < 0

The x-coordinate of the vertex is

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

, which can be rewritten as

b = -2ah

. Since

a < 0

and

h < 0

, their product

ah

is positive. Multiplying this positive product by

-2

yields a negative value for

b

.

5

Find the sign of the product

ab

.

ab > 0

Since both

a

and

b

are negative, their product

ab

must be positive.

Key Concept

Analyzing quadratic coefficients and vertex properties in the coordinate plane.

Question 246Question →

The graph of the function $f$ in the $xy$ -plane is translated $4$ units to the right and $3$ units down to create the graph of the function $g$ . Which of the following equations defines $g(x)$ in terms of $f(x)$ ?

g(x) = f(x + 4) - 3

g(x) = f(x - 4) + 3

g(x) = f(x - 4) - 3

g(x) = f(x + 4) + 3

Show answer & explanation

Answer: $g(x) = f(x - 4) - 3$

Answer

The equation that defines

g(x)

in terms of

f(x)

is

g(x) = f(x - 4) - 3

.

The equation

g(x) = f(x - 4) - 3

correctly represents the translated function. A translation of

4

units to the right is represented by replacing

x

with

x - 4

within the function's input, and a translation of

3

units down is represented by subtracting

3

from the outer function.

Step-by-Step Solution

1

Apply the horizontal translation to the function input.

f(x - 4)

Translating a function

f(x)

horizontally to the right by

h

units is represented by substituting

x

with

x - h

. For a shift of

4

units to the right, we replace

x

with

x - 4

.

2

Apply the vertical translation to the function output.

f(x - 4) - 3

Translating a function vertically down by

k

units is represented by subtracting

k

from the entire function. For a shift of

3

units down, we subtract

3

.

3

Write the final equation for

g(x)

.

g(x) = f(x - 4) - 3

Combining the horizontal and vertical transformations yields the function

g(x)

.

Key Concept

Graph transformations of functions

Question 247Question →

A polynomial $p(x)$ has a remainder of $3$ when divided by $x - 5$ . Which of the following equations must be true?

p(-5) = 3

p(5) = 0

p(5) = 3

p(3) = 0

Show answer & explanation

Answer: $p(5) = 3$

Answer

The equation

p(5) = 3

must be true.

According to the Remainder Theorem, when a polynomial

p(x)

is divided by a linear expression of the form

x - c

, the remainder is equal to

p(c)

. Here, the divisor is

x - 5

, which gives

c = 5

. The remainder is given as

3

. Therefore, evaluating the polynomial at

x = 5

must yield

3

, which is written as

p(5) = 3

.

Step-by-Step Solution

1

Identify the divisor and the remainder from the given problem statement.

The divisor is

x - 5

and the remainder is

3

.

Applying the Remainder Theorem requires identifying the value of

c

in the divisor form

x - c

and the remainder value.

2

Apply the Remainder Theorem to relate the divisor and the remainder to the polynomial function.

Since the divisor is

x - 5

, we have

c = 5

. The Remainder Theorem states that the remainder when

p(x)

is divided by

x - c

is

p(c)

. Therefore,

p(5) = 3

.

This establishes the mathematical relationship directly showing which equation must be true.

Key Concept

The Remainder Theorem

Estimated Time:45s

Question 248Question →

In the polynomial function $p(x) = x^3 - 5x^2 + 2x + k$ , the constant $k$ is chosen such that $p(x)$ is divisible by $x - 4$ . What is the value of $k$ ?

Show answer & explanation

Answer: 8

Answer

The value of

k

is 8.

According to the Factor Theorem, a polynomial

p(x)

is divisible by

x - c

if and only if

p(c) = 0

. Since

p(x)

is divisible by

x - 4

, we must have

p(4) = 0

. Substituting

4

for

x

in the equation for

p(x)

yields

4^3 - 5(4)^2 + 2(4) + k = 0

. Simplifying the numerical terms gives

64 - 80 + 8 + k = 0

, which simplifies further to

-8 + k = 0

. Therefore, the value of the constant

k

is

8

.

Step-by-Step Solution

1

Apply the Factor Theorem to relate the factor

x - 4

to the value of the polynomial.

Since

x - 4

is a factor of

p(x)

,

p(4) = 0

.

By the Factor Theorem, if

x - c

is a factor of a polynomial

p(x)

, then

p(c) = 0

.

2

Substitute

x = 4

into the polynomial expression.

4^3 - 5(4)^2 + 2(4) + k = 0

We substitute

x = 4

into

p(x) = x^3 - 5x^2 + 2x + k

and set the expression to

0

.

3

Simplify the numerical expression to solve for

k

.

64 - 80 + 8 + k = 0 \implies -8 + k = 0 \implies k = 8

Evaluate powers, multiply, and solve the resulting linear equation for

k

.

Key Concept

Factor Theorem

Estimated Time:1m 0s

Question 249Question →

For all $x \neq 2$ , the expression $\frac{3x^2 + kx - 8}{x - 2}$ is equivalent to $3x + 10 + \frac{12}{x - 2}$ , where $k$ is a constant. What is the value of $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of the constant

k

is

4

.

The correct answer is

4

. By rewriting the right-hand side of the equation with a common denominator of

x - 2

, the expression becomes

\frac{(3x + 10)(x - 2) + 12}{x - 2}

. Expanding and simplifying the numerator yields

3x^2 + 4x - 8

. Comparing this to the numerator of the left-hand side,

3x^2 + kx - 8

, shows that the coefficient of the

x

term,

k

, must equal

4

.

Step-by-Step Solution

1

Set up the equation representing the equivalence of the two expressions.

\frac{3x^2 + kx - 8}{x - 2} = 3x + 10 + \frac{12}{x - 2}

This establishes the relationship that must hold for all

x \neq 2

.

2

Find a common denominator for the terms on the right-hand side.

3x + 10 + \frac{12}{x - 2} = \frac{(3x + 10)(x - 2)}{x - 2} + \frac{12}{x - 2}

To combine the terms into a single rational expression, they must have the same denominator.

3

Expand and simplify the numerator on the right-hand side.

(3x + 10)(x - 2) + 12 = 3x^2 + 4x - 8

Expanding the product of binomials and combining like terms yields the simplified numerator.

4

Equate the numerators of the simplified expressions to solve for

k

.

3x^2 + kx - 8 = 3x^2 + 4x - 8

, which means

k = 4

.

Since the denominators are equal, the numerators must be identical for all values of

x \neq 2

, meaning their corresponding coefficients must match.

Key Concept

Equivalent Algebraic Expressions

Alternative Method

An alternative method is to substitute a convenient value for

x

that is not equal to

2

. Substituting

x = 1

into both sides of the equivalence gives:

\frac{3(1)^2 + k(1) - 8}{1 - 2} = 3(1) + 10 + \frac{12}{1 - 2}

. Simplifying this yields

\frac{k - 5}{-1} = 13 - 12

, which simplifies to

-k + 5 = 1

. Solving for

k

results in

k = 4

.

Estimated Time:1m 30s

Question 250Question →

For which of the following values of $k$ does the equation $\frac{x^2 - kx + 9}{x - 3} = 0$ have exactly one real solution?

6

3

-6

-3

Show answer & explanation

Answer: $-6$

Answer

The correct value is

-6

.

For the equation

\frac{x^2 - kx + 9}{x - 3} = 0

to have a solution, we must have

x \neq 3

. Multiplying both sides by the denominator yields the quadratic equation

x^2 - kx + 9 = 0

. For a quadratic equation to have exactly one real solution, its discriminant must equal zero:

(-k)^2 - 4(1)(9) = k^2 - 36 = 0

, which gives

k = 6

or

k = -6

. If

k = 6

, the equation is

(x-3)^2 = 0

, which has the single solution

x = 3

. Since

x = 3

is excluded by the domain restriction, the equation has zero real solutions. If

k = -6

, the equation is

(x+3)^2 = 0

, which has the single solution

x = -3

. Since

-3

is a valid value, this value of

k

yields exactly one real solution.

Step-by-Step Solution

1

Identify the domain restriction of the rational equation.

The denominator

x - 3

cannot be zero, which means

x \neq 3

.

Division by zero is undefined, so

x = 3

cannot be a solution to the equation.

2

Simplify the equation by multiplying both sides by the denominator.

x^2 - kx + 9 = 0

for

x \neq 3

.

A rational expression equals zero when its numerator is zero and its denominator is non-zero.

3

Set the discriminant of the quadratic equation to zero to find the values of

k

that yield exactly one real root.

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

or

k = -6

.

A quadratic equation has exactly one real solution (a double root) when its discriminant is zero.

4

Verify if the roots for each value of

k

are valid under the domain restriction

x \neq 3

.

If

k = 6

,

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

(invalid/extraneous). If

k = -6

,

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

(valid).

The value

k = 6

makes

x = 3

the only solution, which is excluded. Therefore, only

k = -6

yields exactly one valid real solution (

x = -3

).

Key Concept

Solving rational equations that simplify to quadratic equations, using the discriminant to find the number of solutions, and identifying extraneous solutions due to domain restrictions.

Estimated Time:2m 0s

Question 251Question →

If

x

is a real solution to the equation

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

what is the value of

1 - 2x

?

-5

2

-2

0

Show answer & explanation

Answer: 2

Answer

2

The correct answer is 2. Multiplying the entire equation by the least common denominator

(x - 3)(x - 1)

simplifies the rational equation to the quadratic equation

2x^2 - 5x - 3 = 0

. This quadratic factors into

(2x + 1)(x - 3) = 0

, yielding two potential solutions:

x = -1/2

and

x = 3

. However,

x = 3

is an extraneous solution because it results in a denominator of zero in the original equation. Thus, the only valid real solution is

x = -1/2

. Substituting this into the expression

1 - 2x

yields

1 - 2(-1/2) = 2

.

Step-by-Step Solution

1

Factor the quadratic expression in the denominator to identify the least common denominator.

The quadratic in the denominator of the second term factors as

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

. Thus, the least common denominator for all terms in the equation is

(x - 3)(x - 1)

, with the restrictions that

x \neq 3

and

x \neq 1

.

Finding a common denominator allows us to multiply both sides of the equation to clear all rational expressions.

2

Multiply both sides of the equation by the least common denominator

(x - 3)(x - 1)

.

(2x - 5)(x - 1) + (x - 5) = -1(x - 3)

This clears the fractions and converts the rational equation into a polynomial equation.

3

Expand and simplify both sides of the equation.

(2x^2 - 7x + 5) + (x - 5) = -x + 3

2x^2 - 6x = -x + 3

Expanding the terms allows us to combine like terms and set up a quadratic equation.

4

Rearrange the terms to set the quadratic equation equal to zero.

2x^2 - 5x - 3 = 0

Quadratic equations must be set to zero to be solved by factoring or using the quadratic formula.

5

Solve the quadratic equation by factoring.

(2x + 1)(x - 3) = 0

This gives two potential solutions:

x = -1/2

and

x = 3

.

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

6

Check the potential solutions against the original domain restrictions to identify any extraneous solutions.

For

x = 3

, the denominators

x - 3

and

x^2 - 4x + 3

equal zero, so

x = 3

is an extraneous solution. For

x = -1/2

, the denominators are non-zero, so

x = -1/2

is the only valid solution.

Squaring or multiplying by variable expressions can introduce solutions that are undefined in the original equation.

7

Substitute the valid solution

x = -1/2

into the expression

1 - 2x

.

1 - 2\left(-\frac{1}{2}\right) = 1 + 1 = 2

The question asks for the value of the expression

1 - 2x

, not the value of

x

itself.

Key Concept

Solving rational equations by clearing denominators and checking for extraneous solutions.

Question 252Question →

Two different cultures of bacteria, culture $A$ and culture $B$ , begin growing at the same time. The population of culture $A$ increases by $300\%$ every $8$ hours, and the population of culture $B$ increases by $700\%$ every $9$ hours. If both cultures start with the same initial population, after how many hours will the population of culture $B$ be $4$ times the population of culture $A$ ?

9.6

24

2

48

Show answer & explanation

Answer: 24

Answer

The population of culture B will be 4 times the population of culture A after 24 hours.

The correct answer is the option stating 24 hours. The growth model for culture A is

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

and for culture B is

P_B(t) = P_0 \cdot 8^{t/9}

. Setting

P_B(t) = 4 P_A(t)

and converting all terms to base 2 yields

2^{t/3} = 2^{2 + t/4}

. Equating exponents gives

t/3 = 2 + t/4

, which simplifies to

t/12 = 2

and solves to

t = 24

hours.

Step-by-Step Solution

1

Determine the growth factors and set up the population equations for both cultures.

For culture

A

, a

300\%

increase means the population becomes

1 + 3.00 = 4

times its previous value every

8

hours, so

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

. For culture

B

, a

700\%

increase means the population becomes

1 + 7.00 = 8

times its previous value every

9

hours, so

P_B(t) = P_0 \cdot 8^{t/9}

.

Establishing the correct exponential growth models with the appropriate growth factors and time constants is necessary to relate the populations.

2

Set up the equation to find when the population of culture

B

is

4

times that of culture

A

.

P_B(t) = 4 P_A(t) \implies P_0 \cdot 8^{t/9} = 4 \cdot P_0 \cdot 4^{t/8} \implies 8^{t/9} = 4 \cdot 4^{t/8}

This sets up the equation that must be solved for

t

by dividing both sides by the non-zero initial population

P_0

.

3

Express all terms with a common base of

2

.

(2^3)^{t/9} = 2^2 \cdot (2^2)^{t/8} \implies 2^{t/3} = 2^2 \cdot 2^{t/4}

Expressing bases as powers of

2

allows the use of exponent rules to simplify the equation.

4

Apply the product rule of exponents to combine the terms on the right side.

2^{t/3} = 2^{2 + t/4}

The rule

2^a \cdot 2^b = 2^{a+b}

simplifies the multiplication of exponential terms with the same base.

5

Equate the exponents and solve the resulting linear equation for

t

.

t/3 = 2 + t/4 \implies t/3 - t/4 = 2 \implies t/12 = 2 \implies t = 24

Since the bases are equal, their exponents must be equal. Solving the linear equation yields the time in hours.

Key Concept

Solving exponential equations by converting to a common base and applying exponent laws.

Estimated Time:3m 0s

Question 253Question →

If $x$ is the real solution to the equation below, what is the value of $x$ ?

\frac{2x^2 - 30x + 100}{x - 5} + \frac{140}{x + 2} = 10

Show answer & explanation

Answer: 8

Answer

The correct answer is 8.

To solve the rational equation, we first state the domain restrictions as

x \neq 5

and

x \neq -2

because these values make the denominators zero. Next, we factor the numerator of the first term:

2x^2 - 30x + 100 = 2(x^2 - 15x + 50) = 2(x - 5)(x - 10)

. Since

x \neq 5

, we can cancel the common factor of

x - 5

from the numerator and denominator, which simplifies the first term to

2(x - 10) = 2x - 20

. The equation can then be rewritten as

2x - 20 + \frac{140}{x + 2} = 10

. Subtracting 10 from both sides yields

2x - 30 + \frac{140}{x + 2} = 0

. Dividing the entire equation by 2 simplifies the coefficients, giving

x - 15 + \frac{70}{x + 2} = 0

. To clear the fraction, we multiply the entire equation by

x + 2

, leading to

(x - 15)(x + 2) + 70 = 0

. Expanding the product gives

x^2 - 13x - 30 + 70 = 0

, which simplifies to

x^2 - 13x + 40 = 0

. Factoring this quadratic equation yields

(x - 5)(x - 8) = 0

, giving the potential solutions

x = 5

and

x = 8

. However,

x = 5

is extraneous because it is restricted from the domain of the original rational expression. Therefore, the only real solution is 8.

Step-by-Step Solution

1

Find the values of

x

that make any denominator equal to zero to identify domain restrictions.

x \neq 5

and

x \neq -2

Dividing by zero is undefined in real numbers, so these values are excluded from the domain.

2

Factor the numerator of the first term,

2x^2 - 30x + 100

.

2(x - 5)(x - 10)

Factoring allows us to check for common factors in the numerator and denominator.

3

Simplify the first rational term by dividing the numerator and denominator by the common factor

x - 5

, under the assumption that

x \neq 5

.

2(x - 10) = 2x - 20

Reducing the expression simplifies the algebraic structure of the equation.

4

Substitute the simplified expression back into the original equation and subtract 10 from both sides.

2x - 30 + \frac{140}{x + 2} = 0

To consolidate the constant terms and prepare the equation for clearing the remaining fraction.

5

Divide the entire equation by 2.

x - 15 + \frac{70}{x + 2} = 0

Dividing by the common factor simplifies the coefficients before expanding.

6

Multiply the entire equation by

x + 2

to clear the denominator, then expand and combine like terms.

x^2 - 13x + 40 = 0

This transforms the rational equation into a standard quadratic equation.

7

Factor the quadratic equation.

(x - 5)(x - 8) = 0

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

8

Verify the candidate solutions against the domain restrictions identified in Step 1.

The candidate solution

x = 5

is extraneous because it is excluded from the domain of the original equation. The candidate solution

x = 8

is valid.

Extraneous solutions introduced during algebraic steps must be excluded from the final solution set.

Key Concept

Solving rational equations by simplifying terms, finding common denominators, and identifying extraneous solutions.

Question 254Question →

In the $xy$ -plane, the system of equations below has exactly one real solution.

y = -x^2 + 6x + 2

y + 2x = k

If $k$ is a constant, what is the value of $k$ ?

Show answer & explanation

Answer: 18

Answer

18

To find the value of the constant

k

for which the system of equations has exactly one real solution, we can solve the system by substitution. Substituting

y = k - 2x

from the second equation into the first equation gives

k - 2x = -x^2 + 6x + 2

. Rearranging the terms to write this quadratic equation in standard form,

ax^2 + bx + c = 0

, yields

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

. A quadratic equation has exactly one real solution when its discriminant,

b^2 - 4ac

, is equal to zero. Substituting

a = 1

,

b = -8

, and

c = k - 2

into the discriminant formula gives

(-8)^2 - 4(1)(k - 2) = 0

. Simplifying this expression results in

64 - 4k + 8 = 0

, which simplifies further to

72 - 4k = 0

. Solving for

k

yields

k = 18

.

Step-by-Step Solution

1

Rewrite the linear equation to express

y

in terms of

x

and

k

.

y = k - 2x

This allows substituting the expression for

y

into the quadratic equation to eliminate

y

.

2

Substitute the expression for

y

into the quadratic equation.

k - 2x = -x^2 + 6x + 2

To construct a single quadratic equation in terms of

x

whose solutions correspond to the system's intersection points.

3

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

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

To identify the coefficients

a = 1

,

b = -8

, and

c = k - 2

.

4

Set the discriminant of the quadratic equation to zero.

(-8)^2 - 4(1)(k - 2) = 0

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

5

Solve the linear equation for

k

.

k = 18

Simplify the equation:

64 - 4k + 8 = 0 \implies 72 - 4k = 0 \implies 4k = 72 \implies k = 18

.

Key Concept

Nonlinear Systems of Equations

Question 255Question →

The table below shows the population of a colony of bacteria, $P(t)$ , at various times $t$ , in hours.

Time $t$ (hours)	Population $P(t)$
0	120
3	960
6	7,680

If the population can be modeled by an exponential function of the form $P(t) = a \cdot b^t$ , where $a$ and $b$ are constants, what is the value of $b$ ?

2

8

4

\frac{8}{3}

Show answer & explanation

Answer: 2

Answer

2

The correct answer is the value that represents the hourly growth factor of the bacteria. Since the initial population at

t = 0

is 120, we have

a = 120

. Using the population at

t = 3

, we set up the equation

120 \cdot b^3 = 960

. Dividing both sides by 120 yields

b^3 = 8

. Taking the cube root of 8 gives

b = 2

. We can verify this with the third data point:

120 \cdot 2^6 = 120 \cdot 64 = 7,680

, which matches the table.

Step-by-Step Solution

1

Find the value of the constant

a

using the initial value from the table.

a = 120

When

t = 0

, the population

P(0) = 120

. Substituting these values into

P(t) = a \cdot b^t

gives

120 = a \cdot b^0

, and since

b^0 = 1

, we get

a = 120

.

2

Set up an equation to solve for

b

using another data point from the table.

120 \cdot b^3 = 960

When

t = 3

, the population

P(3) = 960

. Substituting

a = 120

,

t = 3

, and

P(3) = 960

into the function gives

960 = 120 \cdot b^3

.

3

Isolate

b^3

and solve for

b

by taking the cube root.

b = 2

Dividing both sides of

120 \cdot b^3 = 960

by

120

gives

b^3 = 8

. Taking the cube root of both sides gives

b = \sqrt[3]{8} = 2

.

Key Concept

Determining parameters of exponential functions from a table of values.

Question 256Question →

If $\frac{\left(\frac{1}{4}\right)^{2x-5}}{8^{3-x}} = \left(\sqrt[3]{2}\right)^{x-5}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

2

By writing all terms with a common base of 2, the equation simplifies to

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

. Equating the exponents yields

-x + 1 = \frac{x-5}{3}

. Solving this linear equation gives

x = 2

.

Step-by-Step Solution

1

Rewrite each base in terms of base 2.

\frac{1}{4} = 2^{-2}

,

8 = 2^3

, and

\sqrt[3]{2} = 2^{\frac{1}{3}}

Expressing all terms with a common base allows the exponents to be directly compared.

2

Substitute the bases back into the left side of the equation and simplify using exponent rules.

\frac{(2^{-2})^{2x-5}}{(2^3)^{3-x}} = \frac{2^{-4x+10}}{2^{9-3x}} = 2^{-x+1}

Applying the power rule

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

and quotient rule

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

simplifies the fraction.

3

Rewrite the right side of the equation using base 2.

(\sqrt[3]{2})^{x-5} = (2^{\frac{1}{3}})^{x-5} = 2^{\frac{x-5}{3}}

Applying the power rule to the fractional exponent converts the radical expression.

4

Equate the simplified left and right sides, then set the exponents equal to each other.

-x + 1 = \frac{x-5}{3}

If two exponential expressions with the same base are equal, their exponents must be equal.

5

Solve the linear equation for

x

.

-3x + 3 = x - 5 \implies 8 = 4x \implies x = 2

Isolating the variable

x

yields the final solution.

Key Concept

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

Question 257Question →

If one of the solutions to the equation $2x^2 - 9x + c = 0$ , where $c$ is a constant, is $x = 2$ , what is the other solution to the equation?

-\frac{13}{2}

-\frac{5}{2}

\frac{5}{2}

7

Show answer & explanation

Answer: $\frac{5}{2}$

Answer

The other solution to the equation is

\frac{5}{2}

.

The correct answer is found by using the relationship between the coefficients of a quadratic equation and the sum of its roots. For any quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a}

. Substituting the values

a = 2

and

b = -9

, we find that the sum of the roots is

-\frac{-9}{2} = \frac{9}{2}

. Since one of the solutions is given as

2

, the other solution must be

\frac{9}{2} - 2 = \frac{5}{2}

. Alternatively, one can find the constant

c

by substituting

x = 2

into the equation to get

2(2)^2 - 9(2) + c = 0

, which yields

c = 10

. Solving the resulting equation

2x^2 - 9x + 10 = 0

by factoring gives

(2x - 5)(x - 2) = 0

, which confirms the other solution is

\frac{5}{2}

.

Step-by-Step Solution

1

Identify the relationship between the coefficients and the sum of the roots for a quadratic equation.

For any quadratic equation in the form

ax^2 + bx + c = 0

, the sum of its roots,

x_1

and

x_2

, is given by the formula

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

.

This relationship allows us to find the second root directly without needing to calculate the value of the constant

c

first.

2

Substitute the given root and the coefficients of the quadratic equation into the sum of the roots formula.

Here,

a = 2

and

b = -9

. The given root is

x_1 = 2

. Substituting these values gives

2 + x_2 = -\frac{-9}{2}

, which simplifies to

2 + x_2 = \frac{9}{2}

.

This sets up a linear equation to solve for the unknown root,

x_2

.

3

Solve for the second root

x_2

.

x_2 = \frac{9}{2} - 2 = \frac{9}{2} - \frac{4}{2} = \frac{5}{2}

.

This isolates the unknown variable to find the final value.

Key Concept

Using the sum of roots formula (

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

) to solve for an unknown solution of a quadratic equation.

Question 258Question →

If

x

is a real solution to the equation

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

what is the value of

5x - 4

?

6

-14

-5

5

Show answer & explanation

Answer: -5

Answer

-5

The correct answer is

-5

. Solving the equation by multiplying both sides by the least common denominator

(x-1)(x-2)

yields the quadratic equation

5x^2 - 9x - 2 = 0

. Factoring this equation gives the potential solutions

x = -1/5

and

x = 2

. However, substituting

x = 2

into the original equation results in division by zero, meaning

x = 2

is extraneous and must be discarded. The only valid solution is

x = -1/5

. Substituting this value into the expression

5x - 4

yields

5(-1/5) - 4 = -5

.

Step-by-Step Solution

1

Identify the least common denominator (LCD) and multiply both sides of the equation to clear the denominators.

The LCD is

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

. Multiplying both sides by the LCD gives:

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

Multiplying by the LCD simplifies the rational equation into a polynomial equation.

2

Expand the terms and collect like terms to form a quadratic equation.

3x^2 - 6x + 2x^2 - 2x = x + 2

5x^2 - 8x = x + 2

5x^2 - 9x - 2 = 0

This puts the equation in the standard quadratic form

ax^2 + bx + c = 0

.

3

Factor the quadratic equation to find potential solutions.

The quadratic factors as:

(5x + 1)(x - 2) = 0

This yields potential solutions

x = -\frac{1}{5}

and

x = 2

.

Factoring allows us to isolate the candidate values of

x

.

4

Check the potential solutions in the original equation to identify any extraneous solutions.

For

x = 2

, the denominators

x - 2

and

x^2 - 3x + 2

become zero, which is undefined. Therefore,

x = 2

is an extraneous solution. For

x = -\frac{1}{5}

, all denominators are non-zero, so it is the only valid solution.

Values that make any denominator in the original equation equal to zero must be excluded.

5

Substitute the valid solution into the target expression

5x - 4

.

5\left(-\frac{1}{5}\right) - 4 = -1 - 4 = -5

Evaluating the expression with the only valid solution gives the final answer.

Key Concept

Solving rational equations by clearing denominators and checking for extraneous solutions.

Estimated Time:2m 30s

Question 259Question →

The function $f$ is defined by $f(x) = x^2 - 6x + c$ , where $c$ is a constant. In the $xy$ -plane, the graph of $f$ has vertex $A$ . The function $g$ is defined by $g(x) = -f(x - 6)$ , and its graph has vertex $B$ . If the distance between points $A$ and $B$ is $10$ , and $c > 10$ , what is the value of $c$ ?

Show answer & explanation

Answer: 13

Answer

The value of

c

is

13

.

Completing the square for

f(x) = x^2 - 6x + c

gives

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

, which shows that vertex

A

is located at

(3, c - 9)

. The transformation

g(x) = -f(x - 6)

translates the graph

6

units to the right and reflects it vertically, giving vertex

B

the coordinates

(3 + 6, -(c - 9)) = (9, 9 - c)

. Using the distance formula, the distance between

A

and

B

is

\sqrt{(9 - 3)^2 + ((9 - c) - (c - 9))^2} = \sqrt{36 + (18 - 2c)^2}

. Setting this distance equal to

10

and squaring both sides gives

36 + (18 - 2c)^2 = 100

, which simplifies to

(18 - 2c)^2 = 64

. Taking the square root of both sides gives

18 - 2c = 8

or

18 - 2c = -8

, yielding solutions of

c = 5

or

c = 13

. Since the question specifies that

c > 10

, the correct value must be

13

.

Step-by-Step Solution

1

Rewrite the function

f(x) = x^2 - 6x + c

in vertex form by completing the square.

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

, which gives the coordinates of vertex

A

as

(3, c - 9)

.

Completing the square reveals the vertex

(h, k)

of a quadratic function in the form

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

.

2

Determine the vertex

B

of the graph of

g(x) = -f(x - 6)

by applying transformations to vertex

A(3, c - 9)

.

The horizontal shift of

f(x - 6)

moves the vertex to

(3 + 6, c - 9) = (9, c - 9)

. The reflection of

-f(x - 6)

negates the

y

-coordinate of the vertex, resulting in

B(9, 9 - c)

.

The transformation

f(x - h)

shifts a graph right by

h

units, and the transformation

-f(x)

reflects it across the

x

-axis.

3

Use the distance formula to set up an equation for the distance between

A(3, c - 9)

and

B(9, 9 - c)

.

\sqrt{(9 - 3)^2 + ((9 - c) - (c - 9))^2} = 10 \implies \sqrt{6^2 + (18 - 2c)^2} = 10

.

The distance

d

between points

(x_1, y_1)

and

(x_2, y_2)

is given by

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

.

4

Solve the radical equation

\sqrt{36 + (18 - 2c)^2} = 10

for

c

.

36 + (18 - 2c)^2 = 100 \implies (18 - 2c)^2 = 64 \implies 18 - 2c = 8

or

18 - 2c = -8

. This yields

c = 5

or

c = 13

.

Squaring both sides eliminates the square root, allowing us to solve the resulting quadratic equation.

5

Apply the constraint

c > 10

to choose the correct value for

c

.

c = 13

.

The question specifies that

c

must be greater than

10

, which excludes

c = 5

.

Key Concept

Using vertex form of quadratic equations to determine vertex coordinates and applying transformations (horizontal shifts and vertical reflections) to find key graphical points.

Question 260Question →

The mass of a radioactive sample, in grams, is modeled by the function $M(t) = M_0 \left(\frac{1}{8}\right)^{\frac{t}{15}}$ , where $M_0$ is the initial mass of the sample and $t$ is the time, in minutes, since the sample started decaying. Which of the following functions correctly models the mass of the sample, in grams, in terms of $s$ seconds after the sample started decaying?

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{5}}

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{2700}}

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

M(s) = M_0 \left(\frac{1}{2}\right)^{12s}

Show answer & explanation

Answer: $M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}$

Answer

The correct function modeling the mass in terms of seconds is

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

.

To express the mass in terms of

s

seconds, we substitute

t = \frac{s}{60}

into the original decay function because the time in minutes is equal to the number of seconds divided by 60. This gives

M(s) = M_0 \left(\frac{1}{8}\right)^{\frac{s}{60 \times 15}} = M_0 \left(\frac{1}{8}\right)^{\frac{s}{900}}

. Next, since the options have a base of

\frac{1}{2}

, we rewrite the base

\frac{1}{8}

as

\left(\frac{1}{2}\right)^3

. Applying the power of a power rule

(x^a)^b = x^{ab}

, we multiply the exponent

\frac{s}{900}

by 3, which yields

M(s) = M_0 \left(\frac{1}{2}\right)^{3 \times \frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

.

Step-by-Step Solution

1

Relate the time variables

t

(in minutes) and

s

(in seconds).

Since 1 minute is equivalent to 60 seconds, the relationship is

t = \frac{s}{60}

.

This substitution allows the function to take time input in seconds instead of minutes.

2

Substitute

t = \frac{s}{60}

into the exponent of the original model.

M(s) = M_0 \left(\frac{1}{8}\right)^{\frac{s/60}{15}} = M_0 \left(\frac{1}{8}\right)^{\frac{s}{900}}

This updates the function variable to seconds and simplifies the fractional exponent.

3

Rewrite the base

\frac{1}{8}

as a power of

\frac{1}{2}

and apply exponent rules.

Since

\frac{1}{8} = \left(\frac{1}{2}\right)^3

, we write

M(s) = M_0 \left(\left(\frac{1}{2}\right)^3\right)^{\frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{3 \times \frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

.

Applying the power of a power rule,

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

, simplifies the function to its final form with a base of

\frac{1}{2}

.

Key Concept

Applying exponent rules to manipulate exponential bases and performing variable substitutions in contextual models.

Estimated Time:2m 0s

Advanced Math

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