Advanced Math Practice Questions — SAT — Page 15

What is the larger solution to the equation below?

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

Show answer & explanation

Answer: 6

Answer

The larger solution to the equation is 6.

To solve the rational equation, we clear the denominators by multiplying both sides by

x(x - 2)

, which yields the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation gives

(x - 6)(x - 1) = 0

, leading to the solutions

x = 6

and

x = 1

. Both values are valid because they do not make any denominator of the original expression equal to zero. The larger of the two values is 6.

Step-by-Step Solution

1

Multiply both sides of the equation by the common denominator

x(x - 2)

to clear the fractions.

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

Clearing denominators simplifies the rational equation into a polynomial equation.

2

Expand both sides of the equation.

x^2 + 3x - 6 = 2x^2 - 4x

Distributing the multiplication allows us to combine like terms.

3

Move all terms to one side of the equation to write it in the standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 7x + 6 = 0

Setting the quadratic expression to zero prepares it for factoring.

4

Factor the quadratic equation.

(x - 6)(x - 1) = 0

Finding two numbers that multiply to

6

and sum to

-7

gives

-6

and

-1

.

5

Identify the values of

x

that satisfy the factored equation and check for extraneous solutions.

x = 6

or

x = 1

Neither solution makes the denominators in the original equation,

x-2

or

x

, equal to zero, so both are valid. The larger of these two solutions is 6.

Key Concept

Solving rational equations by clearing denominators to form a quadratic equation, and verifying solutions against the domain constraints.

Question 282Question →

For all real values of

x

that satisfy the equation below, what is the value of

x + 1

?

\sqrt{5x + 39} - 3 = x

-5

-4

6

7

Show answer & explanation

Answer: $6$

Answer

The correct answer is

6

.

By adding

3

to both sides, we get

\sqrt{5x + 39} = x + 3

. Squaring both sides yields

5x + 39 = x^2 + 6x + 9

. Setting this quadratic equation to zero gives

x^2 + x - 30 = 0

, which factors as

(x + 6)(x - 5) = 0

. The potential solutions are

x = 5

and

x = -6

. Checking these values in the original equation shows that only

x = 5

is valid:

\sqrt{5(5) + 39} - 3 = 8 - 3 = 5

, which satisfies the equation. The value

x = -6

is extraneous since

\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6

. Therefore, the value of

x + 1

is

5 + 1 = 6

.

Step-by-Step Solution

1

Isolate the radical expression on one side of the equation.

\sqrt{5x + 39} = x + 3

Isolating the radical allows us to square both sides to eliminate the square root.

2

Square both sides of the equation.

5x + 39 = (x + 3)^2 \implies 5x + 39 = x^2 + 6x + 9

Squaring a square root removes the radical, allowing us to solve the resulting quadratic equation.

3

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

ax^2 + bx + c = 0

.

x^2 + x - 30 = 0

Subtracting

5x

and

39

from both sides sets the quadratic expression equal to zero.

4

Factor the quadratic equation to find potential solutions.

(x + 6)(x - 5) = 0 \implies x = -6 \text{ or } x = 5

Finding two numbers that multiply to

-30

and add up to

1

gives

6

and

-5

.

5

Substitute both candidate solutions back into the original equation to check for extraneous solutions.

For

x = 5

:

\sqrt{5(5) + 39} - 3 = 8 - 3 = 5

(Valid). For

x = -6

:

\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6

(Extraneous).

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original relation.

6

Use the valid solution to calculate the requested expression.

x + 1 = 5 + 1 = 6

The question asks for the value of

x + 1

.

Key Concept

Solving radical equations and identifying extraneous solutions

Question 283Question →

If $\frac{16^{x+1}}{2^{x-3}} = 64$ , what is the value of $x$ ?

\frac{2}{3}

\frac{5}{3}

-\frac{1}{3}

-\frac{7}{3}

Show answer & explanation

Answer: $-\frac{1}{3}$

Answer

-\frac{1}{3}

The correct answer is

-\frac{1}{3}

. By expressing all terms with base 2, the equation

\frac{16^{x+1}}{2^{x-3}} = 64

becomes

\frac{2^{4x+4}}{2^{x-3}} = 2^6

. Applying exponent division rules yields

2^{(4x+4)-(x-3)} = 2^6

, which simplifies to

2^{3x+7} = 2^6

. Setting the exponents equal gives

3x + 7 = 6

, which simplifies to

x = -\frac{1}{3}

.

Step-by-Step Solution

1

Rewrite all terms in the equation using a common base of 2.

The equation becomes

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

.

Expressing bases 16 and 64 as powers of 2 allows us to apply exponent laws to simplify the equation.

2

Apply the power of a power rule to the numerator, then apply the quotient rule for exponents.

The left side simplifies to

2^{4x+4 - (x-3)} = 2^{3x+7}

.

The expression

(2^4)^{x+1}

becomes

2^{4x+4}

. Using the quotient rule

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

, we subtract the denominator's exponent

(x-3)

from the numerator's exponent

(4x+4)

, distributing the negative sign to get

3x+7

.

3

Set the exponents equal to each other and solve the resulting linear equation.

3x + 7 = 6 \implies 3x = -1 \implies x = -\frac{1}{3}

.

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

Key Concept

Solving exponential equations by rewriting with a common base

Estimated Time:1m 30s

Question 284Question →

The quadratic function $f$ is defined by $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants. In the $xy$ -plane, the graph of $f$ is a parabola with vertex $(3, -5)$ that passes through the point $(0, 4)$ . What is the value of $a + b + c$ ?

11

-9

-1

-7

Show answer & explanation

Answer: -1

Answer

-1

The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex. Substituting the given vertex

(3, -5)

gives

f(x) = a(x - 3)^2 - 5

. Since the graph passes through the point

(0, 4)

, substituting

x = 0

and

f(x) = 4

yields the equation

4 = a(0 - 3)^2 - 5

, which simplifies to

9a = 9

, or

a = 1

. The function is therefore defined by

f(x) = (x - 3)^2 - 5

. The expression

a + b + c

represents the sum of the coefficients of the quadratic function in standard form

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

. Evaluating the function at

x = 1

gives

f(1) = a(1)^2 + b(1) + c = a + b + c

. Substituting

x = 1

into our vertex form equation yields

f(1) = (1 - 3)^2 - 5 = (-2)^2 - 5 = 4 - 5 = -1

. Therefore, the value of

a + b + c

is

-1

.

Step-by-Step Solution

1

Write the quadratic function in vertex form using the given vertex

(3, -5)

.

f(x) = a(x - 3)^2 - 5

, where

a

is a constant.

The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex.

2

Substitute the coordinates of the given point

(0, 4)

into the vertex form equation to solve for

a

.

4 = a(0 - 3)^2 - 5

, which simplifies to

4 = 9a - 5

, and solving for

a

gives

a = 1

.

Since the graph passes through

(0, 4)

, substituting these coordinates into the function's equation must yield a true statement.

3

Find the value of

a + b + c

by evaluating

f(1)

.

f(1) = 1(1 - 3)^2 - 5 = 1(-2)^2 - 5 = 4 - 5 = -1

.

For any quadratic function

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

, evaluating the function at

x = 1

gives

f(1) = a(1)^2 + b(1) + c = a + b + c

.

Key Concept

Quadratic Functions and Graphs

Question 285Question →

In the $xy$ -plane, the graph of a cubic polynomial function $p$ with real coefficients has exactly two $x$ -intercepts, at $(1, 0)$ and $(4, 0)$ . If the graph of $p$ passes through the points $(0, -8)$ and $(2, 2)$ , what is the value of $p(6)$ ?

Show answer & explanation

Answer: 10

Answer

10

A cubic polynomial with real coefficients and exactly two

x

-intercepts at

(1, 0)

and

(4, 0)

must have one root of multiplicity 1 and one root of multiplicity 2. This yields two possible forms:

p(x) = a(x - 1)^2(x - 4)

or

p(x) = a(x - 1)(x - 4)^2

. Substituting the

y

-intercept

(0, -8)

into the first form gives

a = 2

, but the resulting polynomial

p(x) = 2(x - 1)^2(x - 4)

does not pass through

(2, 2)

since

p(2) = -4

. Substituting

(0, -8)

into the second form gives

a = \frac{1}{2}

, and the resulting polynomial

p(x) = \frac{1}{2}(x - 1)(x - 4)^2

correctly passes through

(2, 2)

since

p(2) = 2

. Finally, evaluating this function at

x = 6

yields

p(6) = \frac{1}{2}(6 - 1)(6 - 4)^2 = 10

.

Step-by-Step Solution

1

Determine the possible forms of the cubic polynomial based on its

x

-intercepts.

Two possible cases: Case 1:

p(x) = a(x - 1)^2(x - 4)

or Case 2:

p(x) = a(x - 1)(x - 4)^2

, where

a

is a non-zero constant.

Since the polynomial has degree 3 and exactly two distinct

x

-intercepts, one intercept must be a root of multiplicity 2 (tangent to the

x

-axis) and the other must be a root of multiplicity 1 (crossing the

x

-axis).

2

Solve for the constant

a

in Case 1 using the

y

-intercept

(0, -8)

.

a = 2

, yielding the candidate function

p(x) = 2(x - 1)^2(x - 4)

.

Substituting

x = 0

and

p(0) = -8

into the equation for Case 1 allows us to solve for

a

.

3

Test the point

(2, 2)

in the Case 1 candidate function.

p(2) = -4 \neq 2

, which means Case 1 is incorrect.

The correct function must satisfy all given points, including

(2, 2)

.

4

Solve for the constant

a

in Case 2 using the

y

-intercept

(0, -8)

.

a = \frac{1}{2}

, yielding the candidate function

p(x) = \frac{1}{2}(x - 1)(x - 4)^2

.

Substituting

x = 0

and

p(0) = -8

into the equation for Case 2 allows us to solve for

a

.

5

Test the point

(2, 2)

in the Case 2 candidate function.

p(2) = 2

, which matches the given point.

Confirming that the Case 2 function is consistent with the point

(2, 2)

.

6

Evaluate the correct function at

x = 6

.

p(6) = 10

.

Substituting

x = 6

into the verified function

p(x) = \frac{1}{2}(x - 1)(x - 4)^2

to find the final answer.

Key Concept

Identifying the algebraic form of a polynomial from its

x

-intercepts and multiplicities, and determining unknown coefficients using coordinate points.

Question 286Question →

If the graph of $y = f(x)$ contains the point $(3, 7)$ , and the function $g$ is defined by $g(x) = f(x + 4) - 2$ , what is the value of $g(-1)$ ?

Show answer & explanation

Answer: 5

Answer

5

Since the graph of

y = f(x)

contains the point

(3, 7)

, we have

f(3) = 7

. The function

g

is defined as

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

. To find the value of

g(-1)

, we substitute

x = -1

into the definition of

g

:

g(-1) = f(-1 + 4) - 2 = f(3) - 2

. Substituting

f(3) = 7

gives

g(-1) = 7 - 2 = 5

.

Step-by-Step Solution

1

Translate the point

(3, 7)

on the graph of

f(x)

into function notation.

f(3) = 7

By definition, if a point

(a, b)

is on the graph of

y = f(x)

, then

f(a) = b

.

2

Substitute

x = -1

into the expression for

g(x)

to evaluate

g(-1)

.

g(-1) = f(-1 + 4) - 2

To find

g(-1)

, replace every occurrence of

x

with

-1

in the function definition of

g(x)

.

3

Simplify the input of the function

f

and compute the final value.

g(-1) = f(3) - 2 = 7 - 2 = 5

Simplify

-1 + 4

to

3

, then substitute the known value

f(3) = 7

and subtract

2

.

Key Concept

Evaluating a transformed function using function notation and given coordinate points.

Question 287Question →

In the quadratic equation $x^2 - 8x + k = 0$ , $k$ is a constant. If the difference between the two real solutions to the equation is $2$ , what is the value of $k$ ?

15

-15

12

8

Show answer & explanation

Answer: 15

Answer

15

The correct answer is 15. The sum of the solutions to the quadratic equation

x^2 - 8x + k = 0

is given by

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

. We are given that the difference between the two solutions is

2

. Let the two solutions be

x_1

and

x_2

. We can set up the system of equations:

x_1 + x_2 = 8

and

x_1 - x_2 = 2

. Adding the equations gives

2x_1 = 10

, so

x_1 = 5

. Thus,

x_2 = 3

. The product of the solutions is equal to the constant term

k

. Therefore,

k = x_1 \times x_2 = 5 \times 3 = 15

. Alternatively, using the difference of roots formula,

\frac{\sqrt{b^2 - 4ac}}{a} = 2 \Rightarrow \frac{\sqrt{(-8)^2 - 4(1)(k)}}{1} = 2 \Rightarrow \sqrt{64 - 4k} = 2

. Squaring both sides gives

64 - 4k = 4 \Rightarrow 4k = 60 \Rightarrow k = 15

.

Step-by-Step Solution

1

Find the sum of the two solutions of the quadratic equation.

The sum of the solutions is

8

.

For a quadratic equation in the form

x^2 + bx + c = 0

, the sum of the solutions is equal to

-b

. Here,

-(-8) = 8

.

2

Set up a system of linear equations using the sum and the given difference of the solutions to find the individual solutions.

The solutions are

5

and

3

.

Let the solutions be

x_1

and

x_2

. We have the system:

x_1 + x_2 = 8

and

x_1 - x_2 = 2

. Adding these two equations gives

2x_1 = 10

, so

x_1 = 5

. Substituting

5

back into the first equation gives

5 + x_2 = 8

, so

x_2 = 3

.

3

Use the product of the solutions to determine the value of

k

.

The value of

k

is

15

.

For a quadratic equation in the form

x^2 + bx + c = 0

, the product of the solutions is equal to

c

, which is

k

. Therefore,

k = x_1 \times x_2 = 5 \times 3 = 15

.

Key Concept

Relationship between the roots and coefficients of a quadratic equation (Vieta's Formulas).

Question 288Question →

A polynomial function $p$ of degree 3 has $x$ -intercepts at $(-2, 0)$ with multiplicity 2, and $(3, 0)$ with multiplicity 1. In the $xy$ -plane, the graph of $y = p(x)$ intersects the $y$ -axis at $(0, 24)$ . What is the remainder when $p(x)$ is divided by $x - 1$ ?

8

16

36

-36

Show answer & explanation

Answer: $36$

Answer

The remainder when the polynomial function is divided by

x - 1

is 36.

The correct answer is 36. A polynomial with a root at

x = -2

of multiplicity 2 and a root at

x = 3

of multiplicity 1 has the form

p(x) = a(x + 2)^2(x - 3)

. Since the

y

-intercept is

(0, 24)

, we solve

p(0) = a(2)^2(-3) = 24

to find

a = -2

. Thus, the polynomial is

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

. According to the Remainder Theorem, dividing

p(x)

by

x - 1

leaves a remainder of

p(1)

. Substituting

x = 1

yields

p(1) = -2(3)^2(-2) = 36

.

Step-by-Step Solution

1

Write the general form of the cubic polynomial using its roots and multiplicities.

p(x) = a(x + 2)^2(x - 3)

Since there is an

x

-intercept at

x = -2

with multiplicity 2,

(x + 2)^2

is a factor. Since there is an

x

-intercept at

x = 3

with multiplicity 1,

(x - 3)

is a factor. Here,

a

is a constant coefficient.

2

Determine the value of the constant coefficient

a

using the

y

-intercept.

a = -2

, so

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

The graph intersects the

y

-axis at

(0, 24)

, meaning

p(0) = 24

. Substituting

x = 0

gives

p(0) = a(0 + 2)^2(0 - 3) = -12a

. Setting

-12a = 24

yields

a = -2

.

3

Apply the Remainder Theorem to find the required remainder.

The remainder is equal to

p(1)

.

By the Remainder Theorem, the remainder when a polynomial

p(x)

is divided by

x - c

is

p(c)

. Here, the divisor is

x - 1

, so we evaluate the polynomial at

x = 1

.

4

Calculate the value of

p(1)

.

p(1) = 36

Substituting

x = 1

into

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

gives

p(1) = -2(1 + 2)^2(1 - 3) = -2(9)(-2) = 36

.

Key Concept

Identifying a polynomial from its roots and multiplicities, solving for its leading coefficient using a given point, and applying the Remainder Theorem.

Question 289Question →

The function $f$ is defined for all real numbers, and the graph of $y = f(x)$ in the $xy$ -plane has a single minimum at the point $(5, -2)$ . The function $g$ is defined by $g(x) = -3f(2x - 4) + 7$ . What is the $y$ -coordinate of the maximum point on the graph of $y = g(x)$ ?

Show answer & explanation

Answer: 13

Answer

The correct answer is 13.

The graph of

y = f(x)

has a minimum at

(5, -2)

, which means

f(5) = -2

and

f(x) \ge -2

for all

x

. The function

g(x) = -3f(2x-4) + 7

includes a vertical stretch by a factor of

3

, a vertical reflection across the

x

-axis, and a vertical shift upward by

7

units. Because of the vertical reflection, the minimum value of the original function becomes the maximum value of the transformed function. Applying the vertical transformations to the

y

-coordinate of the minimum point yields

-3(-2) + 7 = 6 + 7 = 13

.

Step-by-Step Solution

1

Identify the minimum point and minimum value of the original function

f(x)

.

f(5) = -2

, and

f(x) \ge -2

for all real numbers

x

.

The problem states that the graph of

y = f(x)

has a single minimum at the point

(5, -2)

.

2

Determine the transformed

x

-coordinate corresponding to the original input of

5

.

2x - 4 = 5 \implies 2x = 9 \implies x = 4.5

.

Setting the argument of the function

f(2x-4)

equal to the original minimum input of

5

allows us to find the corresponding input

x

for the function

g

.

3

Apply the vertical transformations to find the output value of

g(x)

at

x = 4.5

.

g(4.5) = -3f(5) + 7 = -3(-2) + 7 = 6 + 7 = 13

.

Substituting

f(5) = -2

into the definition of

g(x)

gives the vertical transformation of the point.

4

Confirm that the point is indeed the maximum of the transformed function

g(x)

.

Since

f(2x-4) \ge -2

, multiplying by

-3

yields

-3f(2x-4) \le 6

. Adding

7

yields

g(x) \le 13

, confirming that

13

is the maximum value.

Multiplying a function by a negative number reflects its graph vertically, changing a minimum point into a maximum point.

Key Concept

Applying horizontal and vertical transformations to function coordinates, and understanding how vertical reflections affect the extrema (minima and maxima) of a graph.

Question 290Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + bx + c$ , where $b$ and $c$ are constants, has its vertex at $(h, k)$ . The function $g$ is defined by $g(x) = f(x - 3) + 4$ . The graph of $g$ passes through the origin $(0,0)$ , and its vertex lies on the line $y = x$ in the first quadrant. What is the value of $f(0)$ ?

-1

-7

-3

-21

Show answer & explanation

Answer: -7

Answer

-7

The correct answer is

-7

. Representing

f(x)

in vertex form as

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

tells us that translating the function 3 units right and 4 units up shifts the vertex from

(h, k)

to

(h+3, k+4)

. Since this vertex lies on the line

y=x

, we have

h+3 = k+4

, or

k = h-1

. Since the vertex of

g

lies in the first quadrant, its coordinates must be positive, meaning

h > -3

and

k > -4

. Using the fact that the graph of

g

passes through the origin, we have

g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4

. Substituting

x = -3

into the vertex form of

f(x)

gives

-(-3-h)^2 + k = -4 \implies (h+3)^2 - k = 4

. Substituting

k = h-1

yields

(h+3)^2 - (h-1) = 4 \implies h^2 + 5h + 6 = 0

, which factors into

(h+2)(h+3) = 0

. The solution

h = -3

is discarded because it places the vertex of

g

at

(0,0)

, which is not in the first quadrant. Therefore,

h = -2

and

k = -3

. Calculating

f(0)

gives

-h^2 + k = -(-2)^2 + (-3) = -7

.

Step-by-Step Solution

1

Determine the vertex form of

f(x)

and the vertex of

g(x)

.

Since

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

has its vertex at

(h, k)

, its vertex form is

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

. The function

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

represents a horizontal shift of

f

by 3 units to the right and a vertical shift by 4 units up. Thus, the vertex of

g

is

(h + 3, k + 4)

.

Understanding translations allows us to write the coordinates of the new vertex in terms of the original vertex variables.

2

Use the line

y = x

and quadrant constraints to find a relation between

h

and

k

.

Since the vertex of

g

lies on the line

y = x

, we set its coordinates equal:

h + 3 = k + 4

, which simplifies to

k = h - 1

. Additionally, because the vertex is in the first quadrant, we must have

h + 3 > 0

and

k + 4 > 0

.

The geometric placement of the vertex on the line

y = x

constrains its coordinate values.

3

Apply the condition that the graph of

g

passes through the origin.

g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4

. Substituting

x = -3

into the vertex form of

f(x)

gives

-(-3 - h)^2 + k = -4

, which simplifies to

(h + 3)^2 - k = 4

.

The point

(0,0)

lying on the graph of

g

provides an equation to solve for the vertex parameters.

4

Substitute

k = h - 1

into the quadratic equation and solve for

h

.

(h + 3)^2 - (h - 1) = 4 \implies h^2 + 6h + 9 - h + 1 = 4 \implies h^2 + 5h + 6 = 0

. Factoring gives

(h + 2)(h + 3) = 0

, so

h = -2

or

h = -3

.

Substituting the linear relation into the quadratic equation isolates the variable

h

.

5

Filter the solutions using the first quadrant constraint and calculate

f(0)

.

If

h = -3

, then

k = -4

, giving the vertex of

g

at

(0, 0)

, which is not in the first quadrant. If

h = -2

, then

k = -3

, giving the vertex of

g

at

(1, 1)

, which is in the first quadrant. Thus,

h = -2

and

k = -3

. The value of

f(0)

is

f(0) = -(0 - h)^2 + k = -h^2 + k = -(-2)^2 + (-3) = -4 - 3 = -7

.

The quadrant condition uniquely determines the correct vertex coordinates, allowing us to find the y-intercept of the original function.

Key Concept

Quadratic Functions and Graphs

Question 291Question →

If $x$ satisfies the equation below, what is the value of $x - 3$ ?

x - \sqrt{2x + 7} = 4

-2

6

9

-4

Show answer & explanation

Answer: 6

Answer

6

The correct answer is the value

6

. Isolating the radical in

x - \sqrt{2x + 7} = 4

gives

x - 4 = \sqrt{2x + 7}

. Squaring both sides yields

x^2 - 8x + 16 = 2x + 7

, which simplifies to the quadratic equation

x^2 - 10x + 9 = 0

. Factoring this equation gives

(x - 9)(x - 1) = 0

, yielding potential solutions of

x = 9

and

x = 1

. Substituting these potential solutions back into the original equation reveals that

x = 9

is a valid solution because

9 - \sqrt{25} = 4

, whereas

x = 1

is extraneous because

1 - \sqrt{9} = -2 \neq 4

. The value of the expression

x - 3

is

9 - 3 = 6

.

Step-by-Step Solution

1

Isolate the radical term in the equation.

x - 4 = \sqrt{2x + 7}

Before squaring both sides of a radical equation, it is best to isolate the radical term on one side.

2

Square both sides of the equation to eliminate the radical.

(x - 4)^2 = 2x + 7

Squaring a square root removes the radical, allowing us to solve the resulting algebraic equation.

3

Expand the squared binomial and move all terms to one side to set the quadratic equation to zero.

x^2 - 10x + 9 = 0

(x - 4)^2

expands to

x^2 - 8x + 16

. Subtracting

2x

and

7

from both sides gives the standard quadratic form

ax^2 + bx + c = 0

.

4

Factor the quadratic equation.

(x - 9)(x - 1) = 0

Finding two numbers that multiply to

9

and add to

-10

gives

-9

and

-1

. Thus, the potential solutions are

x = 9

and

x = 1

.

5

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

x = 9

is valid, while

x = 1

is extraneous.

Checking

x = 9

:

9 - \sqrt{2(9)+7} = 9 - 5 = 4

(true). Checking

x = 1

:

1 - \sqrt{2(1)+7} = 1 - 3 = -2 \neq 4

(false). Thus,

x = 9

is the only real solution.

6

Calculate the value of the requested expression

x - 3

.

9 - 3 = 6

Since the only valid solution is

x = 9

, we substitute

9

into the expression

x - 3

to find the final answer.

Key Concept

Radical equations require isolating the radical, squaring both sides, solving the resulting equation, and checking for extraneous solutions that do not satisfy the original equation.

Estimated Time:1m 30s

Question 292Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = x^2 - 4x - 5$ intersects the $x$ -axis at the points $(p, 0)$ and $(q, 0)$ and has vertex $(h, k)$ . What is the area of the triangle with vertices at $(p, 0)$ , $(q, 0)$ , and $(h, k)$ ?

Show answer & explanation

Answer: 27

Answer

The area of the triangle is 27.

To find the area of the triangle, we first determine the coordinates of its vertices. The base of the triangle lies on the

x

-axis, with endpoints at the

x

-intercepts of the function

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

. Solving

x^2 - 4x - 5 = 0

by factoring gives

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

, so the intercepts are at

x = -1

and

x = 5

. The distance between these two points is

5 - (-1) = 6

, which is the base of the triangle. The third vertex is the vertex of the parabola. The

x

-coordinate of the vertex is

h = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2

. Substituting

x = 2

into the function gives the

y

-coordinate:

k = f(2) = 2^2 - 4(2) - 5 = -9

. The height of the triangle is the vertical distance from the

x

-axis to the vertex, which is

|-9| = 9

. The area of the triangle is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 9 = 27

.

Step-by-Step Solution

1

Find the

x

-intercepts of the parabola.

The intercepts are

(-1, 0)

and

(5, 0)

.

Setting

f(x) = 0

gives

x^2 - 4x - 5 = 0

. Factoring the quadratic equation yields

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

, which gives

x = 5

and

x = -1

.

2

Calculate the base of the triangle.

The base length is

6

.

The base of the triangle is the segment along the

x

-axis between the two intercepts. The distance between

(-1, 0)

and

(5, 0)

is

5 - (-1) = 6

.

3

Find the vertex

(h, k)

of the parabola.

The vertex is at

(2, -9)

.

The

x

-coordinate of the vertex is the midpoint of the intercepts:

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

. The

y

-coordinate is

k = f(2) = 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9

.

4

Calculate the area of the triangle.

The area is

27

.

The height of the triangle is the distance from the

x

-axis to the vertex, which is

|k| = |-9| = 9

. Using the formula for the area of a triangle,

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 9 = 27

.

Key Concept

Finding the

x

-intercepts and vertex of a quadratic function to solve geometric problems in the coordinate plane.

Question 293Question →

The function $f$ is defined by $f(x) = (x - 3)(x^3 - kx^2 + 5x - 15)$ , where $k$ is a constant. In the $xy$ -plane, the graph of $y = f(x)$ is tangent to the $x$ -axis at the point $(3, 0)$ . What is the value of $k$ ?

Show answer & explanation

Answer: 3

Answer

3

For the graph of a polynomial function to be tangent to the

x

-axis at

(3, 0)

, the root

x = 3

must have an even multiplicity (at least 2). The function is defined as

f(x) = (x - 3)(x^3 - kx^2 + 5x - 15)

. Since there is already one factor of

(x - 3)

explicitly defined, the remaining cubic factor

g(x) = x^3 - kx^2 + 5x - 15

must also have a factor of

(x - 3)

to make the total multiplicity of the root

x = 3

at least 2. According to the Factor Theorem, if

(x - 3)

is a factor of

g(x)

, then

g(3) = 0

. Substituting

x = 3

into

g(x)

gives

3^3 - k(3)^2 + 5(3) - 15 = 0

, which simplifies to

27 - 9k = 0

. Solving this equation for

k

yields

k = 3

.

Step-by-Step Solution

1

Identify the relationship between graph tangency and factor multiplicity.

For the graph of a polynomial function to be tangent to the

x

-axis at a point

(c, 0)

, the factor

(x - c)

must have an even multiplicity of at least 2 in the polynomial's factored form.

An odd multiplicity root causes the graph to cross the

x

-axis, while an even multiplicity root causes the graph to touch the

x

-axis and turn around (tangency).

2

Apply the multiplicity requirement to the given function.

Since

f(x) = (x - 3)(x^3 - kx^2 + 5x - 15)

already contains one factor of

(x - 3)

, the cubic expression

g(x) = x^3 - kx^2 + 5x - 15

must also contain

(x - 3)

as a factor to ensure the total multiplicity of the root

x = 3

is at least 2.

This guarantees that

(x - 3)^2

is a factor of

f(x)

.

3

Apply the Factor Theorem to the cubic expression.

Since

(x - 3)

is a factor of

g(x)

, then

g(3) = 0

.

The Factor Theorem states that a polynomial

P(x)

has a factor

(x - c)

if and only if

P(c) = 0

.

4

Solve for the constant

k

by substituting

x = 3

into

g(x)

.

3^3 - k(3)^2 + 5(3) - 15 = 0 \implies 27 - 9k + 15 - 15 = 0 \implies 27 - 9k = 0 \implies 9k = 27 \implies k = 3

.

Arithmetic simplification yields the value of the constant.

Key Concept

The relationship between polynomial factors, root multiplicities, and the behavior of the graph at

x

-intercepts.

Question 294Question →

A polynomial function $q$ with real coefficients satisfies the equation $q(x) + q(6-x) = 8$ for all real numbers $x$ . In the $xy$ -plane, the graph of $y = q(x)$ has an $x$ -intercept at $(5, 0)$ . What is the remainder when $q(x)$ is divided by $x-1$ ?

8

-8

0

2

Show answer & explanation

Answer: $8$

Answer

The correct answer is

8

, which is the remainder when the polynomial is divided by

x-1

.

To find the remainder when the polynomial

q(x)

is divided by

x-1

, we use the Remainder Theorem, which states that this remainder is equal to

q(1)

. Since the graph of

y = q(x)

has an

x

-intercept at

(5, 0)

, we know that

q(5) = 0

. Substituting

x = 5

into the given equation

q(x) + q(6-x) = 8

yields

q(5) + q(6-5) = 8

, which simplifies to

q(5) + q(1) = 8

. Substituting

q(5) = 0

into this equation gives

0 + q(1) = 8

, so

q(1) = 8

. Geometrically, the equation

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

shows that the graph of

q

is symmetric under a point reflection about

(3, 4)

. Reflecting the point

(5, 0)

about

(3, 4)

gives the point

(1, 8)

, verifying that

q(1) = 8

.

Step-by-Step Solution

1

Translate the given graphical intercept into function notation.

q(5) = 0

An

x

-intercept at

(5, 0)

means that when the input

x

is

5

, the output of the function is

0

.

2

Apply the Remainder Theorem to the divisor

x-1

.

The remainder is equal to

q(1)

.

The Remainder Theorem states that when a polynomial

q(x)

is divided by

x-a

, the remainder is

q(a)

.

3

Substitute

x = 5

into the given functional equation

q(x) + q(6-x) = 8

.

q(5) + q(6-5) = 8 \implies q(5) + q(1) = 8

Substituting

x = 5

allows us to relate the known value

q(5)

directly to the target value

q(1)

.

4

Solve for

q(1)

using the value of

q(5)

.

0 + q(1) = 8 \implies q(1) = 8

Substituting

q(5) = 0

into the equation yields the value of

q(1)

, which is the remainder.

Key Concept

Remainder Theorem and Polynomial Graphs

Alternative Method

Use the point symmetry of the graph. The equation

q(x) + q(6-x) = 8

can be rewritten as

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

. This indicates that the graph of

y = q(x)

has point symmetry about the point

(3, 4)

. Since the graph has an

x

-intercept at

(5, 0)

, point symmetry dictates that the reflected point across

(3, 4)

must also lie on the graph. The midpoint of

(5, 0)

and the symmetric point

(x_0, y_0)

is

(3, 4)

. Solving

\frac{5+x_0}{2} = 3

gives

x_0 = 1

, and

\frac{0+y_0}{2} = 4

gives

y_0 = 8

. Thus,

(1, 8)

lies on the graph, meaning

q(1) = 8

. By the Remainder Theorem, the remainder when

q(x)

is divided by

x-1

is

q(1) = 8

.

Estimated Time:1m 30s

Question 295Question →

An equation is shown below.

\frac{x}{x - 1} - \frac{2}{x + 2} = \frac{6}{x^2 + x - 2}

What is the value of the real solution to the equation?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

The correct answer is 2. Multiplying both sides by the least common denominator

(x - 1)(x + 2)

results in the quadratic equation

x(x + 2) - 2(x - 1) = 6

. Simplifying this equation yields

x^2 + 2 = 6

, which has solutions

x = 2

and

x = -2

. However, substituting

x = -2

into the original equation results in division by zero, making it an extraneous solution. Therefore,

2

is the only valid real solution.

Step-by-Step Solution

1

Multiply the entire equation by the least common denominator,

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

, to clear the denominators.

x(x + 2) - 2(x - 1) = 6

This simplifies the rational equation into a polynomial equation.

2

Expand the terms and simplify the equation.

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

, which simplifies to

x^2 + 2 = 6

.

Distributing the terms allows us to group like terms and solve for the variable.

3

Solve the quadratic equation for

x

.

x^2 = 4

, which gives

x = 2

or

x = -2

.

Subtracting 2 from both sides isolates the squared variable.

4

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

For

x = -2

, the denominator

x + 2

becomes 0, which is undefined. For

x = 2

, all denominators are non-zero.

Solutions that make any denominator in the original equation equal to zero are extraneous and must be excluded.

Key Concept

Solving rational equations by finding a common denominator and checking for extraneous solutions.

Question 296Question →

For the quadratic function $f$ , the table below shows three points that lie on its graph in the $xy$ -plane, where $k$ is a constant.

$x$	$f(x)$
$2$	$0$
$6$	$0$
$1$	$-15$

If the vertex of the graph of $y = f(x)$ is $(4, k)$ , what is the value of $k$ ?

Show answer & explanation

Answer: 12

Answer

The value of

k

is

12

.

The table indicates that the function has

x

-intercepts at

x = 2

and

x = 6

because

f(2) = 0

and

f(6) = 0

. Consequently, the quadratic function can be written in factored form as

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

for some constant

a

. Using the point

(1, -15)

to find

a

, we substitute

x = 1

and

f(1) = -15

, which yields

-15 = a(1 - 2)(1 - 6)

, or

-15 = 5a

, so

a = -3

. Since the vertex of the graph is

(4, k)

, the value of

k

is the function value at the vertex's

x

-coordinate, which is

f(4)

. Evaluating the function gives

k = -3(4 - 2)(4 - 6) = -3(2)(-2) = 12

.

Step-by-Step Solution

1

Write the quadratic function in factored form using the given

x

-intercepts.

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

The table shows that

f(2) = 0

and

f(6) = 0

, meaning the graph has

x

-intercepts at

x = 2

and

x = 6

.

2

Substitute the point

(1, -15)

into the factored equation to find the value of the constant

a

.

a = -3

Substituting

x = 1

and

f(x) = -15

gives

-15 = a(1 - 2)(1 - 6) \implies -15 = 5a \implies a = -3

.

3

Calculate the value of

k

by finding the function value at the vertex

x = 4

.

k = 12

Since the vertex is

(4, k)

, the value of

k

is

f(4)

. Substituting

x = 4

into

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

yields

k = -3(4 - 2)(4 - 6) = -3(2)(-2) = 12

.

Key Concept

Using intercepts and an additional point to determine the equation of a quadratic function, and evaluating it at the vertex.

Question 297Question →

The function $f$ is defined by $f(x) = 2x - 3$ . If the function $g$ is defined by $g(x) = f(x + 4)$ , what is the value of $g(1)$ ?

-9

3

7

13

Show answer & explanation

Answer: 7

Answer

The correct answer is 7.

Since

g(x) = f(x + 4)

, evaluating

g(1)

requires finding

f(1 + 4)

, which is

f(5)

. Substituting

5

into the expression for

f(x)

gives

f(5) = 2(5) - 3 = 10 - 3 = 7

. Therefore, the correct answer is 7.

Step-by-Step Solution

1

Substitute

x = 1

into the definition of

g(x)

to express

g(1)

in terms of

f

.

g(1) = f(1 + 4) = f(5)

To find the value of

g(1)

, we must evaluate the input to the outer function first.

2

Substitute the input value

5

into the expression for

f(x)

.

f(5) = 2(5) - 3

Evaluating

f(5)

requires replacing

x

with

5

in the definition

f(x) = 2x - 3

.

3

Perform the operations to find the final value.

f(5) = 10 - 3 = 7

Simplifying the numerical expression gives the final value of

g(1)

.

Key Concept

Evaluating a transformed function at a given point using function notation.

Estimated Time:45s

Question 298Question →

A projectile is launched from the ground. Its height, in feet, $t$ seconds after launch is modeled by the function $h(t) = -16t^2 + v_0 t$ , where $v_0$ is the initial upward velocity in feet per second. If the projectile reaches its maximum height of $144$ feet, what is the value of $v_0$ ?

Show answer & explanation

Answer: 96

Answer

The correct answer is 96. The initial velocity of the projectile must be 96 feet per second to reach a maximum height of 144 feet.

The maximum height of a projectile modeled by a quadratic function is the

y

-value of its vertex. For

h(t) = -16t^2 + v_0 t

, the time

t

at the vertex is given by

t = -\frac{b}{2a} = -\frac{v_0}{2(-16)} = \frac{v_0}{32}

. Substituting this value of

t

back into the height equation yields the maximum height:

h\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0^2}{1024}\right) + \frac{v_0^2}{32} = -\frac{v_0^2}{64} + \frac{v_0^2}{32} = \frac{v_0^2}{64}

. Given that the maximum height is

144

feet, we set

\frac{v_0^2}{64} = 144

, which simplifies to

v_0^2 = 144 \times 64

. Taking the square root of both sides gives

v_0 = 12 \times 8 = 96

.

Step-by-Step Solution

1

Find the time

t

at which the maximum height occurs using the vertex formula.

t = \frac{v_0}{32}

For a quadratic function of the form

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

, the vertex occurs at

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

. Here,

a = -16

and

b = v_0

, so the vertex is at

t = -\frac{v_0}{2(-16)} = \frac{v_0}{32}

.

2

Substitute the time

t = \frac{v_0}{32}

into the height function to express the maximum height in terms of

v_0

.

h\left(\frac{v_0}{32}\right) = \frac{v_0^2}{64}

Evaluating

h\left(\frac{v_0}{32}\right)

gives

-16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0^2}{1024}\right) + \frac{v_0^2}{32} = -\frac{v_0^2}{64} + \frac{v_0^2}{32} = \frac{v_0^2}{64}

.

3

Set the maximum height expression equal to the given maximum height of

144

feet and solve for

v_0

.

v_0 = 96

Setting

\frac{v_0^2}{64} = 144

yields

v_0^2 = 144 \times 64 = 9216

. Taking the positive square root (since

v_0

represents a positive initial upward velocity) gives

v_0 = \sqrt{144} \times \sqrt{64} = 12 \times 8 = 96

.

Key Concept

Determining the vertex coordinates of a quadratic function to find maximum value in context.

Question 299Question →

If $27^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

To solve the equation

27^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}

, express both sides with the base

3

. Since

27 = 3^3

and

\frac{1}{3} = 3^{-1}

, the equation becomes

(3^3)^{2x - 2} = (3^{-1})^{x - 8}

. Applying the exponent rule

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

yields

3^{6x - 6} = 3^{8 - x}

. Since the bases are now the same, set the exponents equal to each other:

6x - 6 = 8 - x

. Adding

x

to both sides gives

7x - 6 = 8

, and adding

6

to both sides gives

7x = 14

. Dividing by

7

results in

x = 2

.

Step-by-Step Solution

1

Rewrite

27

and

\frac{1}{3}

as powers of

3

.

27 = 3^3

and

\frac{1}{3} = 3^{-1}

This allows both sides of the equation to be expressed with a common base of

3

so their exponents can be equated.

2

Substitute these bases back into the equation and apply exponent rules.

(3^3)^{2x - 2} = (3^{-1})^{x - 8} \implies 3^{3(2x - 2)} = 3^{-(x - 8)} \implies 3^{6x - 6} = 3^{8 - x}

Using the power rule

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

, multiply the outer exponent by the power of the base.

3

Equate the exponents and solve for

x

.

6x - 6 = 8 - x \implies 7x = 14 \implies x = 2

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

Key Concept

Solving exponential equations by finding a common base and applying exponent rules.

Question 300Question →

The quadratic function $f$ is defined by $f(x) = a(x - h)^2 + k$ , where $a$ , $h$ , and $k$ are constants. In the $xy$ -plane, the graph of $y = f(x)$ has a vertex at $(3, -4)$ and passes through the point $(5, 8)$ . If the function $g$ is defined by $g(x) = -2f(x - 1) + 5$ , what is the value of $g(2)$ ?

5

7

-11

13

Show answer & explanation

Answer: -11

Answer

-11

To find the value of

g(2)

, we first determine the equation of the quadratic function

f(x)

. Since the vertex is

(3, -4)

, the vertex form is

f(x) = a(x - 3)^2 - 4

. Substituting the point

(5, 8)

yields

8 = a(5 - 3)^2 - 4

, which simplifies to

12 = 4a

, so

a = 3

. Therefore,

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

. We then substitute

x = 2

into the definition of

g(x)

, obtaining

g(2) = -2f(2 - 1) + 5 = -2f(1) + 5

. Evaluating

f(1)

gives

f(1) = 3(1 - 3)^2 - 4 = 8

. Substituting this back into the expression for

g(2)

gives

-2(8) + 5 = -11

. Thus, the option with value -11 is correct.

Step-by-Step Solution

1

Write the quadratic function

f(x)

in vertex form using the given vertex

(3, -4)

.

f(x) = a(x - 3)^2 - 4

The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex.

2

Substitute the point

(5, 8)

into the vertex form to solve for the constant

a

.

8 = a(5 - 3)^2 - 4 \implies 8 = 4a - 4 \implies 12 = 4a \implies a = 3

. Thus,

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

.

Since the graph of

f

passes through

(5, 8)

, these coordinates must satisfy the function's equation.

3

Substitute

x = 2

into the definition of

g(x)

to express

g(2)

in terms of

f

.

g(2) = -2f(2 - 1) + 5 = -2f(1) + 5

We need to evaluate the inner function transformation

f(x - 1)

at

x = 2

.

4

Evaluate

f(1)

using the formula determined in Step 2.

f(1) = 3(1 - 3)^2 - 4 = 3(-2)^2 - 4 = 3(4) - 4 = 8

To find

g(2)

, we must first compute the value of

f(1)

.

5

Substitute

f(1) = 8

back into the expression for

g(2)

and simplify.

g(2) = -2(8) + 5 = -16 + 5 = -11

This completes the evaluation of the multi-step transformation.

Key Concept

Function Notation and Transformations

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Advanced Math

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Common Mistakes

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Common Mistakes

Alternative Method

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Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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