Advanced Math Practice Questions — SAT

Question 1Question →

When the polynomial $g(x) = x^4 - 2x^3 + ax^2 - 8$ is divided by $x - 3$ , the remainder is $37$ , where $a$ is a constant. What is the value of $a$ ?

Show answer & explanation

Answer: 2

Answer

2

According to the Polynomial Remainder Theorem, dividing a polynomial

g(x)

by a linear divisor

x - c

yields a remainder equal to

g(c)

. In this problem, the divisor is

x - 3

, so we evaluate the polynomial at

x = 3

and set it equal to the given remainder of

37

. Substituting

3

for

x

in

g(x) = x^4 - 2x^3 + ax^2 - 8

yields

3^4 - 2(3)^3 + a(3)^2 - 8 = 37

. Simplifying the numerical expressions gives

81 - 54 + 9a - 8 = 37

, which simplifies to

19 + 9a = 37

. Subtracting

19

from both sides results in

9a = 18

. Dividing both sides by

9

gives the value of the constant

a

as

2

.

Step-by-Step Solution

1

Apply the Polynomial Remainder Theorem

g(3) = 37

The Remainder Theorem states that when a polynomial

g(x)

is divided by

x - c

, the remainder is

g(c)

.

2

Substitute

x = 3

into the polynomial

g(x)

3^4 - 2(3)^3 + a(3)^2 - 8 = 37

This sets the value of the polynomial evaluated at

x = 3

equal to the given remainder of

37

.

3

Simplify the constant terms

19 + 9a = 37

Evaluating the exponents and multiplying:

3^4 = 81

,

2(3^3) = 54

, and

a(3^2) = 9a

. Combining the constant terms gives

81 - 54 - 8 = 19

.

4

Solve the linear equation for

a

a = 2

Subtracting

19

from both sides gives

9a = 18

, and dividing both sides by

9

gives

a = 2

.

Key Concept

Polynomial Remainder Theorem

Question 2Question →

If $\frac{x + 5}{x - 1} = 3$ , what is the value of $x + 1$ ?

2

3

5

4

Show answer & explanation

Answer: 5

Answer

5

The correct answer is

5

. To solve the rational equation

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

, we first multiply both sides of the equation by

x - 1

, which gives

x + 5 = 3(x - 1)

. Distributing the

3

on the right side yields

x + 5 = 3x - 3

. Subtracting

x

from both sides gives

5 = 2x - 3

. Adding

3

to both sides yields

8 = 2x

. Dividing both sides by

2

gives

x = 4

. Substituting this value back into the expression

x + 1

yields

4 + 1 = 5

.

Step-by-Step Solution

1

Multiply both sides of the equation by

x - 1

to clear the fraction.

x + 5 = 3(x - 1)

Multiplying both sides by the denominator clears the fraction and allows us to solve the equation linearly.

2

Distribute the

3

to the terms inside the parentheses on the right side of the equation.

x + 5 = 3x - 3

This step simplifies the equation to prepare for isolating the variable

x

.

3

Isolate the variable

x

by subtracting

x

from both sides and then adding

3

to both sides.

8 = 2x

Grouping like terms on opposite sides of the equation simplifies the process of finding the value of the variable.

4

Divide both sides of the equation by

2

to find the value of

x

.

x = 4

Dividing by the coefficient of

x

isolates the variable completely.

5

Substitute the value of

x

into the expression

x + 1

to find the final answer.

4 + 1 = 5

The question asks for the value of the expression

x + 1

, not just the value of

x

.

Key Concept

Solving rational equations that simplify to linear equations

Question 3Question →

In the quadratic equation $2x^2 - 16x + c = 0$ , $c$ is a constant. If one of the solutions to the equation is $4 - \sqrt{3}$ , what is the value of $c$ ?

Show answer & explanation

Answer: 26

Answer

26

Since the coefficients of the quadratic equation are real, any irrational roots must occur in conjugate pairs. Therefore, if one solution is

4 - \sqrt{3}

, the other solution must be

4 + \sqrt{3}

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

r_1

and

r_2

for a quadratic equation

ax^2 + bx + c = 0

is

\frac{c}{a}

. In this equation,

a = 2

, so the product of the roots is

\frac{c}{2}

. Multiplying the two solutions gives

(4 - \sqrt{3})(4 + \sqrt{3}) = 16 - 3 = 13

. Setting

\frac{c}{2} = 13

and solving for

c

yields

c = 26

.

Step-by-Step Solution

1

Determine the second root of the quadratic equation.

The second root is

4 + \sqrt{3}

.

Since the quadratic equation has real coefficients, the irrational roots must be conjugate pairs.

2

Express the product of the roots using Vieta's formulas.

The product of the roots is

\frac{c}{2}

.

For any quadratic equation

ax^2 + bx + c = 0

, the product of the roots is equal to

\frac{c}{a}

. Here,

a = 2

.

3

Calculate the product of the two roots.

(4 - \sqrt{3})(4 + \sqrt{3}) = 16 - 3 = 13

.

Using the difference of squares formula,

(a-b)(a+b) = a^2 - b^2

.

4

Equate the product of the roots to the expression from step 2 and solve for

c

.

\frac{c}{2} = 13 \Rightarrow c = 26

.

By substituting the calculated product into the formula for the product of the roots, we find the value of the constant.

Key Concept

Using Vieta's formulas and the conjugate root theorem to solve for coefficients of a quadratic equation.

Alternative Method

Alternatively, substitute the given solution

x = 4 - \sqrt{3}

directly into the equation

2x^2 - 16x + c = 0

and solve for

c

. First, calculate

x^2 = (4 - \sqrt{3})^2 = 16 - 8\sqrt{3} + 3 = 19 - 8\sqrt{3}

. Then substitute this into the equation:

2(19 - 8\sqrt{3}) - 16(4 - \sqrt{3}) + c = 0

. Simplifying this yields

38 - 16\sqrt{3} - 64 + 16\sqrt{3} + c = 0 \Rightarrow -26 + c = 0 \Rightarrow c = 26

.

Estimated Time:1m 30s

Question 4Question →

If $5^{x+2} - 5^x = 120\sqrt{5}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 1.5

Answer

1.5 (or 3/2)

By factoring out the common term

5^x

from the left side of the equation, we get

5^x(5^2 - 1) = 120\sqrt{5}

, which simplifies to

24 \cdot 5^x = 120\sqrt{5}

. Dividing both sides by 24 isolates the exponential term:

5^x = 5\sqrt{5}

. Since

5\sqrt{5}

can be written as

5^1 \cdot 5^{0.5} = 5^{1.5}

, we have

5^x = 5^{1.5}

. Equating the exponents gives

x = 1.5

(or

3/2

).

Step-by-Step Solution

1

Factor out

5^x

from the left side of the equation.

5^x(5^2 - 1) = 120\sqrt{5}

To apply exponent rules to rewrite

5^{x+2}

as

5^x \cdot 5^2

and then factor out the common term

5^x

.

2

Simplify the constant term inside the parentheses.

24 \cdot 5^x = 120\sqrt{5}

Evaluating

5^2 - 1 = 25 - 1 = 24

simplifies the coefficient of the exponential expression.

3

Divide both sides of the equation by 24.

5^x = 5\sqrt{5}

To isolate the exponential term

5^x

on one side of the equation.

4

Express the right side as a single power of 5.

5^x = 5^{1.5}

Using exponent rules where

5\sqrt{5} = 5^1 \cdot 5^{0.5} = 5^{1.5}

so that both sides have the same base.

5

Equate the exponents of the common base 5.

x = 1.5

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

Key Concept

Solving exponential equations by factoring and rewriting terms using a common base.

Estimated Time:1m 30s

Question 5Question →

If $27^x = 9^{x + 1}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

By rewriting

27

as

3^3

and

9

as

3^2

, the equation becomes

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

. Applying the power rule of exponents, this simplifies to

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

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

3x = 2x + 2

. Solving for

x

gives

x = 2

.

Step-by-Step Solution

1

Rewrite

27

and

9

as powers of

3

.

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

To solve exponential equations with different bases, it is helpful to express them using a common base.

2

Apply the power of a power rule

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

to simplify the exponents.

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

This simplifies the exponential expressions on both sides of the equation.

3

Set the exponents equal to each other.

3x = 2x + 2

Since the bases are equal (

3 = 3

), the exponents must also be equal for the equation to hold.

4

Solve the linear equation for

x

.

x = 2

Subtracting

2x

from both sides isolates the variable

x

.

Key Concept

Solving exponential equations by expressing both sides with a common base and equating exponents.

Estimated Time:45s

Question 6Question →

What is the sum of all real values of $x$ that satisfy the equation $\sqrt{3x + 40} - x = 4$ ?

-5

3

5

-8

Show answer & explanation

Answer: 3

Answer

The sum of all valid real solutions is 3.

Isolating the radical gives

\sqrt{3x + 40} = x + 4

. Squaring both sides and simplifying results in

x^2 + 5x - 24 = 0

. Factoring gives

(x + 8)(x - 3) = 0

, leading to candidate solutions of 3 and -8. Substituting these back shows that only 3 satisfies the original equation, as -8 results in an extraneous solution. Therefore, the sum of all valid solutions is 3.

Step-by-Step Solution

1

Isolate the radical expression on one side of the equation.

\sqrt{3x + 40} = x + 4

To prepare both sides of the equation for squaring.

2

Square both sides of the equation to eliminate the radical.

3x + 40 = (x + 4)^2

Squaring a square root cancels the radical.

3

Expand the right side and move all terms to one side to form a quadratic equation.

x^2 + 5x - 24 = 0

Expanding

(x+4)^2

yields

x^2 + 8x + 16

. Subtracting

3x

and

40

from both sides sets the quadratic equation to zero.

4

Factor the quadratic equation.

(x + 8)(x - 3) = 0

The factors of -24 that add up to 5 are 8 and -3.

5

Solve for the potential values of

x

.

x = -8

or

x = 3

Setting each factor to zero yields the candidate solutions.

6

Check each candidate solution in the original equation to identify any extraneous solutions.

For

x = 3

,

\sqrt{3(3)+40}-3 = 7-3 = 4

(valid). For

x = -8

,

\sqrt{3(-8)+40}-(-8) = 4+8 = 12 \neq 4

(extraneous).

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

Key Concept

Solving radical equations and checking for extraneous solutions.

Question 7Question →

If $4^{x+2} \cdot 8^{x-1} = 16^{x+1}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

3

To solve the equation

4^{x+2} \cdot 8^{x-1} = 16^{x+1}

, rewrite all bases in terms of base 2:

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

. Simplifying using the power rule yields

2^{2x+4} \cdot 2^{3x-3} = 2^{4x+4}

. Applying the product rule on the left side gives

2^{(2x+4)+(3x-3)} = 2^{5x+1}

. Equating the exponents gives

5x + 1 = 4x + 4

. Solving for

x

results in

x = 3

.

Step-by-Step Solution

1

Express the bases 4, 8, and 16 as powers of 2.

4^{x+2} = (2^2)^{x+2} = 2^{2x+4}

,

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

, and

16^{x+1} = (2^4)^{x+1} = 2^{4x+4}

Writing all parts of the equation with a common base allows the exponents to be compared directly.

2

Combine the terms on the left side by adding their exponents.

2^{2x+4} \cdot 2^{3x-3} = 2^{(2x+4) + (3x-3)} = 2^{5x+1}

According to the product rule of exponents,

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

when the bases are the same.

3

Equate the exponents from both sides of the equation.

5x + 1 = 4x + 4

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

4

Solve the linear equation for

x

.

x = 3

Subtract

4x

and

1

from both sides to isolate the variable

x

.

Key Concept

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

Question 8Question →

In the $xy$ -plane, the line $y = x - 4$ intersects the parabola $y = x^2 - 3x - 4$ at the points $(x_1, y_1)$ and $(x_2, y_2)$ . If $x_1 < x_2$ , what is the value of $x_2$ ?

0

2

4

-4

Show answer & explanation

Answer: 4

Answer

4

The correct answer is 4. Setting the two equations equal to find their points of intersection gives

x^2 - 3x - 4 = x - 4

. Subtracting

x

and adding

4

to both sides yields the simplified quadratic equation

x^2 - 4x = 0

. Factoring out

x

gives

x(x - 4) = 0

, which has the solutions

x = 0

and

x = 4

. Given that

x_1 < x_2

, we have

x_1 = 0

and

x_2 = 4

. Therefore, the value of

x_2

is 4.

Step-by-Step Solution

1

Set the two equations equal to each other to find their points of intersection.

x^2 - 3x - 4 = x - 4

At the intersection points, the y-values of the line and the parabola must be equal.

2

Subtract

x

and add

4

to both sides of the equation to set it to zero.

x^2 - 4x = 0

To solve a quadratic equation, we must rewrite it in standard form:

ax^2 + bx + c = 0

.

3

Factor the quadratic expression by factoring out the greatest common factor, which is

x

.

x(x - 4) = 0

Factoring allows us to use the zero product property to find the individual roots.

4

Solve for the two possible values of

x

.

x = 0

or

x = 4

Setting each factor to zero gives

x = 0

and

x - 4 = 0

, which simplifies to

x = 4

.

5

Compare the two solutions to find the value of

x_2

given the condition

x_1 < x_2

.

x_1 = 0

and

x_2 = 4

Since

0 < 4

, the smaller value is

x_1

and the larger value is

x_2

.

Key Concept

Solving a system of a linear equation and a quadratic equation by substitution.

Estimated Time:1m 0s

Question 9Question →

For all $x > 1$ , which of the following is equivalent to the expression $\frac{2x^2 + 3x - 3}{x - 1} - \frac{x^3 + 2x^2 + x}{x^2 - 1}$ ?

x - 3

x + 3

x + 4

x + 5

Show answer & explanation

Answer: $x + 3$

Answer

x + 3

The correct answer is the expression that results from simplifying both terms and combining them. Simplifying the second term of the expression yields

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

. Subtracting this from the first term gives

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

. Factoring the numerator gives

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

, which simplifies to the expression representing the sum of

x

and

3

since

x-1

cancels out.

Step-by-Step Solution

1

Factor the numerator and denominator of the second term of the expression.

The numerator is

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

. The denominator is

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

.

Factoring allows for the simplification of the rational expression by identifying common factors.

2

Simplify the second term by canceling the common factor

(x+1)

since

x > 1

.

The second term simplifies to

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

.

Since

x > 1

,

x+1 \neq 0

, so we can cancel it from both numerator and denominator.

3

Subtract the simplified second term from the first term.

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

.

With a common denominator of

x-1

, we can combine the numerators directly, being careful to distribute the subtraction sign to both terms in the second numerator.

4

Factor the quadratic expression in the numerator.

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

.

Factoring the quadratic helps to check if there is a common factor with the denominator.

5

Cancel the common factor

(x-1)

to find the final simplified expression.

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

.

Since

x > 1

,

x-1 \neq 0

, allowing us to divide out the common factor.

Key Concept

Simplifying rational expressions and combining algebraic terms over a common denominator

Alternative Method

Instead of simplifying the second term first, a common denominator can be established by multiplying the first term by

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

. This gives:

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

. Factoring by grouping the numerator yields

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

.

Estimated Time:2m 0s

Question 10Question →

\begin{aligned} y &= x^2 - 8x + c \\ y &= 2x - 5 \end{aligned}

In the system of equations above, $c$ is a constant. If the system has two real solutions, $(x_1, y_1)$ and $(x_2, y_2)$ , such that the product of the $y$ -coordinates of the solutions, $y_1 y_2$ , is equal to $5$ , what is the value of $c$ ?

Show answer & explanation

Answer: 15

Answer

The value of the constant

c

is

15

.

Substituting the expression for

y

from the linear equation into the quadratic equation yields the single variable quadratic equation

x^2 - 10x + (c+5) = 0

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

x_1 + x_2 = 10

and the product of the roots is

x_1 x_2 = c+5

. Substituting these relationships into the expanded product of the

y

-coordinates,

y_1 y_2 = (2x_1 - 5)(2x_2 - 5) = 4x_1 x_2 - 10(x_1 + x_2) + 25

, allows us to set up the equation

5 = 4(c+5) - 100 + 25

. Solving for

c

yields

15

. Checking the discriminant of the quadratic equation at

c=15

gives

100 - 80 = 20

, which is positive, confirming the existence of two distinct real solutions.

Step-by-Step Solution

1

Substitute the expression for

y

from the second equation into the first equation.

2x - 5 = x^2 - 8x + c \implies x^2 - 10x + (c+5) = 0

This substitution reduces the system to a single quadratic equation whose roots,

x_1

and

x_2

, represent the

x

-coordinates of the intersection points.

2

Apply Vieta's formulas to the resulting quadratic equation.

x_1 + x_2 = 10

and

x_1 x_2 = c+5

Vieta's formulas state that for a quadratic equation

ax^2 + bx + d = 0

, the sum of the roots is

-\frac{b}{a}

and the product of the roots is \frac{d}{a}.

3

Express the product of the

y

-coordinates,

y_1 y_2

, in terms of

x_1

and

x_2

using the linear relationship.

y_1 y_2 = (2x_1 - 5)(2x_2 - 5) = 4x_1 x_2 - 10(x_1 + x_2) + 25

Since both intersection points lie on the line

y = 2x - 5

, we can substitute

y_1 = 2x_1 - 5

and

y_2 = 2x_2 - 5

and expand.

4

Substitute the Vieta's formulas relations into the product equation and solve for

c

.

5 = 4(c + 5) - 10(10) + 25 \implies 5 = 4c + 20 - 100 + 25 \implies 5 = 4c - 55 \implies 60 = 4c \implies c = 15

By substituting the known values of

(x_1 + x_2)

and

(x_1 x_2)

and setting the product

y_1 y_2

to

5

, we obtain a linear equation in terms of

c

that we can solve directly.

Key Concept

Solving systems of linear-quadratic equations using algebraic substitution and Vieta's formulas.

Question 11Question →

A quadratic function $g$ is defined by $g(x) = 3(x - 5)^2 - 4$ . If the graph of $y = g(x)$ in the $xy$ -plane is translated $6$ units up to produce the graph of $y = f(x)$ , what is the $y$ -coordinate of the vertex of the graph of $y = f(x)$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

The original function

g(x) = 3(x - 5)^2 - 4

is in vertex form

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

, where the vertex is

(h, k)

. Therefore, the vertex of the graph of

g

is

(5, -4)

. Translating a graph upward by

6

units is represented by adding

6

to the function, so

f(x) = g(x) + 6

. This transformation shifts the vertex from

(5, -4)

to

(5, -4 + 6)

, which is

(5, 2)

. The

y

-coordinate of this new vertex is

2

.

Step-by-Step Solution

1

Identify the vertex of the original quadratic function.

The vertex of the graph of

g(x) = 3(x - 5)^2 - 4

is

(5, -4)

.

A quadratic function written in vertex form

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

has its vertex at

(h, k)

.

2

Determine the vertex of the translated function.

The vertex of the graph of

y = f(x)

is

(5, 2)

.

Translating a graph

6

units up increases the

y

-coordinate of all points, including the vertex, by

6

.

Key Concept

Vertex form of a quadratic function and vertical translation of graphs.

Estimated Time:45s

Question 12Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a > 0$ , has vertex $V$ . The graph intersects the $x$ -axis at points $A$ and $B$ . If triangle $VAB$ is an equilateral triangle with an area of $12\sqrt{3}$ , what is the value of $a$ ?

\frac{1}{8}

\frac{1}{2}

1

2

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The correct answer is the value 1/2.

The value 1/2 is correct because the area of the equilateral triangle determines its side length to be

4\sqrt{3}

and its height to be

6

. Since the parabola opens upwards and its base lies on the

x

-axis, the vertex is

V(h, -6)

. The two

x

-intercepts are symmetric about the axis of symmetry

x = h

, meaning they are located at

h \pm 2\sqrt{3}

. Substituting one of these points into the vertex form

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

yields

0 = 12a - 6

, which gives

a = 1/2

.

Step-by-Step Solution

1

Determine the side length and height of the equilateral triangle from the given area.

The side length

s

is

4\sqrt{3}

, and the height of the triangle is

6

.

The area of an equilateral triangle with side length

s

is given by the formula

\text{Area} = \frac{s^2\sqrt{3}}{4}

. Setting this equal to

12\sqrt{3}

yields

s^2 = 48

, which means

s = \sqrt{48} = 4\sqrt{3}

. The height of the triangle is given by

h_{\text{tri}} = \frac{s\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6

.

2

Relate the height and side length of the triangle to the vertex and

x

-intercepts of the parabola.

The vertex of the parabola is

V(h, -6)

and the

x

-intercepts are at

x = h \pm 2\sqrt{3}

.

Because

a > 0

, the parabola opens upwards, meaning the vertex

V(h, k)

lies below the

x

-axis. Since the base of the triangle lies on the

x

-axis, the vertical distance from the vertex to the

x

-axis is the height of the triangle, so

k = -6

. The axis of symmetry

x = h

bisects the base

AB

, so the

x

-intercepts are located at a distance of half the side length,

\frac{s}{2} = 2\sqrt{3}

, to the left and right of

h

.

3

Write the quadratic function in vertex form and substitute one of the

x

-intercepts to solve for

a

.

a = \frac{1}{2}

The vertex form of the quadratic function is

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

. Substituting the

x

-intercept

(h + 2\sqrt{3}, 0)

into this equation gives

0 = a(h + 2\sqrt{3} - h)^2 - 6

, which simplifies to

0 = a(2\sqrt{3})^2 - 6 \implies 12a = 6 \implies a = \frac{1}{2}

.

Key Concept

Using the symmetry properties and vertex form of a quadratic function to relate the geometric features of its graph to its algebraic coefficients.

Alternative Method

Alternatively, you can use the relationship between the roots and coefficients. For any quadratic equation

ax^2 + bx + c = 0

with real roots, the distance between the roots is

\frac{\sqrt{b^2 - 4ac}}{|a|}

. Since the triangle is equilateral, the height is

\frac{\sqrt{3}}{2}

times this distance. The vertex is at

(-\frac{b}{2a}, -\frac{b^2-4ac}{4a})

. Setting the absolute value of the vertex

y

-coordinate equal to the height allows you to solve for

a

directly when combined with the area formula.

Estimated Time:3m 0s

Question 13Question →

If $\frac{8}{x - 3} = 2$ , what is the value of $x$ ?

Show answer & explanation

Answer: 7

Answer

7

To solve the equation

\frac{8}{x - 3} = 2

, we clear the denominator by multiplying both sides of the equation by

x - 3

. This gives

8 = 2(x - 3)

. Distributing the

2

results in

8 = 2x - 6

. Adding

6

to both sides yields

14 = 2x

. Dividing both sides by

2

gives the solution

x = 7

.

Step-by-Step Solution

1

Multiply both sides of the equation by

x - 3

to eliminate the fraction.

8 = 2(x - 3)

To clear the denominator and simplify the equation.

2

Distribute the 2 to the terms inside the parentheses.

8 = 2x - 6

To expand the right-hand side of the equation.

3

Add 6 to both sides of the equation.

14 = 2x

To isolate the term with the variable

x

.

4

Divide both sides by 2 to solve for

x

.

x = 7

To find the final value of the variable.

Key Concept

Solving a rational equation that simplifies to a linear equation by multiplying by the common denominator.

Question 14Question →

A sample of a radioactive isotope decays such that its mass, in grams, is modeled by the function $M(t) = A \cdot d^{\frac{t}{12}}$ , where $t$ is the time in hours since the measurement began, and $A$ and $d$ are positive constants. The mass of the sample decreases by $75\%$ every $24$ hours. If the mass of the sample is $15$ grams when $t = 36$ , what was the initial mass, in grams, of the sample?

Show answer & explanation

Answer: 120

Answer

The initial mass of the sample was 120 grams.

The correct answer is 120. A decrease of

75\%

over 24 hours means the mass at

t+24

is

0.25

times the mass at

t

. According to the function,

M(t+24) = A \cdot d^{(t+24)/12} = A \cdot d^{t/12} \cdot d^2 = M(t) \cdot d^2

. Equating the two yields

d^2 = 0.25

, so

d = 0.5

. Substituting

t = 36

and

M(36) = 15

gives

15 = A \cdot (0.5)^{36/12} = A \cdot (0.5)^3 = 0.125A

. Solving for

A

gives

A = 120

.

Step-by-Step Solution

1

Relate the 24-hour decay rate to the exponent in the function to set up an equation for

d

.

d^2 = 0.25

Every 24 hours (

t

increases by 24), the mass decreases by

75\%

, so it becomes

25\%

(

0.25

) of its previous value. The exponent increases by

24/12 = 2

, multiplying the mass by

d^2

.

2

Solve for the decay base

d

.

d = 0.5

Since

d

is a positive constant, taking the square root of

0.25

gives

0.5

.

3

Substitute the given mass at

t = 36

into the model to solve for the initial mass

A

.

A = 120

Plugging

t = 36

and

M(36) = 15

into

M(t) = A \cdot (0.5)^{t/12}

gives

15 = A \cdot (0.5)^3

, which simplifies to

15 = 0.125A

.

Key Concept

Interpreting and solving exponential decay functions with fractional exponents

Alternative Method

Instead of solving for

d

first, recognize that

36

hours is exactly

1.5

intervals of

24

hours. Since the mass is multiplied by

0.25

(or

\frac{1}{4}

) every

24

hours, after

36

hours it will be multiplied by

(\frac{1}{4})^{1.5} = (\frac{1}{4})^{3/2} = \frac{1}{8}

of its initial value. Therefore,

15 = A \cdot \frac{1}{8} \implies A = 120

.

Estimated Time:3m 0s

Question 15Question →

If the expression $\frac{3x^2 - bx + 10}{x - 2}$ is equivalent to $3x - c - \frac{12}{x - 2}$ for all $x \neq 2$ , where $b$ and $c$ are positive constants, what is the value of $b$ ?

Show answer & explanation

Answer: 17

Answer

17

Multiplying both sides of the equivalence by

x - 2

results in

3x^2 - bx + 10 = (3x - c)(x - 2) - 12

. Expanding the right side gives

3x^2 - (6 + c)x + (2c - 12)

. By equating the coefficients of corresponding terms, we get

2c - 12 = 10

, which solves to

c = 11

. Substituting this into the

x

-coefficient equivalence

-b = -(6 + c)

gives

b = 6 + 11 = 17

.

Step-by-Step Solution

1

Multiply both sides of the expression by the denominator

x - 2

to equate the numerators.

3x^2 - bx + 10 = (3x - c)(x - 2) - 12

Since the rational expressions are equivalent for all

x \neq 2

, their numerators must be equal when written over a common denominator.

2

Expand and simplify the right side of the equation.

3x^2 - bx + 10 = 3x^2 - (6 + c)x + (2c - 12)

Expanding

(3x - c)(x - 2)

yields

3x^2 - 6x - cx + 2c

, and subtracting

12

gives the simplified polynomial expression.

3

Equate the constant terms to solve for

c

.

10 = 2c - 12 \Rightarrow c = 11

For the polynomials to be equivalent, their constant terms must be equal.

4

Equate the coefficients of the

x

terms to solve for

b

.

b = 6 + c \Rightarrow b = 6 + 11 = 17

Equating the coefficients of

x

gives

-b = -(6 + c)

, which simplifies to

b = 6 + c

.

Key Concept

Equivalence of rational and polynomial expressions via coefficient comparison

Alternative Method

Alternatively, you can evaluate the equivalence at a convenient value of

x

. For instance, substituting

x = 0

into the expression gives

-5 = -c + 6

, yielding

c = 11

. Then, evaluating the equation at another convenient value such as

x = 1

allows you to solve for

b

directly using the now-known value of

c

.

Estimated Time:1m 30s

Question 16Question →

What is the set of all real solutions to the equation below?

\sqrt{3x+10} = x+2

\{-3, 2\}

\{2\}

\{-3\}

\{-2, 3\}

Show answer & explanation

Answer: $\{2\}$

Answer

The set containing only 2

The correct answer is the set containing only 2. Squaring both sides of the equation yields

3x + 10 = x^2 + 4x + 4

. Rearranging the terms to set the equation to zero gives

x^2 + x - 6 = 0

. Factoring the quadratic expression gives

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

, indicating potential solutions of

x = -3

and

x = 2

. Substituting

x = 2

back into the original equation gives

\sqrt{16} = 4

, which is valid. Substituting

x = -3

back into the original equation gives

\sqrt{1} = -1

, which is invalid because the principal square root is non-negative. Thus,

x = -3

is extraneous and must be discarded.

Step-by-Step Solution

1

Square both sides of the equation to eliminate the radical.

3x + 10 = (x + 2)^2

, which expands to

3x + 10 = x^2 + 4x + 4

.

Squaring both sides is the standard algebraic method to isolate the term under a square root.

2

Move all terms to one side to form a quadratic equation equal to zero.

x^2 + x - 6 = 0

.

Setting the quadratic expression to zero allows us to solve it by factoring.

3

Factor the quadratic equation.

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

, which gives the potential solutions

x = -3

and

x = 2

.

Finding the factors helps us determine the roots of the equation.

4

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

For

x = 2

:

\sqrt{3(2)+10} = \sqrt{16} = 4

and

2+2 = 4

, which is true. For

x = -3

:

\sqrt{3(-3)+10} = \sqrt{1} = 1

and

-3+2 = -1

, which is false.

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

Key Concept

Identifying extraneous solutions in radical equations

Estimated Time:1m 30s

Question 17Question →

If $x > 2$ and satisfies the equation

x - \frac{8}{x^2 - 2x} = 5 - \frac{4}{x - 2}

what is the value of $x$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

To solve the rational equation, we first determine the domain restrictions:

x \neq 0

and

x \neq 2

. We then clear the denominators by multiplying the entire equation by the least common denominator

x(x - 2)

, which simplifies the equation to the cubic form

x^3 - 7x^2 + 14x - 8 = 0

. Factoring the cubic polynomial yields

(x - 1)(x - 2)(x - 4) = 0

. The potential solutions are

x = 1

,

x = 2

, and

x = 4

. Since

x = 2

is an extraneous solution and the constraint requires

x > 2

,

x = 4

is the only valid solution.

Step-by-Step Solution

1

Determine the domain restrictions of the rational terms in the equation.

x \neq 0

and

x \neq 2

Division by zero is undefined, so the denominators

x(x-2)

and

x-2

cannot be zero.

2

Multiply the entire equation by the least common denominator

x(x - 2)

to eliminate the fractions.

x^2(x - 2) - 8 = 5x(x - 2) - 4x

Multiplying by the common denominator converts the rational equation into an equivalent polynomial equation within the defined domain.

3

Expand the terms and collect them on one side to form a standard polynomial equation.

x^3 - 7x^2 + 14x - 8 = 0

Simplifying the expressions allows us to find the roots of the polynomial.

4

Factor the cubic polynomial using the rational root theorem or synthetic division.

(x - 1)(x - 2)(x - 4) = 0

Since

x = 1

makes the polynomial zero, we factor out

(x-1)

to get the remaining quadratic factor

x^2 - 6x + 8

, which factors into

(x - 2)(x - 4)

.

5

Identify the valid solution based on the domain restriction and the given inequality constraint.

x = 4

The value

x = 2

is extraneous because it makes the original equation undefined. The value

x = 1

is rejected because the problem specifies

x > 2

.

Key Concept

Solving rational equations by clearing denominators, factoring polynomials, identifying extraneous solutions, and applying inequality constraints.

Question 18Question →

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

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

Show answer & explanation

Answer: 6

Answer

The only real solution to the equation is 6.

Substituting

x = 6

into the original equation yields

\sqrt{3(6) - 2} - \sqrt{6 + 3} = \sqrt{16} - \sqrt{9} = 4 - 3 = 1

, which is a true statement. Therefore, the only real solution is 6.

Step-by-Step Solution

1

Isolate one of the radical terms

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

Preparing the equation to square both sides.

2

Square both sides of the equation

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

Eliminating one of the square root radicals.

3

Isolate the remaining radical term and simplify

x - 3 = \sqrt{x + 3}

Simplifying the equation by isolating the second radical and dividing both sides by 2.

4

Square both sides again to eliminate the second radical

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

Converting the radical equation into a polynomial equation.

5

Write in standard quadratic form and factor

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

Setting the quadratic equation to zero and factoring to find potential solutions.

6

Verify solutions in the original equation to check for extraneous solutions

x = 6

is valid;

x = 1

is extraneous

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

Key Concept

Solving equations containing radical expressions and identifying extraneous solutions.

Estimated Time:2m 30s

Question 19Question →

The graph of the function $y = f(x)$ in the $xy$ -plane passes through the point $(3, -4)$ . If the function $g$ is defined by $g(x) = f(x + 2) - 1$ , which of the following points must lie on the graph of $y = g(x)$ ?

(1, -5)

(5, -5)

(1, -3)

(5, -3)

Show answer & explanation

Answer: $(1, -5)$

Answer

(1, -5)

The correct answer is the point

(1, -5)

. Since the point

(3, -4)

lies on the graph of

f

, we know that

f(3) = -4

. For the function

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

, substituting

x = 1

gives

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

. Substituting

-4

for

f(3)

yields

g(1) = -4 - 1 = -5

. Thus, the point

(1, -5)

must lie on the graph of

g

.

Step-by-Step Solution

1

Identify the given function value from the point on the graph of

f

.

Since the graph of

y = f(x)

passes through

(3, -4)

, we have

f(3) = -4

.

Any point

(x, y)

on the graph of a function satisfies the equation

y = f(x)

.

2

Determine the input variable

x

for the function

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

that corresponds to the known input of

3

for

f

.

Set the argument of

f

in

g(x)

equal to

3

:

x + 2 = 3

, which simplifies to

x = 1

.

This allows us to substitute the known value

f(3)

into the expression for

g(x)

.

3

Evaluate

g(1)

using the value of

f(3)

.

g(1) = f(1 + 2) - 1 = f(3) - 1 = -4 - 1 = -5

.

Substituting

f(3) = -4

into the simplified expression gives the output value of

g

at

x = 1

.

4

State the resulting point on the graph of

y = g(x)

.

The point is

(1, -5)

.

An input of

x = 1

yields an output of

y = -5

for the function

g

.

Key Concept

Function transformations and their effects on individual coordinate points.

Question 20Question →

The amount of a radioactive isotope remaining after $t$ days is modeled by the function $A(t) = A_0(0.64)^{\frac{t}{2}}$ , where $A_0$ is the initial amount of the isotope. If the function is rewritten in the form $A(t) = A_0(1 - r)^t$ , where $r$ is the daily decay rate, what is the value of $r$ ?

0.18

0.20

0.36

0.59

Show answer & explanation

Answer: 0.20

Answer

The daily decay rate

r

is

0.20

.

The correct value of

r

is

0.20

. By rewriting the given expression using exponent rules, we obtain

(0.64)^{t/2} = ((0.64)^{1/2})^t = (0.8)^t

. Setting the base equal to the target decay rate form gives

1 - r = 0.8

, which simplifies to

r = 0.20

.

Step-by-Step Solution

1

Apply the power of a power exponent rule,

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

, to rewrite the exponential term.

A(t) = A_0\left(0.64^{1/2}\right)^t

This isolates the variable

t

as the exponent, allowing us to find the daily decay factor.

2

Evaluate the base

0.64^{1/2}

.

\sqrt{0.64} = 0.8

, so the expression becomes

A(t) = A_0(0.8)^t

.

Calculating the square root of

0.64

gives the daily decay factor of

0.8

.

3

Equate the daily decay factor

0.8

to the target form

1 - r

and solve for

r

.

1 - r = 0.8 \implies r = 0.20

Solving the equation gives the value of the daily decay rate.

Key Concept

Rewriting exponential equations by manipulating bases and exponents

Alternative Method

We can solve for

r

by choosing a specific value for

t

, such as

t = 2

. After 2 days, the remaining amount is

A(2) = A_0(0.64)^1 = 0.64A_0

. Using the target equation form, the remaining amount after 2 days is

A_0(1 - r)^2

. Setting the two expressions equal to each other gives

A_0(1 - r)^2 = 0.64A_0

. Dividing both sides by

A_0

gives

(1 - r)^2 = 0.64

. Taking the square root of both sides gives

1 - r = 0.8

, which yields

r = 0.20

.

Estimated Time:1m 30s

Advanced Math

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