Advanced Math Practice Questions — SAT — Page 4

Question 61Question →

For all $x \neq -2.5$ , the expression $\frac{6x^2 + 17x + 8}{2x + 5}$ is equivalent to $3x + 1 + \frac{a}{2x + 5}$ , where $a$ is a constant. What is the value of $a$ ?

Show answer & explanation

Answer: 3

Answer

The value of

a

is

3

.

Multiplying both sides of the equivalent relation by

2x + 5

yields

6x^2 + 17x + 8 = (3x + 1)(2x + 5) + a

. Expanding the right side gives

6x^2 + 17x + 5 + a

. Since the expressions are equivalent, the constant terms must be equal:

8 = 5 + a

, which simplifies to

a = 3

.

Step-by-Step Solution

1

Multiply both sides of the equation by

2x + 5

to clear the denominators.

6x^2 + 17x + 8 = (3x + 1)(2x + 5) + a

This clears the rational expressions so we can work with polynomials directly.

2

Expand the expression

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

using the distributive property.

6x^2 + 17x + 8 = 6x^2 + 15x + 2x + 5 + a

To write the right side as a polynomial in standard form.

3

Combine like terms on the right side of the equation.

6x^2 + 17x + 8 = 6x^2 + 17x + (5 + a)

To group coefficients of like powers of

x

for easy comparison.

4

Equate the constant terms from both sides of the equation to solve for

a

.

8 = 5 + a

, which gives

a = 3

.

For the two polynomial expressions to be equivalent for all values of

x

, their corresponding coefficients and constants must be equal.

Key Concept

Rewriting rational expressions by clearing denominators or polynomial long division to find equivalent expressions.

Question 62Question →

In the system of equations below, $k$ is a constant.

y = 3x^2 - 5x + 4

y = 7x + k

If the system has exactly one real solution $(x, y)$ , what is the value of $k$ ?

8

-8

16

40

Show answer & explanation

Answer: $-8$

Answer

-8

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

3x^2 - 5x + 4 = 7x + k

. Subtracting

7x

and

k

from both sides gives the standard quadratic equation

3x^2 - 12x + (4 - k) = 0

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

b^2 - 4ac

, must equal

0

. Substituting

a = 3

,

b = -12

, and

c = 4 - k

into the discriminant formula gives

(-12)^2 - 4(3)(4 - k) = 0

. Simplifying this equation yields

144 - 12(4 - k) = 0

, which simplifies to

144 - 48 + 12k = 0

, or

96 + 12k = 0

. Solving for

k

gives

k = -8

.

Step-by-Step Solution

1

Equate the two equations to eliminate

y

and form a single quadratic equation.

3x^2 - 5x + 4 = 7x + k \Rightarrow 3x^2 - 12x + (4 - k) = 0

To find the points of intersection of the system, we set the expressions for

y

equal to each other.

2

Identify the coefficients of the quadratic equation in standard form

ax^2 + bx + c = 0

.

a = 3

,

b = -12

, and

c = 4 - k

These coefficients are needed to calculate the discriminant of the quadratic equation.

3

Set the discriminant

\Delta = b^2 - 4ac

equal to zero and solve for

k

.

(-12)^2 - 4(3)(4 - k) = 0 \Rightarrow 144 - 12(4 - k) = 0 \Rightarrow 12(4 - k) = 144 \Rightarrow 4 - k = 12 \Rightarrow k = -8

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

Key Concept

Solving nonlinear systems of equations by setting them equal to each other and using the discriminant of the resulting quadratic equation to determine the number of solutions.

Estimated Time:1m 30s

Question 63Question →

What are the solutions to the quadratic equation $x^2 - 4x - 12 = 0$ ?

x = -6

and

x = 2

x = 6

and

x = -2

x = 6

and

x = 2

x = -6

and

x = -2

Show answer & explanation

Answer: $x = 6$ and $x = -2$

Answer

The solutions are

x = 6

and

x = -2

.

The correct answer is

x = 6

and

x = -2

. Factoring the quadratic equation

x^2 - 4x - 12 = 0

involves finding two numbers whose product is

-12

and whose sum is

-4

. These numbers are

-6

and

2

. Rewriting the equation as

(x - 6)(x + 2) = 0

and setting each factor to zero yields the solutions

x = 6

and

x = -2

.

Step-by-Step Solution

1

Identify the quadratic equation to factor.

The equation is

x^2 - 4x - 12 = 0

.

To find the solutions, we need to rewrite the quadratic expression in factored form.

2

Find two numbers that multiply to

-12

and add up to

-4

.

The numbers are

-6

and

2

.

These numbers satisfy both conditions:

(-6) \times 2 = -12

and

-6 + 2 = -4

.

3

Write the quadratic equation in factored form and solve for

x

.

(x - 6)(x + 2) = 0

, which means

x - 6 = 0

or

x + 2 = 0

. Thus,

x = 6

or

x = -2

.

Applying the zero product property allows us to solve for the individual values of

x

.

Key Concept

Solving quadratic equations by factoring

Question 64Question →

A linear equation and a quadratic equation form a system, as shown.

y = x^2 + 5

y = 3x + 5

If the ordered pair $(x, y)$ is a solution to the system where $x$ is positive, what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

Equating the two expressions for

y

gives the equation

x^2 + 5 = 3x + 5

. Subtracting 5 from both sides results in

x^2 = 3x

. Subtracting

3x

from both sides gives the quadratic equation

x^2 - 3x = 0

, which can be factored as

x(x - 3) = 0

. This yields two solutions for

x

:

0

and

3

. Since the question specifies that

x

is positive, the correct value is 3.

Step-by-Step Solution

1

Equate the equations

x^2 + 5 = 3x + 5

Since both equations are solved for

y

, we can substitute the quadratic expression into the linear equation.

2

Simplify the equation

x^2 - 3x = 0

Subtracting 5 from both sides and then subtracting

3x

from both sides collects all terms on one side.

3

Factor and solve

x(x - 3) = 0

, so

x = 0

or

x = 3

Factoring out the greatest common factor,

x

, allows us to apply the zero product property to find the individual roots.

4

Apply the constraint

x = 3

The problem states that

x

must be positive, which excludes the solution

x = 0

.

Key Concept

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

Question 65Question →

If the expression $\frac{10x^2 + x - 3}{2x - 1}$ is equivalent to $ax + b$ for all $x \neq 0.5$ , where $a$ and $b$ are constants, what is the value of $a + b$ ?

Show answer & explanation

Answer: 8

Answer

The value of

a + b

is 8.

To find the equivalent expression, factor the numerator:

10x^2 + x - 3 = (2x - 1)(5x + 3)

. For all

x \neq 0.5

, the denominator

2x - 1

is non-zero, so we can divide out the common factor

(2x - 1)

to get

5x + 3

. Matching

5x + 3

with

ax + b

gives

a = 5

and

b = 3

. The sum of these constants is

5 + 3 = 8

.

Step-by-Step Solution

1

Factor the numerator of the expression

10x^2 + x - 3 = (2x - 1)(5x + 3)

Factoring the quadratic trinomial allows us to identify common factors that can be simplified.

2

Simplify the rational expression

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

(for

x \neq 0.5

)

Since

x \neq 0.5

, the term

2x - 1

is non-zero and can be canceled from both the numerator and the denominator.

3

Identify the values of

a

and

b

a = 5

and

b = 3

By comparing the simplified expression

5x + 3

to the form

ax + b

, the coefficients of corresponding terms must be equal.

4

Calculate the sum of

a

and

b

5 + 3 = 8

The question asks for the sum of the constants

a

and

b

.

Key Concept

Simplifying rational expressions by factoring and coefficient matching

Question 66Question →

For $x > 1$ , which of the following expressions is equivalent to $\frac{x - x^{-1}}{x^{1/2} - x^{-1/2}}$ ?

x^{1/2} - x^{-1/2}

x + 1

x^{1/2} + x^{-1/2}

x^{3/2} + x^{1/2}

Show answer & explanation

Answer: $x^{1/2} + x^{-1/2}$

Answer

x^{1/2} + x^{-1/2}

The expression

x^{1/2} + x^{-1/2}

is correct because the numerator

x - x^{-1}

can be factored using the difference of squares identity,

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

, where

a = x^{1/2}

and

b = x^{-1/2}

. This yields

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

. Substituting this back into the original fraction and canceling the common factor of

x^{1/2} - x^{-1/2}

in both the numerator and denominator simplifies the expression directly to

x^{1/2} + x^{-1/2}

.

Step-by-Step Solution

1

Recognize the numerator

x - x^{-1}

as a difference of squares in terms of the base variables

x^{1/2}

and

x^{-1/2}

.

Write

x - x^{-1}

as

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

.

Since the denominator is

x^{1/2} - x^{-1/2}

, factoring the numerator as a difference of squares allows us to identify a common factor that can be canceled.

2

Substitute the factored numerator back into the original expression and cancel the common factor of

x^{1/2} - x^{-1/2}

from the numerator and the denominator.

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

For all

x > 1

, the term

x^{1/2} - x^{-1/2}

is non-zero, so we can divide both the numerator and the denominator by this common term to simplify the expression.

Key Concept

Factoring algebraic expressions using the difference of squares identity with fractional exponents.

Alternative Method

Convert the fractional and negative exponents into standard algebraic fractions:

\frac{x - \frac{1}{x}}{\sqrt{x} - \frac{1}{\sqrt{x}}}

. Find common denominators for both the numerator and denominator to write the expression as

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

. Next, multiply the numerator by the reciprocal of the denominator:

\frac{(x - 1)(x + 1)}{x} \cdot \frac{\sqrt{x}}{x - 1}

. Cancel the common factor of

x - 1

to get

\frac{(x + 1)\sqrt{x}}{x}

. Distributing

\sqrt{x}

and dividing each term by

x

gives

\frac{x\sqrt{x}}{x} + \frac{\sqrt{x}}{x} = \sqrt{x} + \frac{1}{\sqrt{x}}

, which is equivalent to

x^{1/2} + x^{-1/2}

.

Estimated Time:1m 30s

Question 67Question →

In the $xy$ -plane, the line $y = mx$ , where $m$ is a positive constant, is tangent to the parabola $y = x^2 + 9$ . What is the value of $m$ ?

3

6

12

36

Show answer & explanation

Answer: 6

Answer

The value of

m

is 6.

Equating the equations of the line and the parabola yields the quadratic equation

x^2 - mx + 9 = 0

. For the line to be tangent to the parabola, this system must have exactly one real solution, meaning the discriminant

b^2 - 4ac

must equal

0

. Substituting

a = 1

,

b = -m

, and

c = 9

into the discriminant formula gives

(-m)^2 - 4(1)(9) = 0

, which simplifies to

m^2 - 36 = 0

. Solving for the positive constant

m

gives

6

.

Step-by-Step Solution

1

Set the equation of the line equal to the equation of the parabola to find their intersection points.

mx = x^2 + 9

The intersection points of the system correspond to the values of

x

where both equations have the same

y

-value.

2

Rewrite the equation in standard quadratic form

ax^2 + bx + c = 0

by subtracting

mx

from both sides.

x^2 - mx + 9 = 0

This allows the identification of the coefficients

a

,

b

, and

c

to compute the discriminant.

3

Identify the coefficients and set the discriminant

b^2 - 4ac

equal to 0.

(-m)^2 - 4(1)(9) = 0

A line is tangent to a parabola if and only if the system has exactly one real solution, which corresponds to a discriminant of zero.

4

Solve the equation for the positive constant

m

.

m^2 - 36 = 0 \implies m^2 = 36 \implies m = 6

Taking the square root of both sides gives

m = \pm 6

. Since

m

is specified as a positive constant, we choose

m = 6

.

Key Concept

Solving nonlinear systems of equations where a line is tangent to a parabola by setting the discriminant of the resulting quadratic equation to zero.

Question 68Question →

A circle and a line are graphed in the $xy$ -plane. The circle is defined by the equation $x^2 + y^2 = 25$ , and the line is defined by the equation $y = 3$ . If the line intersects the circle at the point $(x, 3)$ , where $x > 0$ , what is the value of $x$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

Substituting

y = 3

into the circle equation

x^2 + y^2 = 25

yields

x^2 + 3^2 = 25

. Simplifying this gives

x^2 + 9 = 25

, which simplifies to

x^2 = 16

. Taking the square root of both sides gives

x = 4

or

x = -4

. Since it is given that

x > 0

, the value of

x

must be

4

.

Step-by-Step Solution

1

Substitute the value of

y = 3

into the circle's equation.

x^2 + 3^2 = 25

Since the line is

y = 3

, any point of intersection must satisfy this y-coordinate. Substituting it into the circle's equation allows us to solve for the x-coordinate.

2

Simplify the equation and isolate

x^2

.

x^2 = 16

Squaring

3

gives

9

, and subtracting

9

from both sides of the equation

x^2 + 9 = 25

isolates

x^2

.

3

Solve for

x

and apply the constraint

x > 0

.

x = 4

Taking the square root of both sides of

x^2 = 16

gives

x = 4

or

x = -4

. Since the problem states

x > 0

, the only valid solution is

4

.

Key Concept

Solving systems of nonlinear equations using substitution

Question 69Question →

If the expression $\frac{4x^2 + kx - 7}{2x - 1}$ is equivalent to $2x + 5 - \frac{2}{2x - 1}$ for all $x \neq \frac{1}{2}$ , where $k$ is a constant, what is the value of $k$ ?

Show answer & explanation

Answer: 8

Answer

8

The correct answer is 8. Combining the expression

2x + 5 - \frac{2}{2x - 1}

into a single fraction requires finding a common denominator of

2x - 1

. Multiplying the linear term

2x + 5

by

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

and subtracting 2 yields

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

. Expanding and simplifying the numerator gives

4x^2 + 8x - 7

. Equating this to the numerator of the original expression,

4x^2 + kx - 7

, shows that the coefficient of the linear term,

k

, must be equal to 8.

Step-by-Step Solution

1

Multiply the linear expression by the denominator to prepare for combining the terms.

(2x + 5)(2x - 1) = 4x^2 + 8x - 5

To combine all terms under a single common denominator of

2x - 1

, the non-fractional terms must be multiplied by the denominator.

2

Subtract the numerator of the fractional term from the expanded product.

(4x^2 + 8x - 5) - 2 = 4x^2 + 8x - 7

This completes the subtraction of the fraction over the common denominator, resulting in a single rational expression.

3

Compare the resulting numerator to the numerator of the original expression to find the value of the constant.

k = 8

For the two rational expressions to be equivalent, their numerators must be equal for all values of

x

. Thus, the coefficient of

x

in both expressions must match.

Key Concept

Equivalence of rational expressions through finding a common denominator

Question 70Question →

A system of equations consists of the equations $y = -x^2 + 2x + 7$ and $y = 6x + k$ , where $k$ is a constant. If the system has two distinct real solutions, what is the greatest integer value of $k$ ?

Show answer & explanation

Answer: 10

Answer

The correct answer is 10. The greatest integer value of the constant that allows the system to have two distinct real solutions is 10.

To find the number of solutions to the system, equate the two equations:

-x^2 + 2x + 7 = 6x + k

. Rearranging this equation into standard quadratic form gives

x^2 + 4x + (k - 7) = 0

. For the system to have two distinct real solutions, the discriminant of this quadratic equation must be strictly greater than zero. The discriminant is calculated as

b^2 - 4ac = 4^2 - 4(1)(k - 7) = 16 - 4k + 28 = 44 - 4k

. Setting this greater than zero yields

44 - 4k > 0

, which simplifies to

k < 11

. The greatest integer value of

k

that is strictly less than 11 is 10.

Step-by-Step Solution

1

Equate the expressions for

y

from both equations.

-x^2 + 2x + 7 = 6x + k

At the points of intersection, the

y

-values of both equations must be equal.

2

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 + 4x + (k - 7) = 0

Standard form is required to calculate the discriminant of the quadratic equation.

3

Write the expression for the discriminant

\Delta = b^2 - 4ac

using the coefficients from the quadratic equation.

\Delta = 4^2 - 4(1)(k - 7) = 44 - 4k

The discriminant determines the number of real solutions to the quadratic equation.

4

Set the discriminant to be strictly greater than 0 and solve the inequality for

k

.

44 - 4k > 0 \implies k < 11

For the system to have two distinct real solutions, the discriminant must be positive.

5

Determine the greatest integer value of

k

that satisfies the inequality

k < 11

.

10

The largest integer strictly less than 11 is 10.

Key Concept

Using the discriminant of a quadratic equation derived from a nonlinear system to determine the number of real solutions.

Question 71Question →

\begin{aligned} y &= x^2 - 7 \\ y &= 2x + 1 \end{aligned}

If $(x, y)$ is a solution to the system of equations above and $y > 0$ , what is the value of $y$ ?

4

5

-3

9

Show answer & explanation

Answer: 9

Answer

9

Substituting the expression for

y

from the second equation into the first equation yields

2x + 1 = x^2 - 7

. Subtracting

2x

and

1

from both sides results in the standard form quadratic equation

x^2 - 2x - 8 = 0

. Factoring this equation gives

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

, which yields the solutions

x = 4

and

x = -2

. Substituting these

x

-values back into the linear equation

y = 2x + 1

gives the corresponding

y

-values:

y = 2(4) + 1 = 9

and

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

. Since the system specifies the constraint

y > 0

, the correct value of

y

must be

9

.

Step-by-Step Solution

1

Substitute the expression for

y

from the linear equation into the quadratic equation.

2x + 1 = x^2 - 7

This eliminates the variable

y

and leaves a single quadratic equation in terms of

x

.

2

Rearrange the quadratic equation into standard form

ax^2 + bx + c = 0

.

x^2 - 2x - 8 = 0

Moving all terms to one side allows us to solve the quadratic equation by factoring.

3

Factor the quadratic equation.

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

, which gives

x = 4

or

x = -2

.

Finding the roots of the quadratic equation provides the possible

x

-coordinates of the solution points.

4

Substitute the

x

-values back into the linear equation

y = 2x + 1

to find the corresponding

y

-values.

For

x = 4

,

y = 2(4) + 1 = 9

. For

x = -2

,

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

.

This gives the complete coordinate pairs

(4, 9)

and

(-2, -3)

for the system's solutions.

5

Apply the constraint

y > 0

to identify the correct solution.

Since

9 > 0

and

-3 < 0

, the correct value of

y

is

9

.

Only the solution

(4, 9)

satisfies the condition that

y

must be positive.

Key Concept

Solving linear-quadratic systems of equations by substitution and evaluating solutions under given constraints.

Question 72Question →

For all $x > 0$ , which of the following expressions is equivalent to $\frac{x^2 - 8}{x^{4/3} + 2x^{2/3} + 4} + \frac{x - 4x^{1/3}}{x^{2/3} + 2x^{1/3}}$ ?

x^{2/3} + x^{1/3} + 4

x^{4/3} + x^{2/3} - 4

x^{2/3} + x^{1/3} - 4

x^{2/3} + x^{1/3} - 6

Show answer & explanation

Answer: $x^{2/3} + x^{1/3} - 4$

Answer

The expression

x^{2/3} + x^{1/3} - 4

The correct answer shows the sum of the simplified terms, which is

x^{2/3} + x^{1/3} - 4

. First, the numerator of the first term,

x^2 - 8

, can be written as a difference of cubes:

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

. Dividing by the denominator leaves

x^{2/3} - 2

. Second, the second term can be factored by extracting

x^{1/3}

from both the numerator and denominator, leaving

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

. Factoring the numerator as a difference of squares,

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

, and dividing by the denominator leaves

x^{1/3} - 2

. Adding the two simplified parts,

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

, gives

x^{2/3} + x^{1/3} - 4

.

Step-by-Step Solution

1

Simplify the first term,

\frac{x^2 - 8}{x^{4/3} + 2x^{2/3} + 4}

x^{2/3} - 2

Rewrite

x^2 - 8

as

(x^{2/3})^3 - 2^3

and expand using the difference of cubes identity:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

, where

a = x^{2/3}

and

b = 2

. The term

x^{4/3} + 2x^{2/3} + 4

in the numerator and denominator cancels out.

2

Simplify the second term,

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

x^{1/3} - 2

Factor out

x^{1/3}

from both the numerator and denominator to get

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

. Then rewrite

x^{2/3} - 4

as

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

and expand using the difference of squares identity:

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

, where

a = x^{1/3}

and

b = 2

. The term

x^{1/3} + 2

in the numerator and denominator cancels out.

3

Sum the two simplified terms

x^{2/3} + x^{1/3} - 4

Add the two simplified expressions:

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

.

Key Concept

Simplification of rational expressions involving fractional exponents, difference of cubes, and difference of squares.

Estimated Time:2m 0s

Question 73Question →

For all $x > 0$ , the expression $\left( \frac{x^{4/3} + 4x^{2/3} + 16}{x^{2/3} + 2x^{1/3} + 4} + 2x^{1/3} \right)^3 - x^2 - 12x^{4/3} - 48x^{2/3}$ is equivalent to a constant $C$ . What is the value of $C$ ?

Show answer & explanation

Answer: 64

Answer

The constant value is 64.

The expression inside the parentheses simplifies to

x^{2/3} + 4

after factoring the numerator as

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

and canceling the common factor in the denominator. Cubing

x^{2/3} + 4

yields

x^2 + 12x^{4/3} + 48x^{2/3} + 64

. Subtracting the remaining terms

x^2 + 12x^{4/3} + 48x^{2/3}

from this expansion results in the constant value 64.

Step-by-Step Solution

1

Substitute

u = x^{1/3}

into the expression to simplify the fractional exponents.

The terms become

x^{1/3} = u

,

x^{2/3} = u^2

,

x^{4/3} = u^4

, and

x^2 = u^6

. The expression inside the parentheses is rewritten as

\frac{u^4 + 4u^2 + 16}{u^2 + 2u + 4} + 2u

.

Using a temporary variable

u

simplifies the algebraic factoring and manipulation of terms with fractional exponents.

2

Factor the numerator

u^4 + 4u^2 + 16

by completing the square.

u^4 + 4u^2 + 16 = (u^2 + 4)^2 - 4u^2 = (u^2 + 2u + 4)(u^2 - 2u + 4)

.

Expressing the quartic polynomial as a difference of squares allows it to be factored into two quadratic polynomials.

3

Simplify the rational expression and add

2u

.

\frac{(u^2 + 2u + 4)(u^2 - 2u + 4)}{u^2 + 2u + 4} + 2u = (u^2 - 2u + 4) + 2u = u^2 + 4

.

Canceling the common factor

u^2 + 2u + 4

in the numerator and denominator simplifies the expression inside the parentheses to

u^2 + 4

.

4

Substitute

u = x^{1/3}

back into

u^2 + 4

and cube the expression.

(x^{2/3} + 4)^3 = (x^{2/3})^3 + 3(x^{2/3})^2(4) + 3(x^{2/3})(16) + 64 = x^2 + 12x^{4/3} + 48x^{2/3} + 64

.

Applying the binomial expansion formula

(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3

expands the cubed expression.

5

Subtract the remaining terms from the expanded expression.

(x^2 + 12x^{4/3} + 48x^{2/3} + 64) - x^2 - 12x^{4/3} - 48x^{2/3} = 64

.

Subtracting the variable terms cancels them out entirely, leaving the constant value 64.

Key Concept

Factoring quartic polynomials using the difference of squares and simplifying rational expressions with fractional exponents.

Question 74Question →

If $3^{x - 2} = 81$ , what is the value of $x$ ?

2

4

6

29

Show answer & explanation

Answer: 6

Answer

6

To solve the equation

3^{x - 2} = 81

, we first express the number 81 as a power of 3, which is

3^4

. This gives us

3^{x - 2} = 3^4

. Since the bases are equal, we can set their exponents equal to each other, resulting in the equation

x - 2 = 4

. Solving for

x

by adding 2 to both sides gives the correct value of 6.

Step-by-Step Solution

1

Express both sides of the equation with a common base of 3.

3^{x - 2} = 3^4

Since 81 is equal to

3 \times 3 \times 3 \times 3

, it can be written as

3^4

.

2

Set the exponents equal to each other because the bases are now the same.

x - 2 = 4

If

b^y = b^z

for a positive base

b \neq 1

, then

y = z

.

3

Solve the linear equation for

x

by adding 2 to both sides of the equation.

x = 6

Adding 2 to both sides isolates the variable

x

.

Key Concept

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

Question 75Question →

A circle in the $xy$ -plane is defined by the equation $(x - 2)^2 + (y + 1)^2 = 10$ . The line $y = 3x + k$ , where $k$ is a constant, is tangent to the circle. If $k < 0$ , what is the value of $k$ ?

-3

-5

-17

3

Show answer & explanation

Answer: -17

Answer

The correct value of

k

is

-17

.

Substituting the line equation

y = 3x + k

into the circle equation yields

(x - 2)^2 + (3x + k + 1)^2 = 10

. Expanding and writing this in standard form gives

10x^2 + (6k + 2)x + (k^2 + 2k - 5) = 0

. For the line to be tangent to the circle, the quadratic equation must have exactly one real solution, meaning its discriminant must be

0

. Setting the discriminant

\Delta = (6k+2)^2 - 4(10)(k^2 + 2k - 5)

to

0

and simplifying gives

-4k^2 - 56k + 204 = 0

. Dividing by

-4

yields

k^2 + 14k - 51 = 0

, which factors as

(k+17)(k-3)=0

. Since

k < 0

, the value of

k

must be

-17

.

Step-by-Step Solution

1

Substitute the linear equation

y = 3x + k

into the circle equation.

(x - 2)^2 + (3x + k + 1)^2 = 10

To find the coordinates where the line and the circle intersect.

2

Expand both squared terms and simplify the equation to standard quadratic form

Ax^2 + Bx + C = 0

.

10x^2 + (6k + 2)x + (k^2 + 2k - 5) = 0

To write the system as a single quadratic equation in terms of

x

.

3

Set the discriminant of the quadratic equation to zero.

(6k + 2)^2 - 4(10)(k^2 + 2k - 5) = 0

which simplifies to

-4k^2 - 56k + 204 = 0

Since the line is tangent to the circle, there must be exactly one intersection point, which means the quadratic equation must have exactly one real solution.

4

Divide the simplified equation by

-4

and solve for

k

.

k^2 + 14k - 51 = 0 \implies (k + 17)(k - 3) = 0 \implies k = -17 \text{ or } k = 3

To find the values of

k

that make the line tangent to the circle.

5

Apply the given constraint

k < 0

.

k = -17

The problem specifies that

k

must be a negative value.

Key Concept

Solving nonlinear systems of equations involving circles and lines by substitution and using the discriminant to determine tangency.

Estimated Time:2m 30s

Question 76Question →

For all $x > 2$ , which of the following is equivalent to the expression $\frac{3x^2 - 5x - 2}{x^2 - 4} - \frac{x - 1}{x + 2}$ ?

\frac{2x}{x + 2}

\frac{2x + 2}{x + 2}

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

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

Show answer & explanation

Answer: $\frac{2x + 2}{x + 2}$

Answer

The expression is equivalent to

\frac{2x + 2}{x + 2}

.

The correct answer is obtained by first factoring the first term:

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

. Canceling the common factor of

x - 2

yields

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

. Subtracting the second term gives

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

.

Step-by-Step Solution

1

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

The numerator factors as

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

, and the denominator factors as

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

.

This allows common factors in the numerator and denominator to be identified and canceled.

2

Simplify the first term by canceling the common factor

(x - 2)

for

x > 2

.

The first term simplifies to

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

.

For

x > 2

,

x - 2 \neq 0

, so we can divide both numerator and denominator by

x - 2

to simplify the fraction.

3

Subtract the second term from the simplified first term.

The expression becomes

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

.

Since both fractions have the same denominator,

x + 2

, their numerators can be subtracted directly.

4

Distribute the negative sign in the numerator and combine like terms.

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

.

Distributing the subtraction to both terms in the parenthesis

(x - 1)

gives

-x + 1

. Combining

3x - x

yields

2x

, and

1 + 1

yields

2

.

Key Concept

Simplifying rational expressions by factoring and performing algebraic operations with common denominators.

Estimated Time:1m 30s

Question 77Question →

In the quadratic equation $x^2 - kx + 9 = 0$ , $k$ is a positive constant. If the equation has exactly one real solution, what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

6

For a quadratic equation

ax^2 + bx + c = 0

to have exactly one real solution, its discriminant must equal zero (

b^2 - 4ac = 0

). In the equation

x^2 - kx + 9 = 0

, the coefficients are

a = 1

,

b = -k

, and

c = 9

. Setting the discriminant to zero gives

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

, which simplifies to

k^2 - 36 = 0

. Solving this equation yields

k^2 = 36

, so

k = 6

or

k = -6

. Since

k

is a positive constant, we reject the negative root, leaving

k = 6

.

Step-by-Step Solution

1

Identify the condition for a quadratic equation to have exactly one real solution.

The discriminant of the quadratic equation must be equal to 0, which is represented by the formula

b^2 - 4ac = 0

.

The discriminant determines the number of real solutions of a quadratic equation. If the discriminant is 0, there is exactly one real solution.

2

Identify the coefficients

a

,

b

, and

c

of the given equation

x^2 - kx + 9 = 0

.

a = 1

,

b = -k

, and

c = 9

.

These coefficients are required to compute the value of the discriminant.

3

Set the discriminant equal to 0 and simplify the equation.

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

.

Substituting the coefficients into the discriminant formula sets up the algebraic relationship to find

k

.

4

Solve for the positive constant

k

.

k^2 = 36 \Rightarrow k = 6

or

k = -6

. Since

k

must be positive,

k = 6

.

Solving the equation gives two possible values, but the negative solution is discarded because the problem specifies

k

is a positive constant.

Key Concept

Discriminant of a quadratic equation

Alternative Method

Alternatively, a quadratic equation has exactly one real solution if it can be written as a perfect square trinomial in the form

(x - d)^2 = 0

, which expands to

x^2 - 2dx + d^2 = 0

. Comparing this with

x^2 - kx + 9 = 0

, we get

d^2 = 9

and

2d = k

. Since

d^2 = 9

,

d

can be

3

or

-3

. Given that

k

is positive and

k = 2d

,

d

must also be positive, meaning

d = 3

. Substituting this back gives

k = 2(3) = 6

.

Estimated Time:45s

Question 78Question →

In the quadratic equation $3x^2 - kx + 12 = 0$ , $k$ is a constant. The equation has two distinct real solutions, and the difference between these two solutions is less than $2$ . Which of the following describes all possible values of $k$ ?

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

-6\sqrt{5} < k < 6\sqrt{5}

-12\sqrt{2} < k < -12

or

12 < k < 12\sqrt{2}

k < -12

or

k > 12

Show answer & explanation

Answer: $-6\sqrt{5} < k < -12$ or $12 < k < 6\sqrt{5}$

Answer

The possible values of

k

are

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

The correct answer is found by combining two constraints. First, for the quadratic equation to have two distinct real solutions, the discriminant must be strictly positive:

(-k)^2 - 4(3)(12) > 0 \Rightarrow k^2 > 144

, which means

|k| > 12

. Second, the difference between the roots of

ax^2 + bx + c = 0

is

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

. Here, the difference is

\frac{\sqrt{k^2 - 144}}{3}

. Setting this difference to be less than 2 gives

\frac{\sqrt{k^2 - 144}}{3} < 2 \Rightarrow \sqrt{k^2 - 144} < 6 \Rightarrow k^2 - 144 < 36 \Rightarrow k^2 < 180

, which means

|k| < 6\sqrt{5}

. Combining these inequalities yields

12 < |k| < 6\sqrt{5}

, which translates to

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

.

Step-by-Step Solution

1

Find the condition for the quadratic equation to have two distinct real solutions.

k^2 > 144

, which means

k < -12

or

k > 12

.

For the equation

3x^2 - kx + 12 = 0

to have two distinct real solutions, its discriminant

\Delta = b^2 - 4ac

must be strictly positive. Here,

\Delta = (-k)^2 - 4(3)(12) = k^2 - 144 > 0

.

2

Express the difference between the two solutions using the quadratic formula.

|x_1 - x_2| = \frac{\sqrt{k^2 - 144}}{3}

.

The solutions to the quadratic equation are given by

x = \frac{k \pm \sqrt{k^2 - 144}}{6}

. The difference between these solutions is

x_1 - x_2 = \frac{(k + \sqrt{k^2 - 144}) - (k - \sqrt{k^2 - 144})}{6} = \frac{2\sqrt{k^2 - 144}}{6} = \frac{\sqrt{k^2 - 144}}{3}

.

3

Apply the condition that the difference between the solutions is less than 2.

k^2 < 180

, which means

-6\sqrt{5} < k < 6\sqrt{5}

.

We set the difference expression to be less than 2:

\frac{\sqrt{k^2 - 144}}{3} < 2 \Rightarrow \sqrt{k^2 - 144} < 6

. Squaring both sides gives

k^2 - 144 < 36 \Rightarrow k^2 < 180

. Since

\sqrt{180} = 6\sqrt{5}

, this yields

-6\sqrt{5} < k < 6\sqrt{5}

.

4

Combine the inequalities from Step 1 and Step 3 to find the final overlapping range for

k

.

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

.

We must satisfy both

k^2 > 144

(for real solutions) and

k^2 < 180

(for the difference to be less than 2). This gives

144 < k^2 < 180

, which corresponds to the union of intervals

-6\sqrt{5} < k < -12

and

12 < k < 6\sqrt{5}

.

Key Concept

Using the discriminant and quadratic formula to analyze properties of roots under inequality constraints.

Question 79Question →

A circle in the $xy$ -plane is defined by the equation $(x - h)^2 + (y - k)^2 = 16$ , where $h$ and $k$ are constants. The center $(h, k)$ of the circle lies on the line $y = x$ . A second line, which passes through the origin and has a slope of $-\frac{3}{4}$ , is tangent to the circle at exactly one point $(x, y)$ . If this point of tangency lies in a quadrant where $x > 0$ and $y < 0$ , what is the value of $h$ ?

20

-\frac{20}{7}

\frac{20}{7}

\frac{80}{7}

Show answer & explanation

Answer: $\frac{20}{7}$

Answer

\frac{20}{7}

The correct option is

\frac{20}{7}

. By setting the center of the circle to

(h, h)

and the equation of the line to

3x + 4y = 0

, we find that the distance from the center to the line is

\frac{|7h|}{5}

. Since the line is tangent to the circle, this distance must equal the radius, which is

4

. This gives

h = \pm\frac{20}{7}

. Finding the point of tangency shows that

x = \frac{4}{25}h

and

y = -\frac{3}{25}h

. For the point of tangency to lie in Quadrant IV (

x > 0

and

y < 0

),

h

must be positive, which yields

h = \frac{20}{7}

.

Step-by-Step Solution

1

Express the center of the circle and the equation of the tangent line in terms of the given parameters.

Since the center

(h, k)

lies on the line

y = x

, we have

k = h

. The circle has radius

R = \sqrt{16} = 4

and is centered at

(h, h)

. The tangent line passes through the origin with slope

-\frac{3}{4}

, so its equation is

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

, which simplifies to

3x + 4y = 0

.

Setting up the algebraic expressions for both geometric entities is necessary to relate them using coordinate geometry formulas.

2

Apply the tangency condition using the point-to-line distance formula.

The distance from the center

(h, h)

to the line

3x + 4y = 0

must equal the radius

4

. Thus:

\frac{|3h + 4h|}{\sqrt{3^2 + 4^2}} = 4 \implies \frac{|7h|}{5} = 4 \implies |7h| = 20

.

A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line equals the radius.

3

Solve for the possible values of

h

.

h = \pm\frac{20}{7}

.

Solving the absolute value equation yields two symmetric possibilities for the x-coordinate of the circle's center.

4

Determine the relationship between the center

h

and the coordinates of the point of tangency

(x, y)

to apply the quadrant constraint.

The radius connecting the center

(h, h)

to the point of tangency

(x, y)

is perpendicular to the tangent line. Since the tangent line has a slope of

-\frac{3}{4}

, the perpendicular radius line has a slope of

\frac{4}{3}

. Its equation is:

y - h = \frac{4}{3}(x - h) \implies y = \frac{4}{3}x - \frac{1}{3}h

.

The intersection of the perpendicular radius line and the tangent line will locate the exact point of tangency.

5

Solve the system of equations for the point of tangency

(x, y)

in terms of

h

.

Equating the tangent line and the perpendicular line:

-\frac{3}{4}x = \frac{4}{3}x - \frac{1}{3}h \implies -\frac{9}{12}x = \frac{16}{12}x - \frac{4}{12}h \implies -\frac{25}{12}x = -\frac{4}{12}h \implies x = \frac{4}{25}h

. Substituting back:

y = -\frac{3}{4}\left(\frac{4}{25}h\right) = -\frac{3}{25}h

.

This yields the coordinates of the tangency point as a function of the parameter

h

.

6

Apply the quadrant constraint (

x > 0

and

y < 0

) to choose the correct sign of

h

.

We require

x = \frac{4}{25}h > 0

and

y = -\frac{3}{25}h < 0

. Both inequalities are satisfied if and only if

h > 0

. Therefore,

h = \frac{20}{7}

.

This filters out the extraneous geometric solution that lies in Quadrant II.

Key Concept

Solving systems of nonlinear equations representing circles and lines by utilizing geometric relations, distance formulas, and quadrant constraints.

Estimated Time:3m 0s

Question 80Question →

If $(2x - 3)(x + 4) = 0$ and $x > 0$ , what is the value of $x$ ?

-4

\frac{3}{2}

-\frac{3}{2}

4

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

The value of

x

is

\frac{3}{2}

.

To find the solutions to the equation

(2x - 3)(x + 4) = 0

, we set each factor equal to zero. Setting

2x - 3 = 0

gives

x = \frac{3}{2}

, and setting

x + 4 = 0

gives

x = -4

. The problem states that

x > 0

, meaning the value of

x

must be positive. Therefore, the only valid solution is

\frac{3}{2}

.

Step-by-Step Solution

1

Set each factor of the quadratic equation equal to zero using the zero product property.

2x - 3 = 0

or

x + 4 = 0

If the product of two factors is zero, then at least one of the individual factors must equal zero.

2

Solve each linear equation to find the possible values of

x

.

x = \frac{3}{2}

or

x = -4

Isolating

x

in each equation gives the roots of the quadratic equation.

3

Apply the given constraint

x > 0

to identify the correct solution.

x = \frac{3}{2}

is the only valid solution because

\frac{3}{2} > 0

and

-4 \ngtr 0

.

The problem restricts the solution to values of

x

that are strictly greater than zero, so the negative root must be discarded.

Key Concept

Solving factored quadratic equations with inequality constraints

Advanced Math

Answer

Step-by-Step Solution

Key Concept

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