Advanced Math Practice Questions — SAT — Page 7

Question 121Question →

The graph of the quadratic function $g$ in the $xy$ -plane is a parabola with vertex $(2, -5)$ . Which of the following equations could define the function $g$ ?

g(x) = (x + 2)^2 - 5

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

g(x) = (x - 2)^2 + 5

g(x) = (x + 2)^2 + 5

Show answer & explanation

Answer: $g(x) = (x - 2)^2 - 5$

Answer

The equation

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

could define the function

g

.

The vertex form of a quadratic function is given by

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

, where the point

(h, k)

is the vertex of the parabola. We are given that the vertex is

(2, -5)

, which means

h = 2

and

k = -5

. Substituting these values into the vertex form equation gives

g(x) = a(x - 2)^2 - 5

. Letting

a = 1

yields the equation

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

.

Step-by-Step Solution

1

Identify the vertex form of a quadratic function.

The vertex form of a quadratic function is written as

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

, where

(h, k)

is the vertex of the parabola and

a

is a non-zero constant.

Recognizing the vertex form allows us to directly substitute the coordinates of the given vertex.

2

Substitute the given vertex coordinates into the vertex form equation.

Given the vertex is

(2, -5)

, we substitute

h = 2

and

k = -5

into the vertex form:

g(x) = a(x - 2)^2 - 5

. If we assume

a = 1

, this simplifies to

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

.

This matches one of the given choices to find the equation that could define the function.

Key Concept

Vertex form of a quadratic function

Estimated Time:45s

Question 122Question →

For a constant $k$ , the circle $(x - 5)^2 + (y - 5)^2 = 18$ and the line $y = kx$ are graphed in the $xy$ -plane. For how many integer values of $k$ will the circle and the line intersect at exactly two points?

5

6

7

8

Show answer & explanation

Answer: 6

Answer

6

The correct answer is 6. By substituting the line equation

y = kx

into the circle equation, we obtain the quadratic equation

(k^2 + 1)x^2 - 10(k + 1)x + 32 = 0

. For the circle and line to intersect at exactly two points, this equation must have two distinct real roots, meaning its discriminant must be positive. This leads to the inequality

7k^2 - 50k + 7 < 0

, which factors as

(7k - 1)(k - 7) < 0

. The solution is the interval

\frac{1}{7} < k < 7

. The integer values of

k

in this range are

1, 2, 3, 4, 5,

and

6

, giving a total of 6 integers.

Step-by-Step Solution

1

Substitute the line equation

y = kx

into the circle equation

(x - 5)^2 + (y - 5)^2 = 18

.

(x - 5)^2 + (kx - 5)^2 = 18

Substitution reduces the system of equations to a single quadratic equation in

x

, representing the

x

-coordinates of the intersection points.

2

Expand the terms and write the equation in standard quadratic form

Ax^2 + Bx + C = 0

.

(k^2 + 1)x^2 - 10(k + 1)x + 32 = 0

Putting the equation in standard form is necessary to analyze its discriminant.

3

Set the discriminant

\Delta = b^2 - 4ac

strictly greater than zero to ensure exactly two distinct real solutions.

[-10(k+1)]^2 - 4(k^2 + 1)(32) > 0 \Rightarrow 7k^2 - 50k + 7 < 0

A quadratic equation has two distinct real roots if and only if its discriminant is positive.

4

Solve the quadratic inequality

7k^2 - 50k + 7 < 0

by factoring the quadratic expression.

\frac{1}{7} < k < 7

The roots of

(7k - 1)(k - 7) = 0

are

k = \frac{1}{7}

and

k = 7

. Since the leading coefficient is positive, the expression is negative between these roots.

5

Identify and count the integer values of

k

that lie within the interval

(\frac{1}{7}, 7)

.

The integers are

1, 2, 3, 4, 5, 6

, which gives a total of 6 values.

The integers that are strictly greater than

\frac{1}{7}

and strictly less than

7

are the whole numbers from

1

to

6

.

Key Concept

Solving systems consisting of a circle and a line algebraically by utilizing the quadratic discriminant to find the condition for two real intersection points.

Question 123Question →

The system of equations below consists of a linear equation and a quadratic equation:

\begin{aligned} y - 2x &= 5 \\ y &= x^2 - 3x - 1 \end{aligned}

Let

(x_1, y_1)

and

(x_2, y_2)

represent the two distinct real coordinate points where the graphs of these equations intersect. If

y_1 > y_2

, what is the value of

x_1 - x_2

?

Show answer & explanation

Answer: 7

Answer

7

Solving the system algebraically by expressing

y

as

2x + 5

and substituting it into the quadratic equation leads to

x^2 - 5x - 6 = 0

. Factoring this equation yields

x = 6

and

x = -1

. Evaluating these values in the linear equation gives the coordinates

(6, 17)

and

(-1, 3)

. Since

17 > 3

, the point

(6, 17)

corresponds to

(x_1, y_1)

and

(-1, 3)

corresponds to

(x_2, y_2)

. The difference

x_1 - x_2

is equal to

6 - (-1) = 7

.

Step-by-Step Solution

1

Solve for

y

in the linear equation to express it in terms of

x

.

y = 2x + 5

This allows for substitution into the quadratic equation to eliminate one variable.

2

Substitute

2x + 5

for

y

in the quadratic equation.

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

To create a single-variable quadratic equation in terms of

x

.

3

Rearrange the equation by subtracting

2x

and

5

from both sides to set it equal to zero.

x^2 - 5x - 6 = 0

Setting the quadratic equation to zero is the standard first step to find its roots.

4

Factor the quadratic trinomial.

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

, which gives

x = 6

and

x = -1

.

Factoring finds the

x

-coordinates of the points of intersection.

5

Substitute the

x

-values back into the linear equation

y = 2x + 5

to determine the corresponding

y

-values.

For

x = 6

,

y = 2(6) + 5 = 17

. For

x = -1

,

y = 2(-1) + 5 = 3

. The intersection points are

(6, 17)

and

(-1, 3)

.

This defines the full coordinate pairs of the system's solutions.

6

Apply the condition

y_1 > y_2

to identify

(x_1, y_1)

and

(x_2, y_2)

.

Since

17 > 3

, the first solution is

(x_1, y_1) = (6, 17)

and the second solution is

(x_2, y_2) = (-1, 3)

.

This aligns the coordinate values with the variables defined in the problem constraint.

7

Calculate the value of

x_1 - x_2

.

6 - (-1) = 7

Subtracting a negative value is equivalent to adding its absolute value, yielding the final required answer.

Key Concept

Nonlinear Systems of Equations

Question 124Question →

In the system of equations below,

k

is a constant.

\begin{aligned} x^2 + y^2 + 8y &= 9 \\ x^2 + y &= k \end{aligned}

If the system has exactly 3 distinct real solutions

(x, y)

, what is the value of

k

?

Show answer & explanation

Answer: 1

Answer

The value of

k

is 1.

Substituting

x^2 = k - y

into

x^2 + y^2 + 8y = 9

gives

y^2 + 7y + k - 9 = 0

. For the system to have exactly 3 distinct real solutions, the vertex of the parabola

y = -x^2 + k

must lie on the circle, which corresponds to the root

y = k

. Substituting

y = k

into the quadratic equation yields

k^2 + 8k - 9 = 0

, which gives

k = 1

or

k = -9

. For

k = 1

, the roots of the quadratic are

y = 1

and

y = -8

. The root

y = 1

yields 1 real solution,

(0, 1)

, and the root

y = -8

yields 2 real solutions,

(3, -8)

and

(-3, -8)

, for a total of 3 real solutions. For

k = -9

, the roots are

y = -9

and

y = 2

. The root

y = 2

does not yield any real solutions for

x

because

2 > -9

, so the system has only 1 real solution. Thus,

k = 1

.

Step-by-Step Solution

1

Express

x^2

in terms of

y

and

k

from the second equation.

x^2 = k - y

(with constraint

y \le k

for real

x

)

To prepare for substitution into the first equation and establish the boundary condition for real solutions.

2

Substitute

x^2 = k - y

into the first equation.

y^2 + 7y + k - 9 = 0

To create a single quadratic equation in terms of

y

.

3

Analyze the conditions on the roots of the quadratic equation to get exactly 3 distinct real solutions.

One root must equal

k

and the other must be less than

k

.

A root

y = k

yields 1 real solution for

x

(

x = 0

), while a root

y < k

yields 2 real solutions (

x = \pm\sqrt{k-y}

).

4

Find the candidate values of

k

by setting

y = k

in the quadratic equation.

k^2 + 8k - 9 = 0 \implies k = 1

or

k = -9

To find the values of

k

where the parabola's vertex lies on the circle.

5

Verify which candidate value of

k

satisfies all conditions.

For

k = 1

, the roots are

y = 1

and

y = -8 < 1

(3 solutions). For

k = -9

, the roots are

y = -9

and

y = 2 > -9

(1 solution). Therefore,

k = 1

.

To ensure the second root is strictly less than

k

, guaranteeing exactly 3 solutions.

Key Concept

Analyzing the number of solutions in a nonlinear system of equations using algebraic substitution and boundary constraints.

Estimated Time:3m 0s

Question 125Question →

In the quadratic equation $3x^2 - kx + 24 = 0$ , $k$ is a positive constant. If one of the roots of the equation is twice the other root, what is the value of $k$ ?

6

12

18

36

Show answer & explanation

Answer: 18

Answer

18

The correct answer is 18. Let the roots of the equation be

r

and

2r

. The product of the roots is given by

\frac{c}{a} = \frac{24}{3} = 8

. Therefore,

2r^2 = 8

, which means

r^2 = 4

. Since

k > 0

, the roots must be positive, so

r = 2

and the two roots are 2 and 4. The sum of the roots is

2 + 4 = 6

. According to Vieta's formulas, the sum of the roots is also equal to

-\frac{b}{a} = \frac{k}{3}

. Setting the two expressions for the sum of the roots equal to each other gives

\frac{k}{3} = 6

, which simplifies to

k = 18

.

Step-by-Step Solution

1

Define the roots and set up the product of the roots using Vieta's formulas.

Let the roots be

r

and

2r

. The product of the roots is

r \times 2r = 2r^2

. From the equation

3x^2 - kx + 24 = 0

, the product of the roots is also

\frac{c}{a} = \frac{24}{3} = 8

.

This establishes a relationship between the given ratio of the roots and the coefficients of the equation.

2

Solve for the root variable

r

.

Setting

2r^2 = 8

gives

r^2 = 4

. Since

k

is a positive constant, the sum of the roots must be positive, which means the roots themselves must be positive. Thus,

r = 2

, and the roots are

2

and

4

.

Finding the actual values of the roots is necessary to calculate their sum.

3

Use Vieta's formula for the sum of the roots to find

k

.

The sum of the roots is

2 + 4 = 6

. According to Vieta's formulas, the sum of the roots is

-\frac{b}{a} = \frac{k}{3}

. Setting them equal gives

\frac{k}{3} = 6

, which simplifies to

k = 18

.

This directly isolates and solves for the unknown constant

k

.

Key Concept

Using Vieta's formulas to relate the roots of a quadratic equation

ax^2 + bx + c = 0

to its coefficients: the sum of the roots is

-\frac{b}{a}

and the product of the roots is

\frac{c}{a}

.

Question 126Question →

For all $x > 2$ , which of the following is equivalent to the expression $\frac{3x^2 - 12}{x^2 - 2x} - \frac{2x^2 + 2x - 12}{x^2 + 3x}$ ?

\frac{x+2}{x}

\frac{x+10}{x}

\frac{x-12}{x}

\frac{x+14}{x}

Show answer & explanation

Answer: $\frac{x+10}{x}$

Answer

\frac{x+10}{x}

The correct answer is obtained by factoring both rational expressions and simplifying them before subtracting. The first term factors into

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

, which simplifies to

\frac{3(x+2)}{x}

. The second term factors into

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

, which simplifies to

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

. Subtracting these two expressions yields

\frac{3(x+2) - 2(x-2)}{x} = \frac{3x+6-2x+4}{x} = \frac{x+10}{x}

.

Step-by-Step Solution

1

Factor the numerators and denominators of both rational expressions.

The expression becomes

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

.

Factoring helps identify common binomial factors in the numerator and denominator.

2

Simplify both terms by canceling common factors.

For

x > 2

, the expression simplifies to

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

.

Since

x > 2

, the terms

x-2

and

x+3

are non-zero and can be canceled.

3

Combine the simplified terms over the common denominator

x

.

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

Both terms share the common denominator

x

.

4

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

\frac{3x + 6 - 2x + 4}{x} = \frac{x+10}{x}

Distributing subtraction to the terms inside

(2x-4)

yields

-2x + 4

.

Key Concept

Simplifying and subtracting rational expressions by factoring and finding a common denominator

Question 127Question →

The function $f$ is defined by $f(x) = a \cdot 3^x$ , where $a$ is a constant. If $f(2) = 45$ , what is the value of $f(1)$ ?

Show answer & explanation

Answer: 15

Answer

15

To find the value of

f(1)

, we first determine the value of the constant

a

. We are given that

f(2) = 45

, so substituting

x = 2

into the function definition

f(x) = a \cdot 3^x

gives

45 = a \cdot 3^2

. Simplifying

3^2

to

9

yields

45 = 9a

, which means

a = 5

. Now that we know

a = 5

, we can write the function as

f(x) = 5 \cdot 3^x

. To find

f(1)

, we substitute

x = 1

into this equation, yielding

f(1) = 5 \cdot 3^1 = 15

.

Step-by-Step Solution

1

Substitute the given point

(2, 45)

into the function equation to solve for

a

.

a = 5

Since

f(2) = 45

, we have

45 = a \cdot 3^2 = 9a

, which gives

a = 5

.

2

Evaluate the function at

x = 1

using the value of

a = 5

.

f(1) = 15

Substituting

a = 5

and

x = 1

into

f(x) = a \cdot 3^x

gives

f(1) = 5 \cdot 3^1 = 15

.

Key Concept

Evaluating and solving exponential functions given initial conditions or points.

Question 128Question →

In the $xy$ -plane, the graph of the linear equation $3x + 4y = k$ , where $k$ is a positive constant, is tangent to the circle with equation $x^2 + y^2 - 2x - 4y = 20$ . What is the value of $k$ ?

Show answer & explanation

Answer: 36

Answer

The value of the constant

k

is

36

.

To find the value of

k

where the line

3x + 4y = k

is tangent to the circle, we first rewrite the circle's equation

x^2 + y^2 - 2x - 4y = 20

in standard form by completing the square:

(x^2 - 2x + 1) + (y^2 - 4y + 4) = 20 + 1 + 4

, which simplifies to

(x - 1)^2 + (y - 2)^2 = 25

. This is a circle centered at

(1, 2)

with a radius of

5

. A line is tangent to a circle if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle. Using the distance formula

\frac{|Ax_0 + By_0 - C|}{\sqrt{A^2 + B^2}}

for the point

(1, 2)

and the line

3x + 4y - k = 0

, we set up the equation:

\frac{|3(1) + 4(2) - k|}{\sqrt{3^2 + 4^2}} = 5

. This simplifies to

\frac{|11 - k|}{5} = 5

, or

|11 - k| = 25

. Solving the absolute value equation gives

11 - k = 25 \Rightarrow k = -14

and

11 - k = -25 \Rightarrow k = 36

. Since the problem specifies that

k

is a positive constant, the correct answer is

36

.

Step-by-Step Solution

1

Rewrite the circle's equation by completing the square for both variables.

(x - 1)^2 + (y - 2)^2 = 25

To identify the center and radius of the circle.

2

Identify the center and radius from the standard form of the circle's equation.

Center is

(1, 2)

and radius is

5

.

The standard form of a circle is

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

.

3

Apply the condition for tangency: the perpendicular distance from the center

(1, 2)

to the line

3x + 4y - k = 0

must equal the radius

5

.

\frac{|3(1) + 4(2) - k|}{\sqrt{3^2 + 4^2}} = 5

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

4

Simplify the distance equation and solve the resulting absolute value equation for

k

.

|11 - k| = 25

, yielding

k = -14

or

k = 36

.

To find all mathematically possible values of the constant

k

.

5

Select the positive value for

k

as specified by the problem constraints.

k = 36

The problem states that

k

is a positive constant.

Key Concept

The relationship between a line and a circle in a nonlinear system, specifically using the distance from the center to a tangent line to solve for an unknown constant.

Question 129Question →

If $3^{x+1} = k$ , where $k > 0$ , which of the following is equivalent to $27^{x-1}$ ?

\frac{k^3}{729}

\frac{k^3}{81}

\frac{k^3}{27}

\frac{k^3}{9}

Show answer & explanation

Answer: $\frac{k^3}{729}$

Answer

The expression \frac{k^3}{729}

To express

27^{x-1}

in terms of

k

, we can rewrite both the given equation and the target expression using base

3

. The given equation

3^{x+1} = k

can be adjusted to find the value of

3^{x-1}

by dividing both sides by

3^2

, which yields

3^{x-1} = \frac{k}{9}

. The target expression

27^{x-1}

can be rewritten as

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

. Substituting

3^{x-1} = \frac{k}{9}

into this expression gives

\left(\frac{k}{9}\right)^3 = \frac{k^3}{729}

.

Step-by-Step Solution

1

Express the given equation in terms of a simpler base

3

exponent.

3^{x-1} = \frac{k}{9}

We start with the given equation

3^{x+1} = k

. To relate this to the exponent

x-1

, we divide both sides of the equation by

3^2 = 9

:

3^{x-1} = \frac{3^{x+1}}{3^2} = \frac{k}{9}

.

2

Rewrite the target expression

27^{x-1}

with base

3

.

27^{x-1} = (3^{x-1})^3

Since

27 = 3^3

, we can rewrite the target expression as

27^{x-1} = (3^3)^{x-1} = 3^{3(x-1)} = (3^{x-1})^3

.

3

Substitute the expression for

3^{x-1}

from Step 1 into the expression from Step 2.

\frac{k^3}{729}

Substituting

3^{x-1} = \frac{k}{9}

into

(3^{x-1})^3

gives

\left(\frac{k}{9}\right)^3 = \frac{k^3}{9^3} = \frac{k^3}{729}

.

Key Concept

Manipulating exponential equations by expressing terms with a common base and applying exponent rules.

Question 130Question →

If $2^{2x+2} - 2^{2x} = 12$ for some real number $x$ , what is the value of $2^{4x}$ ?

Show answer & explanation

Answer: 16

Answer

16

By applying the product rule of exponents, the expression

2^{2x+2}

can be rewritten as

2^{2x} \cdot 2^2

, or

4 \cdot 2^{2x}

. Substituting this into the given equation yields

4 \cdot 2^{2x} - 2^{2x} = 12

. Factoring out

2^{2x}

gives

2^{2x}(4 - 1) = 12

, which simplifies to

3 \cdot 2^{2x} = 12

. Dividing both sides of the equation by 3 results in

2^{2x} = 4

. Since

2^{4x}

can be written as

(2^{2x})^2

, substituting 4 for

2^{2x}

gives

4^2 = 16

. Alternatively, solving

2^{2x} = 4

gives

2x = 2

, which means

x = 1

. Substituting

x = 1

into

2^{4x}

yields

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

.

Step-by-Step Solution

1

Rewrite

2^{2x+2}

using exponent properties.

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

To express both exponential terms with the same base power, allowing them to be factored or combined.

2

Substitute this back into the equation and factor out the common term

2^{2x}

.

2^{2x}(4 - 1) = 12 \implies 3 \cdot 2^{2x} = 12

To isolate the exponential expression.

3

Solve for

2^{2x}

by dividing both sides of the equation by 3.

2^{2x} = 4

To find the value of the exponential term.

4

Express

2^{4x}

in terms of

2^{2x}

and evaluate.

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

To find the final requested value using the exponent rule

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

.

Key Concept

Factoring exponential equations and applying power of a power exponent rules

Question 131Question →

A projectile is launched from a platform. Its height $h(t)$ , in meters, $t$ seconds after launch is modeled by the function $h(t) = -5t^2 + bt + 12$ , where $b$ is a positive constant. If the projectile reaches a maximum height of $32$ meters, what is the value of $b$ ?

Show answer & explanation

Answer: 20

Answer

The correct answer is 20.

To find the maximum height of the projectile, we locate the vertex of the quadratic function

h(t) = -5t^2 + bt + 12

. The

t

-coordinate of the vertex is given by

t = -\frac{B}{2A} = -\frac{b}{2(-5)} = \frac{b}{10}

. Substituting

t = \frac{b}{10}

into

h(t)

and setting the height to

32

meters gives

-5\left(\frac{b}{10}\right)^2 + b\left(\frac{b}{10}\right) + 12 = 32

. Simplifying this equation results in

-\frac{b^2}{20} + \frac{b^2}{10} = 20

, which reduces to

\frac{b^2}{20} = 20

. Solving for

b

gives

b^2 = 400

. Because

b

must be a positive constant,

b = 20

.

Step-by-Step Solution

1

Identify the time

t

at which the maximum height occurs using the vertex formula

t = -\frac{B}{2A}

for the quadratic function

h(t) = At^2 + Bt + C

.

t = \frac{b}{10}

The vertex of a downward-opening parabola represents its maximum value. For

h(t) = -5t^2 + bt + 12

, the coefficients are

A = -5

and

B = b

.

2

Substitute the time

t = \frac{b}{10}

back into the height function

h(t)

and set the expression equal to the maximum height of

32

meters.

-5\left(\frac{b}{10}\right)^2 + b\left(\frac{b}{10}\right) + 12 = 32

At the maximum height, the height of the projectile is

32

meters, which corresponds to the value of the function at the vertex time.

3

Simplify the equation to solve for

b^2

.

\frac{b^2}{20} = 20

Squaring the fraction yields

-5\left(\frac{b^2}{100}\right) + \frac{b^2}{10} + 12 = 32

. Simplifying the coefficients leads to

-\frac{b^2}{20} + \frac{b^2}{10} = 20

, which simplifies to

\frac{b^2}{20} = 20

.

4

Solve the equation for

b

.

b = 20

Multiplying both sides by

20

gives

b^2 = 400

. Taking the square root of both sides gives

b = \pm 20

. Since

b

is specified to be a positive constant,

b = 20

.

Key Concept

Finding the vertex of a quadratic function to determine maximum values in a real-world context.

Estimated Time:2m 0s

Question 132Question →

For a constant $k$ , the quadratic equation $x^2 - 2kx + k^2 - 4k + 3 = 0$ has two distinct real solutions. If both solutions are positive, which of the following describes all possible values of $k$ ?

k > 3

or

\frac{3}{4} < k < 1

k > 3

k > \frac{3}{4}

k < 1

or

k > 3

Show answer & explanation

Answer: $k > 3$ or $\frac{3}{4} < k < 1$

Answer

k > 3

or

\frac{3}{4} < k < 1

The correct answer is the option that includes both the interval between

\frac{3}{4}

and

1

and the interval greater than

3

. To ensure the quadratic equation has two distinct real solutions, the discriminant must be positive, which yields

k > \frac{3}{4}

. To ensure both solutions are positive, their sum and product must both be positive. The sum

2k > 0

requires

k > 0

. The product

k^2 - 4k + 3 > 0

requires

k < 1

or

k > 3

. The intersection of all these inequalities is

\frac{3}{4} < k < 1

or

k > 3

.

Step-by-Step Solution

1

Determine the condition for two distinct real solutions using the discriminant.

k > \frac{3}{4}

For the quadratic equation

x^2 - 2kx + k^2 - 4k + 3 = 0

to have two distinct real solutions, its discriminant

\Delta

must be strictly greater than

0

. Here,

a = 1

,

b = -2k

, and

c = k^2 - 4k + 3

. Thus,

\Delta = (-2k)^2 - 4(1)(k^2 - 4k + 3) = 4k^2 - 4k^2 + 16k - 12 = 16k - 12

. Setting

\Delta > 0

gives

16k - 12 > 0

, which simplifies to

k > \frac{3}{4}

.

2

Set up conditions for both solutions to be positive using the sum and product of the roots.

2k > 0

and

k^2 - 4k + 3 > 0

If a quadratic equation has two positive solutions, their sum must be positive and their product must also be positive. By Vieta's formulas, the sum of the solutions is

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

and the product of the solutions is

\frac{c}{a} = k^2 - 4k + 3

.

3

Solve the inequalities for the sum and product of the roots.

k > 0

and either

k < 1

or

k > 3

The sum condition

2k > 0

simplifies to

k > 0

. The product condition

k^2 - 4k + 3 > 0

factors as

(k-1)(k-3) > 0

, which is satisfied when

k < 1

or

k > 3

.

4

Find the intersection of all the derived inequalities.

\frac{3}{4} < k < 1

or

k > 3

We must find the values of

k

that satisfy all three conditions:

k > \frac{3}{4}

,

k > 0

, and (

k < 1

or

k > 3

). Combining

k > \frac{3}{4}

and

k > 0

yields

k > \frac{3}{4}

. Combining

k > \frac{3}{4}

with the constraint that

k

must be less than

1

or greater than

3

gives the combined solution of

\frac{3}{4} < k < 1

or

k > 3

.

Key Concept

Using the discriminant and Vieta's formulas to determine the signs of the roots of a quadratic equation dependent on a parameter.

Alternative Method

Alternatively, you can write the solutions using the quadratic formula:

x = \frac{2k \pm \sqrt{16k - 12}}{2} = k \pm \sqrt{4k - 3}

. For both roots to be positive and distinct, we first need the term under the radical to be positive, so

4k - 3 > 0 \implies k > \frac{3}{4}

. Then, we need the smaller root to be positive:

k - \sqrt{4k - 3} > 0 \implies k > \sqrt{4k - 3}

. Since

k > \frac{3}{4} > 0

, we can square both sides:

k^2 > 4k - 3 \implies k^2 - 4k + 3 > 0 \implies (k - 1)(k - 3) > 0

. This yields

k < 1

or

k > 3

. Combining this with

k > \frac{3}{4}

gives the same result:

\frac{3}{4} < k < 1

or

k > 3

.

Estimated Time:3m 0s

Question 133Question →

If $3^{x+4} = 9^x$ , what is the value of $x$ ?

4

2

-4

\frac{4}{3}

Show answer & explanation

Answer: 4

Answer

4

To solve the equation

3^{x+4} = 9^x

, we rewrite the base 9 as

3^2

, which gives

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

. Applying the exponent rule

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

, we get

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

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

x+4 = 2x

. Subtracting

x

from both sides yields the solution

4

.

Step-by-Step Solution

1

Express both sides of the equation with a common base.

Since

9 = 3^2

, the equation

3^{x+4} = 9^x

can be rewritten as

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

. Using the power of a power rule, this becomes

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

.

To solve an exponential equation, it is helpful to have the same base on both sides so that the exponents can be equated.

2

Set the exponents equal to each other.

x + 4 = 2x

Since the bases are equal and positive (and not equal to 1), their exponents must be equal.

3

Solve the linear equation for

x

.

4 = x

(or

x = 4

)

Subtract

x

from both sides of the equation to isolate the variable.

Key Concept

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

Estimated Time:45s

Question 134Question →

A toy rocket is launched vertically upward from ground level. Its height, $h$ , in meters, $t$ seconds after launch is modeled by the equation $h(t) = -5t^2 + 40t$ . How many seconds after launch does the rocket reach its maximum height?

8

4

80

20

Show answer & explanation

Answer: 4

Answer

4 seconds

The height of the rocket is modeled by the quadratic function

h(t) = -5t^2 + 40t

. Because the coefficient of

t^2

is negative (

a = -5

), the graph of this function is a parabola that opens downward, meaning its vertex represents the maximum height. The time

t

at which the vertex occurs is given by

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

. Substituting

a = -5

and

b = 40

gives

t = -\frac{40}{2(-5)} = 4

seconds.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation.

a = -5

,

b = 40

, and

c = 0

The equation is given in standard quadratic form

h(t) = at^2 + bt + c

.

2

Apply the vertex formula to find the time at which the maximum height occurs.

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

For a downward-opening parabola, the maximum value occurs at the vertex.

3

Substitute the values of

a

and

b

and simplify.

t = -\frac{40}{2(-5)} = 4

This calculation yields the time in seconds when the rocket reaches its maximum height.

Key Concept

Finding the vertex of a quadratic function to determine the maximum or minimum value in context.

Alternative Method

Find the times when the rocket is at ground level by solving

h(t) = 0

. Factoring

-5t^2 + 40t = 0

gives

-5t(t - 8) = 0

, so the rocket is on the ground at

t = 0

seconds and

t = 8

seconds. Since a parabola is symmetric, the maximum height must occur exactly halfway between the launch and landing times, which is

\frac{0 + 8}{2} = 4

seconds.

Estimated Time:45s

Question 135Question →

A circle in the $xy$ -plane is defined by the equation $(x - 3)^2 + (y - 2)^2 = 13$ . The line $y = 3x$ intersects the circle at the origin $(0, 0)$ and at a second point $P$ . What is the $y$ -coordinate of point $P$ ?

\frac{9}{5}

\frac{11}{5}

\frac{27}{5}

-\frac{9}{5}

Show answer & explanation

Answer: $\frac{27}{5}$

Answer

The y-coordinate of point P is

\frac{27}{5}

.

Substituting the linear relationship

y = 3x

into the circle equation yields a quadratic equation in terms of

x

. Solving this equation gives the

x

-coordinates of the intersection points as

x = 0

and

x = \frac{9}{5}

. Since point

P

is not the origin, its

x

-coordinate is

\frac{9}{5}

. Substituting this back into

y = 3x

gives the

y

-coordinate of point

P

as

\frac{27}{5}

.

Step-by-Step Solution

1

Substitute the linear equation

y = 3x

into the equation of the circle.

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

Substitution reduces the system of two equations with two variables to a single quadratic equation with one variable.

2

Expand and simplify the equation.

10x^2 - 18x = 0

Expanding

(x - 3)^2

to

x^2 - 6x + 9

and

(3x - 2)^2

to

9x^2 - 12x + 4

, then combining like terms and subtracting 13 from both sides, isolates the quadratic terms.

3

Solve for

x

by factoring.

2x(5x - 9) = 0

, which gives

x = 0

or

x = \frac{9}{5}

.

The roots of the factored equation represent the x-coordinates of the two intersection points.

4

Find the y-coordinate of the second point

P

using

y = 3x

.

y = 3 \left(\frac{9}{5}\right) = \frac{27}{5}

Since point

P

is not the origin, its x-coordinate must be

\frac{9}{5}

. Substituting this into the linear equation gives its corresponding y-coordinate.

Key Concept

Solving nonlinear systems of equations involving a circle and a line by substitution.

Question 136Question →

The quadratic equation $3x^2 - 12x + c = 0$ , where $c$ is a constant, has two real solutions, $r$ and $s$ . If $\frac{1}{r} + \frac{1}{s} = \frac{3}{2}$ , what is the value of $c$ ?

-8

8

18

24

Show answer & explanation

Answer: 8

Answer

8

The correct answer is 8. By Vieta's formulas, the sum of the solutions is

r + s = -\frac{-12}{3} = 4

and their product is

rs = \frac{c}{3}

. The given equation

\frac{1}{r} + \frac{1}{s} = \frac{3}{2}

can be rewritten as

\frac{r+s}{rs} = \frac{3}{2}

. Substituting the expressions for the sum and product gives

\frac{4}{c/3} = \frac{12}{c} = \frac{3}{2}

. Solving this equation yields

c = 8

. Checking the discriminant of the resulting quadratic equation

3x^2 - 12x + 8 = 0

gives

\Delta = (-12)^2 - 4(3)(8) = 48

, which is positive and confirms the existence of two real solutions.

Step-by-Step Solution

1

Use Vieta's formulas to express the sum and product of the solutions in terms of the coefficients of the quadratic equation.

The sum of the solutions is

r + s = -\frac{b}{a} = -\frac{-12}{3} = 4

, and the product of the solutions is

rs = \frac{c}{a} = \frac{c}{3}

.

Vieta's formulas relate the sum and product of the roots of a quadratic equation

ax^2 + bx + c = 0

directly to its coefficients.

2

Express the given algebraic relation

\frac{1}{r} + \frac{1}{s}

in terms of

r+s

and

rs

by finding a common denominator.

\frac{1}{r} + \frac{1}{s} = \frac{s + r}{rs} = \frac{r + s}{rs}

.

Finding a common denominator allows us to rewrite the sum of reciprocals using the sum and product of the solutions.

3

Substitute the sum and product from Step 1 into the rewritten equation and solve for

c

.

Substituting the values gives

\frac{4}{\frac{c}{3}} = \frac{3}{2}

, which simplifies to

\frac{12}{c} = \frac{3}{2}

. Solving for

c

yields

3c = 24

, or

c = 8

.

This substitution turns the given relation into a single linear equation in terms of the variable

c

.

4

Verify that

c = 8

yields two real solutions by checking the discriminant of the quadratic equation.

The discriminant is

\Delta = b^2 - 4ac = (-12)^2 - 4(3)(8) = 144 - 96 = 48

. Since

\Delta > 0

, there are indeed two real solutions.

The problem states that the equation has two real solutions, so we must confirm that

c = 8

satisfies this condition.

Key Concept

Applying Vieta's formulas and algebraic manipulation to solve for constants in a quadratic equation.

Estimated Time:2m 0s

Question 137Question →

\begin{aligned} y &= x^2 - x - 7 \\ 2x - y &= -3 \end{aligned}

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

18

-3

-2

0

Show answer & explanation

Answer: -3

Answer

The correct answer is -3.

The correct answer is

-3

. To find this, substitute the expression for

y

from the first equation into the second equation, which yields

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

. Distributing the negative sign and combining like terms gives

-x^2 + 3x + 7 = -3

. Setting the equation to zero results in

x^2 - 3x - 10 = 0

. Factoring the quadratic expression gives

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

, which means

x = 5

or

x = -2

. Since the question specifies that

x < 0

, we select

x = -2

. Substituting

x = -2

back into the linear equation gives

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

. Finally, calculating

x + y

gives

-2 + (-1) = -3

.

Step-by-Step Solution

1

Substitute the expression for

y

from the first equation into the second equation.

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

This allows us to eliminate

y

and solve for

x

in a single variable quadratic equation.

2

Distribute the negative sign and simplify the equation.

-x^2 + 3x + 7 = -3

Expanding the parentheses correctly preserves the signs of the quadratic terms.

3

Rearrange the terms to set the quadratic equation to zero.

x^2 - 3x - 10 = 0

Writing the equation in standard form (

ax^2 + bx + c = 0

) allows us to factor it.

4

Factor the quadratic expression.

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

Finding factors that multiply to

-10

and add to

-3

gives us the solutions for

x

.

5

Solve for

x

and apply the constraint

x < 0

.

x = -2

The equation yields

x = 5

or

x = -2

. The constraint

x < 0

means we must choose

x = -2

.

6

Substitute

x = -2

back into one of the original equations to solve for

y

.

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

We need the corresponding

y

-value to find the value of

x + y

.

7

Calculate the value of

x + y

.

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

This provides the final required value specified in the question.

Key Concept

Solving a system of one linear equation and one quadratic equation using substitution.

Question 138Question →

\begin{aligned} x - y &= 3 \\ y &= x^2 - 7x + 12 \end{aligned}

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

Show answer & explanation

Answer: 10

Answer

10

By substituting the linear equation

y = x - 3

into the quadratic equation, we obtain a single quadratic equation

x^2 - 8x + 15 = 0

. Factoring this equation yields the solutions

x = 3

and

x = 5

. The corresponding

y

-coordinates are

y = 0

and

y = 2

, respectively. Since the problem specifies that

y > 0

, we choose the solution

(5, 2)

. The product of the coordinates is

5 \cdot 2 = 10

.

Step-by-Step Solution

1

Express

y

in terms of

x

from the linear equation.

y = x - 3

This allows for substitution into the second equation.

2

Substitute

y = x - 3

into the quadratic equation.

x - 3 = x^2 - 7x + 12

To create a single quadratic equation in terms of

x

.

3

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 8x + 15 = 0

Subtracting

x

and adding

3

to both sides enables factoring.

4

Factor the quadratic expression.

(x - 3)(x - 5) = 0

To find the roots of the equation, which are

x = 3

and

x = 5

.

5

Find the corresponding

y

-values for each

x

-value.

For

x = 3

,

y = 3 - 3 = 0

. For

x = 5

,

y = 5 - 3 = 2

.

To obtain the complete coordinate pairs of the intersection points.

6

Apply the constraint

y > 0

to select the correct solution pair.

The solution

(5, 2)

is selected since its

y

-coordinate is greater than

0

.

The other solution,

(3, 0)

, has

y = 0

, which does not satisfy the constraint

y > 0

.

7

Calculate the value of

xy

for the chosen solution.

xy = 5 \cdot 2 = 10

To find the final requested value.

Key Concept

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

Estimated Time:1m 30s

Question 139Question →

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

Show answer & explanation

Answer: 2.5

Answer

2.5 (or 5/2)

The correct answer is 2.5 (or 5/2). By expressing 4 as

2^2

and 8 as

2^3

, the equation is rewritten as

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

. Applying the exponent power rule simplifies this to

2^{2x-2} = 2^3

. Equating the exponents yields the linear equation

2x-2 = 3

. Solving for

x

gives

2x = 5

, which results in

x = 2.5

or

5/2

.

Step-by-Step Solution

1

Rewrite both sides of the equation with a common base of 2.

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

Expressing both bases as powers of 2 allows us to equate the exponents later.

2

Apply the power of a power exponent rule

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

to the left side.

2^{2x-2} = 2^3

Multiplying the exponent 2 by the exponent

(x-1)

simplifies the expression to

2(x-1) = 2x-2

.

3

Equate the exponents and solve the linear equation for

x

.

x = 2.5

Since the bases are identical, their exponents must be equal, giving

2x-2 = 3

, which simplifies to

2x = 5

.

Key Concept

Solving exponential equations by finding a common base.

Question 140Question →

The function $f$ is defined by $f(x) = x^2 - 10x + 29$ . For what value of $x$ does $f(x)$ reach its minimum value?

Show answer & explanation

Answer: 5

Answer

The function reaches its minimum value at

x = 5

.

A quadratic function defined by

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

where

a > 0

reaches its minimum value at its vertex. The

x

-coordinate of the vertex is calculated using the formula

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

. For

f(x) = x^2 - 10x + 29

, the coefficients are

a = 1

and

b = -10

. Substituting these values into the formula gives

x = -\frac{-10}{2(1)} = 5

.

Step-by-Step Solution

1

Identify the coefficients of the quadratic function in standard form

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

.

For

f(x) = x^2 - 10x + 29

, the coefficients are

a = 1

and

b = -10

.

To use the vertex formula

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

.

2

Apply the vertex formula to calculate the

x

-coordinate where the function reaches its minimum.

x = 5

.

Since the coefficient of

x^2

is positive (

a = 1

), the parabola opens upward, and its minimum value occurs at the vertex.

Key Concept

Finding the x-coordinate of the vertex of a quadratic function

Advanced Math

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