Nonlinear Systems of Equations Practice Questions — SAT

Question 1Question →

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 2Question →

\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 3Question →

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

x - y = 4

x^2 - y^2 = 40

Show answer & explanation

Answer: 7

Answer

The value of

x

is 7.

The equation

x^2 - y^2 = 40

can be factored as

(x-y)(x+y) = 40

. Substituting

x-y = 4

yields

4(x+y) = 40

, which simplifies to

x+y = 10

. Adding the two linear equations

x-y = 4

and

x+y = 10

eliminates

y

, resulting in

2x = 14

, or

x = 7

.

Step-by-Step Solution

1

Factor the second equation using the difference of squares identity.

(x-y)(x+y) = 40

To rewrite the quadratic expression in a form where the linear equation can be substituted.

2

Substitute the first equation

x-y = 4

into the factored expression.

4(x+y) = 40

, which simplifies to

x+y = 10

To find a simpler linear relation for the sum of the variables.

3

Add the equations

x-y = 4

and

x+y = 10

.

2x = 14

To eliminate the variable

y

and solve for

x

directly.

4

Solve for

x

by dividing by 2.

x = 7

To isolate the variable and find the final value of

x

.

Key Concept

Solving a nonlinear system of equations by factoring a difference of squares and substituting.

Alternative Method

Express

x

from the first equation as

x = y + 4

and substitute it into the second equation:

(y+4)^2 - y^2 = 40

. Expanding and simplifying yields

y^2 + 8y + 16 - y^2 = 40

, which simplifies to

8y + 16 = 40 \implies 8y = 24 \implies y = 3

. Substituting

y = 3

back into

x = y + 4

gives

x = 7

.

Estimated Time:1m 15s

Question 4Question →

A system of two equations is shown below.

y = x^2 - 4

y = x - 2

Which of the following ordered pairs $(x, y)$ is a solution to the system?

(2, 0)

(0, -2)

(-2, 0)

(1, -3)

Show answer & explanation

Answer: $(2, 0)$

Answer

The ordered pair

(2, 0)

is the correct solution to the system.

The ordered pair

(2, 0)

is the correct solution because substituting these values into both equations in the system results in true statements. For the quadratic equation,

0 = 2^2 - 4

simplifies to

0 = 0

. For the linear equation,

0 = 2 - 2

simplifies to

0 = 0

.

Step-by-Step Solution

1

Substitute the

x

-value and

y

-value from the candidate solution

(2, 0)

into the first equation,

y = x^2 - 4

.

0 = 2^2 - 4 \implies 0 = 4 - 4 \implies 0 = 0

, which is a true statement.

An ordered pair must satisfy the first equation to be a candidate solution for the system.

2

Substitute the

x

-value and

y

-value from the candidate solution

(2, 0)

into the second equation,

y = x - 2

.

0 = 2 - 2 \implies 0 = 0

, which is also a true statement.

An ordered pair must satisfy all equations in the system simultaneously to be a valid solution.

Key Concept

Verifying a solution to a nonlinear system of equations by substituting the coordinate values into both equations.

Question 5Question →

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

y = x^2

y = x + 6

Show answer & explanation

Answer: 3

Answer

The correct answer is

3

.

Substituting

y = x^2

into

y = x + 6

gives the quadratic equation

x^2 - x - 6 = 0

. Factoring this expression yields

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

, which gives solutions of

x = 3

and

x = -2

. Since the system requires

x > 0

, the only valid solution is

3

.

Step-by-Step Solution

1

Substitute the expression for

y

from the first equation into the second equation.

x^2 = x + 6

To eliminate the variable

y

and solve for

x

directly.

2

Subtract

x

and

6

from both sides to write the quadratic equation in standard form.

x^2 - x - 6 = 0

Setting the quadratic expression equal to zero allows it to be factored.

3

Factor the quadratic trinomial.

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

Finding factors whose product is

-6

and whose sum is

-1

helps find the roots.

4

Solve for

x

and apply the constraint

x > 0

.

x = 3

The equation has solutions

x = 3

and

x = -2

. Because

x

must be greater than

0

, we discard the negative solution.

Key Concept

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

Question 6Question →

The graphs of the equations $y = x^2 - 10$ and $y = 2x - 2$ intersect at the point $(x, y)$ in the first quadrant. What is the value of $y$ ?

Show answer & explanation

Answer: 6

Answer

The value of

y

is 6.

Equating the equations gives

x^2 - 10 = 2x - 2

. Moving all terms to one side yields

x^2 - 2x - 8 = 0

, which factors as

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

. Since the point is in the first quadrant, both coordinates must be positive, so we use

x = 4

. Substituting

x = 4

into the linear equation gives

y = 2(4) - 2 = 6

.

Step-by-Step Solution

1

Equate the two expressions for

y

x^2 - 10 = 2x - 2

Since both equations define

y

, their right-hand sides must be equal at the points of intersection.

2

Rewrite the equation in standard quadratic form

ax^2 + bx + c = 0

x^2 - 2x - 8 = 0

Subtract

2x

and add

2

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

3

Factor the quadratic equation

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

Find two integers that multiply to

-8

and add to

-2

, which are

-4

and

2

.

4

Solve for the possible values of

x

x = 4

or

x = -2

Set each factor equal to zero and solve.

5

Determine the positive

x

-coordinate and find

y

y = 6

For the point to be in the first quadrant, both coordinates must be positive. Thus, we select

x = 4

and substitute it into the linear equation:

y = 2(4) - 2 = 6

.

Key Concept

Solving a system of linear and quadratic equations via substitution.

Alternative Method

Instead of factoring, the quadratic formula can be used to solve

x^2 - 2x - 8 = 0

:

x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}

. This yields

x = 4

and

x = -2

. Then substitute the positive root to find

y

.

Estimated Time:1m 30s

Question 7Question →

A system of equations is shown below.

y = x^2 - 2x

y = 3

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

2

3

6

4

Show answer & explanation

Answer: 6

Answer

6

To solve the system of equations, substitute the expression for

y

from the second equation into the first equation:

3 = x^2 - 2x

. Subtracting 3 from both sides results in the quadratic equation

x^2 - 2x - 3 = 0

. Factoring the quadratic yields

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

, which gives the possible values of

x

as

3

and

-1

. Since the problem specifies that

x > 0

, the value of

x

is 3. Given that

y = 3

, the value of

x + y

is

3 + 3 = 6

.

Step-by-Step Solution

1

Substitute the value of

y

from the second equation into the first equation.

3 = x^2 - 2x

Since both equations are equal to

y

, their right-hand sides must be equal to each other.

2

Rearrange the equation to set it equal to zero.

x^2 - 2x - 3 = 0

This puts the equation into standard quadratic form so that it can be factored.

3

Factor the quadratic equation.

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

Finding factors of -3 that add up to -2 helps isolate the solutions for

x

.

4

Find the solutions for

x

and apply the given constraint.

x = 3

(since

x > 0

)

The factors give

x = 3

and

x = -1

. The constraint

x > 0

excludes

x = -1

.

5

Calculate the value of

x + y

.

3 + 3 = 6

Substitute the value of

x = 3

and the given value of

y = 3

to find the final sum.

Key Concept

Solving nonlinear systems of equations using substitution and solving quadratic equations by factoring.

Question 8Question →

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 9Question →

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 10Question →

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 11Question →

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 12Question →

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 13Question →

\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 14Question →

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 15Question →

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 16Question →

The graphs of the equations $y - 2x = 3$ and $y = x^2$ intersect at two points in the $xy$ -plane. If $(x, y)$ represents an intersection point with a positive $x$ -coordinate, what is the value of $y$ ?

1

3

6

9

Show answer & explanation

Answer: 9

Answer

9

Substituting

y = x^2

into the first equation yields

x^2 - 2x = 3

. Setting the quadratic equation to zero gives

x^2 - 2x - 3 = 0

, which factors as

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

. This gives the solutions

x = 3

and

x = -1

. Because the question specifies a positive

x

-coordinate,

x

must be 3. Substituting

x = 3

back into

y = x^2

yields

y = 9

. Thus, the correct answer is 9.

Step-by-Step Solution

1

Substitute

y = x^2

into the equation

y - 2x = 3

.

x^2 - 2x = 3

This eliminates the variable

y

and creates a single quadratic equation in terms of

x

to find the

x

-coordinates of the intersection points.

2

Rewrite the quadratic equation in standard form.

x^2 - 2x - 3 = 0

Subtracting 3 from both sides sets the quadratic equation to 0, which is necessary for factoring.

3

Factor the quadratic equation to find its solutions.

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

, which gives

x = 3

or

x = -1

.

Factoring allows us to find the roots, which represent the

x

-coordinates of the intersection points.

4

Apply the constraint that the

x

-coordinate must be positive.

x = 3

The question specifies that

x > 0

, so we discard

x = -1

.

5

Substitute

x = 3

back into

y = x^2

to find the corresponding value of

y

.

y = 3^2 = 9

This determines the

y

-coordinate of the intersection point with the positive

x

-coordinate.

Key Concept

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

Question 17Question →

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

x - y = k

x^2 - 3xy + y^2 = 5

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

Show answer & explanation

Answer: 2

Answer

The value of

k

is 2.

Substituting

y = x - k

into the second equation yields

-x^2 + kx + k^2 - 5 = 0

, which can be rewritten in standard form as

x^2 - kx + (5 - k^2) = 0

. For this quadratic equation to have exactly one real solution, its discriminant must equal 0:

b^2 - 4ac = (-k)^2 - 4(1)(5 - k^2) = 5k^2 - 20 = 0

. Solving

5k^2 - 20 = 0

gives

k^2 = 4

, and since

k

must be positive,

k = 2

.

Step-by-Step Solution

1

Rearrange the first equation to express

y

in terms of

x

.

y = x - k

This allows for substitution into the second equation to eliminate

y

.

2

Substitute

y = x - k

into the second equation and expand.

x^2 - 3x(x - k) + (x - k)^2 = 5 \implies -x^2 + kx + k^2 - 5 = 0

To create a single quadratic equation in terms of

x

.

3

Multiply by

-1

to write the quadratic equation in standard form

ax^2 + bx + c = 0

.

x^2 - kx + (5 - k^2) = 0

Standard form makes it easier to identify the coefficients

a = 1

,

b = -k

, and

c = 5 - k^2

.

4

Set the discriminant

b^2 - 4ac

equal to 0.

(-k)^2 - 4(1)(5 - k^2) = 0

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

5

Simplify the discriminant equation and solve for

k

.

5k^2 - 20 = 0 \implies k^2 = 4 \implies k = 2

(since

k

must be positive)

To find the positive constant

k

that satisfies the condition.

Key Concept

Determining the number of solutions to a nonlinear system by substituting and setting the discriminant of the resulting quadratic equation to zero.

Alternative Method

Alternatively, one can rewrite the second equation by grouping:

x^2 - 3xy + y^2 = (x - y)^2 - xy = 5

. Since

x - y = k

, we have

k^2 - xy = 5

, so

xy = k^2 - 5

. We now have a system of

x - y = k

and

xy = k^2 - 5

. Substituting

y = x - k

gives

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

, leading to

x^2 - kx + (5 - k^2) = 0

, which can be solved using the discriminant as shown in the primary method.

Estimated Time:2m 30s

Question 18Question →

A parabola and a line intersect at exactly one point in the $xy$ -plane. The parabola is defined by the equation $y = -x^2 + 6x - 2$ and the line is defined by the equation $y = 2x + k$ , where $k$ is a constant. What is the value of $k$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

To find the value of

k

where the parabola and the line intersect at exactly one point, we equate the two equations:

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

. Rearranging this into standard quadratic form yields

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

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

b^2 - 4ac

, must be equal to zero. Substituting

a = 1

,

b = -4

, and

c = k + 2

into the discriminant formula gives

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

, which simplifies to

16 - 4k - 8 = 0

, or

8 - 4k = 0

. Solving for

k

gives

k = 2

.

Step-by-Step Solution

1

Equate the equations of the parabola and the line to set up an equation for the x-coordinates of their intersection points.

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

At the points of intersection, the y-values of both equations are equal.

2

Rearrange the equation into the standard quadratic form

ax^2 + bx + c = 0

.

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

This allows us to identify the coefficients

a = 1

,

b = -4

, and

c = k + 2

to apply the quadratic discriminant.

3

Set the discriminant

b^2 - 4ac

equal to zero.

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

A quadratic system has exactly one real solution (tangency) if and only if the discriminant of the resulting quadratic equation is zero.

4

Solve the linear equation for

k

.

k = 2

Simplifying the expression yields

16 - 4k - 8 = 0

, which simplifies to

8 - 4k = 0

, giving

k = 2

.

Key Concept

Nonlinear Systems of Equations

Question 19Question →

y = x^2 - 5x + 8

y = 2x + 2

The system of equations above has two real solutions, $(x_1, y_1)$ and $(x_2, y_2)$ . If $y_1 > y_2$ , what is the value of $x_1 - x_2$ ?

-5

5

7

10

Show answer & explanation

Answer: 5

Answer

The value of the difference between the two x-coordinates is 5.

To solve the system, substitute

y = 2x + 2

into

y = x^2 - 5x + 8

to obtain

2x + 2 = x^2 - 5x + 8

. Subtracting

2x + 2

from both sides gives the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation yields

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

, which gives the x-coordinates

x = 6

and

x = 1

. Substituting these back into

y = 2x + 2

gives the corresponding y-coordinates:

y = 14

when

x = 6

, and

y = 4

when

x = 1

. Thus, the two solutions are

(6, 14)

and

(1, 4)

. Because

y_1 > y_2

, we must have

(x_1, y_1) = (6, 14)

and

(x_2, y_2) = (1, 4)

. The value of

x_1 - x_2

is therefore

6 - 1 = 5

.

Step-by-Step Solution

1

Substitute the expression for

y

from the second equation into the first equation.

2x + 2 = x^2 - 5x + 8

This substitution reduces the system of two equations to a single quadratic equation in terms of

x

.

2

Rearrange the quadratic equation into standard form by subtracting

2x

and

2

from both sides.

x^2 - 7x + 6 = 0

Putting the equation in standard form is necessary to factor it and find its roots.

3

Factor the quadratic equation to find the two possible values of

x

.

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

, which gives

x = 6

or

x = 1

.

Factoring allows us to find the x-coordinates of the points where the two graphs intersect.

4

Substitute the x-values back into the linear equation

y = 2x + 2

to find their corresponding y-values.

For

x = 6

,

y = 2(6) + 2 = 14

. For

x = 1

,

y = 2(1) + 2 = 4

. The two solutions are

(6, 14)

and

(1, 4)

.

Finding the y-values helps identify which coordinate pair corresponds to

(x_1, y_1)

and which to

(x_2, y_2)

using the given condition.

5

Apply the condition

y_1 > y_2

to assign the variables and calculate

x_1 - x_2

.

Since

14 > 4

, the solution with the larger y-value is

(x_1, y_1) = (6, 14)

and the other is

(x_2, y_2) = (1, 4)

. Thus,

x_1 - x_2 = 6 - 1 = 5

.

This calculation yields the final requested value.

Key Concept

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

Question 20Question →

\begin{aligned} y &= x^2 - 5 \\ y &= 4x \end{aligned}

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

Show answer & explanation

Answer: 20

Answer

20

Substituting

y = 4x

into the equation

y = x^2 - 5

yields

4x = x^2 - 5

. Rearranging this equation into standard form gives

x^2 - 4x - 5 = 0

. Factoring the quadratic expression gives

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

, which means

x = 5

or

x = -1

. Substituting these values back into

y = 4x

gives the coordinates of the two solutions:

(5, 20)

and

(-1, -4)

. Since the question specifies that

y > 0

, the correct value is the positive

y

-value, which is 20.

Step-by-Step Solution

1

Substitute the expression for

y

from the second equation into the first equation.

4x = x^2 - 5

This eliminates

y

and creates a single equation in terms of

x

.

2

Subtract

4x

from both sides to write the quadratic equation in standard form.

x^2 - 4x - 5 = 0

Standard quadratic form

ax^2 + bx + c = 0

is required to solve by factoring.

3

Factor the quadratic equation.

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

Factoring helps find the values of

x

that satisfy the equation.

4

Solve for

x

by setting each factor to zero.

x = 5

or

x = -1

Applying the zero product property yields the possible

x

-coordinates of the solutions.

5

Calculate the corresponding

y

-values using the equation

y = 4x

.

For

x = 5

,

y = 20

. For

x = -1

,

y = -4

.

This determines the coordinates of the intersection points, which are

(5, 20)

and

(-1, -4)

.

6

Apply the constraint

y > 0

to identify the correct value of

y

.

y = 20

Since

-4

is not greater than zero, the only valid solution is

(5, 20)

, giving

y = 20

.

Key Concept

Solving a system of linear and quadratic equations using substitution.

Nonlinear Systems of Equations

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

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Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes