Intermediate Algebra Practice Questions — ACT — Page 14

Question 261Question →

A circular search-and-rescue radar zone centered at a local station is modeled by the equation $x^2 + y^2 = 25$ in the standard $(x, y)$ coordinate plane, where coordinates are measured in miles. A rescue helicopter flies along a straight path modeled by the line $y = 2x - 5$ . What is the distance, in miles, the helicopter travels through the radar zone?

2\sqrt{5}

4\sqrt{5}

8

10

10\sqrt{2}

Show answer & explanation

Answer: $4\sqrt{5}$

Answer

The distance the helicopter travels through the radar zone is

4\sqrt{5}

miles.

To find the distance the helicopter travels through the radar zone, we must determine the distance between the two points of intersection of the circular boundary

x^2 + y^2 = 25

and the line

y = 2x - 5

. Substituting the expression for

y

into the circular equation yields

x^2 + (2x-5)^2 = 25

. Expanding the binomial correctly gives

x^2 + 4x^2 - 20x + 25 = 25

, which simplifies to

5x^2 - 20x = 0

. Factoring this equation as

5x(x-4) = 0

gives

x = 0

and

x = 4

. Substituting these values back into the linear equation yields the points of intersection

(0, -5)

and

(4, 3)

. The distance between these two points is

\sqrt{(4-0)^2 + (3 - (-5))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}

miles.

Step-by-Step Solution

1

Substitute the linear equation

y = 2x - 5

into the circular equation

x^2 + y^2 = 25

.

x^2 + (2x - 5)^2 = 25

To find the coordinates of the intersection points where the helicopter's path meets the boundary of the radar zone.

2

Expand the binomial

(2x - 5)^2

and simplify the quadratic equation.

x^2 + 4x^2 - 20x + 25 = 25 \implies 5x^2 - 20x = 0

To collect like terms and put the equation in a solvable quadratic form.

3

Factor the quadratic equation

5x^2 - 20x = 0

to solve for

x

.

5x(x - 4) = 0 \implies x = 0 \text{ or } x = 4

To determine the

x

-coordinates of the two intersection points.

4

Determine the corresponding

y

-coordinates by substituting the

x

-values into the linear equation

y = 2x - 5

.

For

x = 0

:

y = 2(0) - 5 = -5 \implies (0, -5)

. For

x = 4

:

y = 2(4) - 5 = 3 \implies (4, 3)

.

To obtain the exact coordinate pairs for the entry and exit points.

5

Use the distance formula to calculate the distance between the two points

(0, -5)

and

(4, 3)

.

d = \sqrt{(4 - 0)^2 + (3 - (-5))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}

To compute the straight-line distance traveled by the helicopter through the radar zone.

Key Concept

Solving systems of linear and circular equations by substitution and finding the distance between their intersection points.

Question 262Question →

A right triangle has legs of length $x - 0.5$ inches and $2x$ inches, and a hypotenuse of length $2x + 0.5$ inches. What is the value of $x$ ?

1

1.5

2

3

4

Show answer & explanation

Answer: 3

Answer

3

Applying the Pythagorean theorem to the right triangle yields

(x - 0.5)^2 + (2x)^2 = (2x + 0.5)^2

. Expanding the terms gives

x^2 - x + 0.25 + 4x^2 = 4x^2 + 2x + 0.25

. Subtracting

4x^2

and

0.25

from both sides simplifies the equation to

x^2 - x = 2x

. Subtracting

2x

from both sides gives the standard quadratic equation

x^2 - 3x = 0

. Factoring this expression gives

x(x - 3) = 0

, which yields solutions

x = 0

and

x = 3

. Since the side length

x - 0.5

must be positive,

x

must be greater than

0.5

. Therefore, the only valid solution is

3

.

Step-by-Step Solution

1

Set up the equation using the Pythagorean theorem,

a^2 + b^2 = c^2

, with the given side lengths.

(x - 0.5)^2 + (2x)^2 = (2x + 0.5)^2

The Pythagorean theorem relates the legs and hypotenuse of any right triangle.

2

Expand each squared term algebraically.

(x^2 - x + 0.25) + 4x^2 = 4x^2 + 2x + 0.25

Expanding the binomials allows us to combine like terms and simplify the equation.

3

Subtract

4x^2

and

0.25

from both sides of the equation.

x^2 - x = 2x

Simplifying the equation makes it easier to solve.

4

Move all terms to the left side to write the quadratic equation in standard form.

x^2 - 3x = 0

A quadratic equation must be set to zero to be solved by factoring.

5

Factor the quadratic expression.

x(x - 3) = 0

Factoring allows us to find the roots of the equation.

6

Solve for

x

and choose the value that makes all side lengths positive.

x = 3

(since

x = 0

is not a valid length because a side length

x - 0.5

must be greater than

0

)

Only a positive value of

x

greater than

0.5

yields physically possible side lengths for the triangle.

Key Concept

Setting up and solving quadratic equations using algebraic expansion and the Pythagorean theorem

Estimated Time:1m 30s

Question 263Question →

A projectile is launched vertically upward from an initial height of $5$ meters. Its height, $h(t)$ in meters, $t$ seconds after launch is given by the function $h(t) = -4.9t^2 + 19.6t + 5$ . To the nearest tenth of a second, how many seconds after launch does the projectile reach a height of $15$ meters on its way down?

Show answer & explanation

Answer: 3.4

Answer

To the nearest tenth of a second, the projectile reaches a height of

15

meters on its way down at

3.4

seconds.

The correct answer is

3.4

seconds. Setting the height equation

h(t) = 15

yields

-4.9t^2 + 19.6t - 10 = 0

. Solving this quadratic equation via the quadratic formula gives two solutions:

t \approx 0.6

seconds and

t \approx 3.4

seconds. The projectile travels upward first, passing the

15

-meter mark at

0.6

seconds, and then descends, passing the

15

-meter mark again at

3.4

seconds.

Step-by-Step Solution

1

Set up the quadratic equation by setting the height function

h(t)

equal to

15

.

-4.9t^2 + 19.6t + 5 = 15

This allows us to find the specific values of time

t

when the height of the projectile is exactly

15

meters.

2

Rearrange the quadratic equation into the standard form

at^2 + bt + c = 0

by subtracting

15

from both sides.

-4.9t^2 + 19.6t - 10 = 0

Writing the equation in standard form is necessary before applying the quadratic formula.

3

Substitute the coefficients

a = -4.9

,

b = 19.6

, and

c = -10

into the quadratic formula.

t = \frac{-19.6 \pm \sqrt{(19.6)^2 - 4(-4.9)(-10)}}{2(-4.9)}

Since the quadratic equation has non-integer decimal coefficients, using the quadratic formula is the most reliable method to solve for the roots.

4

Simplify the discriminant and calculate the two values of

t

.

t \approx 0.6

and

t \approx 3.4

The discriminant is

19.6^2 - 196 = 188.16

. Taking the square root gives

\sqrt{188.16} \approx 13.72

, resulting in two real roots.

5

Determine which root corresponds to the projectile's motion on the way down.

t \approx 3.4

seconds

The smaller root (

0.6

seconds) represents the first time the projectile reaches

15

meters while ascending. The larger root (

3.4

seconds) represents the time the projectile passes

15

meters while descending.

Key Concept

Solving quadratic equations with decimal coefficients using the quadratic formula and interpreting the physical context of the roots.

Question 264Question →

What are the solutions for $x$ in the quadratic equation $0.2x^2 - 0.6x + 2.2 = 0$ ?

\frac{3 \pm \sqrt{35}}{2}

\frac{3 \pm \sqrt{55}}{2}

\frac{3 \pm i\sqrt{35}}{2}

\pm\frac{i\sqrt{35}}{2}

\frac{15 \pm 2i\sqrt{10}}{10}

Show answer & explanation

Answer: $\frac{3 \pm i\sqrt{35}}{2}$

Answer

The correct answer is

\frac{3 \pm i\sqrt{35}}{2}

.

The correct answer is the pair of complex solutions

\frac{3 \pm i\sqrt{35}}{2}

. By multiplying the equation

0.2x^2 - 0.6x + 2.2 = 0

by

5

, we get the equivalent equation with integer coefficients,

x^2 - 3x + 11 = 0

. Applying the quadratic formula with

a = 1

,

b = -3

, and

c = 11

yields a discriminant of

9 - 44 = -35

. Since the discriminant is negative, the solutions are complex:

\frac{3 \pm i\sqrt{35}}{2}

.

Step-by-Step Solution

1

Multiply both sides of the quadratic equation by

5

to eliminate the decimal coefficients.

The equation

0.2x^2 - 0.6x + 2.2 = 0

becomes

x^2 - 3x + 11 = 0

.

Working with integer coefficients simplifies algebraic manipulation and reduces the risk of calculation errors.

2

Identify the coefficients

a

,

b

, and

c

in the standard quadratic form

ax^2 + bx + c = 0

.

a = 1

,

b = -3

, and

c = 11

.

These values are required to apply the quadratic formula.

3

Substitute the coefficients into the quadratic formula,

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

.

x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(11)}}{2(1)} = \frac{3 \pm \sqrt{9 - 44}}{2} = \frac{3 \pm \sqrt{-35}}{2}

.

The quadratic formula provides the exact solutions for any quadratic equation.

4

Simplify the radical using the imaginary unit,

i = \sqrt{-1}

.

\sqrt{-35} = \sqrt{35} \cdot \sqrt{-1} = i\sqrt{35}

, so the solutions are

x = \frac{3 \pm i\sqrt{35}}{2}

.

Standard mathematical notation represents the square root of a negative number using

i

.

Key Concept

Using the quadratic formula to solve quadratic equations with decimal coefficients and complex roots.

Question 265Question →

For the quadratic equation $0.4x^2 + bx - 4.8 = 0$ , where $b$ is a constant, the sum of the two solutions is equal to the product of the two solutions. What is the value of $b$ ?

Show answer & explanation

Answer: 4.8

Answer

The value of the constant

b

is

4.8

.

According to Vieta's formulas, the sum of the solutions to the quadratic equation

ax^2 + bx + c = 0

is given by

-\frac{b}{a}

and their product is given by

\frac{c}{a}

. For the given equation

0.4x^2 + bx - 4.8 = 0

, the product of the solutions is

\frac{-4.8}{0.4} = -12

. Setting the sum equal to the product yields the equation

-\frac{b}{0.4} = -12

. Multiplying both sides by

-0.4

isolates

b

, giving

b = 4.8

.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation.

a = 0.4

,

b = b

, and

c = -4.8

.

To apply formulas relating the coefficients to the solutions.

2

Express the sum and product of the solutions using Vieta's formulas.

Sum of solutions is

-\frac{b}{0.4}

and product of solutions is

\frac{-4.8}{0.4} = -12

.

To establish the mathematical relationship given in the problem.

3

Equate the sum and product of the solutions and solve for the constant

b

.

-\frac{b}{0.4} = -12 \implies b = -12 \times (-0.4) = 4.8

.

The problem states that the sum of the two solutions is equal to their product.

Key Concept

Sum and Product of Roots (Vieta's Formulas)

Question 266Question →

A circle and a line are graphed in the standard $(x, y)$ coordinate plane. The equations of the circle and the line are given by:

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

y = x + 1

The line intersects the circle at two points, $P$ and $Q$ . What is the distance between $P$ and $Q$ ?

2\sqrt{10}

3\sqrt{2}

\sqrt{22}

7\sqrt{2}

\sqrt{5}

Show answer & explanation

Answer: $3\sqrt{2}$

Answer

The distance between the intersection points is

3\sqrt{2}

.

To find the points of intersection, substitute the linear equation into the circle's equation. This results in the quadratic equation

x^2 - 5x + 4 = 0

, which yields the solutions

x = 1

and

x = 4

. Substituting these values into the linear equation gives the points

(1, 2)

and

(4, 5)

. The distance between these points is computed using the distance formula, which gives

3\sqrt{2}

.

Step-by-Step Solution

1

Substitute the equation of the line into the equation of the circle.

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

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

x

.

2

Expand the squared binomials and simplify the quadratic equation.

(x^2 - 6x + 9) + (x^2 - 4x + 4) = 5 \implies 2x^2 - 10x + 13 = 5 \implies 2x^2 - 10x + 8 = 0 \implies x^2 - 5x + 4 = 0

Expanding the terms allows us to combine like terms and set the quadratic equation to zero.

3

Factor the quadratic equation to find the

x

-coordinates of the intersection points.

(x-1)(x-4) = 0 \implies x = 1 \text{ or } x = 4

Factoring is the most direct method to solve the simplified quadratic equation.

4

Find the corresponding

y

-coordinates by substituting the

x

-values back into the linear equation

y = x + 1

.

\text{For } x = 1: y = 1 + 1 = 2 \implies P(1, 2)

\text{For } x = 4: y = 4 + 1 = 5 \implies Q(4, 5)

This determines the coordinates of the two intersection points.

5

Use the distance formula to find the distance between the two points

P(1, 2)

and

Q(4, 5)

.

d = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}

The distance formula calculates the straight-line distance between the two coordinates.

Key Concept

Solving systems of linear and quadratic equations by substitution and finding the distance between intersection points.

Alternative Method

Find the distance geometrically: The center of the circle is

(3, 3)

and the radius is

r = \sqrt{5}

. The distance

d

from the center to the line

x - y + 1 = 0

is

d = \frac{|3 - 3 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}

. Using a right triangle formed by the radius, the distance from the center, and half of the chord length

h

, we have

h = \sqrt{r^2 - d^2} = \sqrt{5 - \frac{1}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}

. The total distance between the intersection points is the full chord length,

2h = 2 \times \frac{3}{\sqrt{2}} = 3\sqrt{2}

.

Estimated Time:3m 0s

Question 267Question →

For the imaginary unit $i$ , where $i^2 = -1$ , which of the following complex numbers is equal to $\frac{8 + i}{3 + 2i}$ ?

\frac{22}{13} - i

\frac{26}{5} - \frac{13}{5}i

2 + i

2 - i

\frac{8}{3} + \frac{1}{2}i

Show answer & explanation

Answer: 2 - i

Answer

The complex number

2 - i

To divide two complex numbers, we multiply both the numerator and denominator by the complex conjugate of the denominator, which is

3 - 2i

. Expanding the numerator gives

(8 + i)(3 - 2i) = 24 - 16i + 3i - 2i^2 = 26 - 13i

because

i^2 = -1

. Expanding the denominator yields

(3 + 2i)(3 - 2i) = 9 - 4i^2 = 9 + 4 = 13

. Dividing the terms of the numerator by the denominator gives

\frac{26 - 13i}{13} = 2 - i

.

Step-by-Step Solution

1

Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator.

\frac{8 + i}{3 + 2i} \cdot \frac{3 - 2i}{3 - 2i} = \frac{(8 + i)(3 - 2i)}{(3 + 2i)(3 - 2i)}

Multiplying by the conjugate rationalizes the denominator, converting it into a real number.

2

Expand the numerator and the denominator using binomial multiplication.

Numerator:

(8 + i)(3 - 2i) = 24 - 16i + 3i - 2i^2

Denominator:

(3 + 2i)(3 - 2i) = 9 - 6i + 6i - 4i^2 = 9 - 4i^2

Distribute each term in the first binomial to each term in the second binomial.

3

Substitute

i^2 = -1

and simplify both expressions.

Numerator:

24 - 13i - 2(-1) = 24 - 13i + 2 = 26 - 13i

Denominator:

9 - 4(-1) = 9 + 4 = 13

The definition of the imaginary unit is

i^2 = -1

.

4

Divide each term of the simplified numerator by the simplified denominator.

\frac{26 - 13i}{13} = \frac{26}{13} - \frac{13i}{13} = 2 - i

Separate the real and imaginary parts to write the complex number in standard form

a + bi

.

Key Concept

Division of complex numbers using the complex conjugate of the denominator.

Alternative Method

Instead of dividing directly, let the result be

x + yi

. Then

(x + yi)(3 + 2i) = 8 + i

. Expanding this gives

(3x - 2y) + (2x + 3y)i = 8 + i

. Equating the real and imaginary parts gives the system of equations

3x - 2y = 8

and

2x + 3y = 1

. Solving this system yields

x = 2

and

y = -1

, which corresponds to the complex number

2 - i

.

Estimated Time:1m 30s

Question 268Question →

A small drone's path in a vertical plane is modeled by the equation $y = 3x^2 - 4x + 2$ , where $x$ is the horizontal distance in meters and $y$ is the height in meters. A laser beam travels along a straight line in the same plane such that the sum of twice its horizontal distance and its height is a constant $c$ , where both are in meters. The laser beam intersects the drone's path at two distinct points. If the distance between these two intersection points is $\frac{5\sqrt{2}}{3}$ meters, what is the value of $2c$ ?

6

10

5

-\frac{1}{3}

\frac{5}{3}

Show answer & explanation

Answer: $5$

Answer

The correct value of

2c

is

5

.

The correct value of

2c

is

5

. Substituting the linear equation

y = c - 2x

into the quadratic equation

y = 3x^2 - 4x + 2

yields

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

. The distance between the intersection points is

d = |x_2 - x_1|\sqrt{5} = \frac{5\sqrt{2}}{3}

, which simplifies to

(x_2 - x_1)^2 = \frac{10}{9}

. Using the identity

(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2

and Vieta's formulas, we find the equation

\frac{4}{9} - \frac{8-4c}{3} = \frac{10}{9}

. Solving this equation yields

c = \frac{5}{2}

, and thus

2c = 5

.

Step-by-Step Solution

1

Set up the system of equations.

The drone's path is

y = 3x^2 - 4x + 2

and the laser's path is

2x + y = c \implies y = -2x + c

.

This represents the mathematical formulation of both paths in the vertical plane.

2

Equate the equations to find the x-coordinates of the intersection points.

3x^2 - 4x + 2 = -2x + c \implies 3x^2 - 2x + (2-c) = 0

.

The intersection points satisfy both equations, so we can solve for

x

by substitution.

3

Express the distance between the intersection points

P(x_1, y_1)

and

Q(x_2, y_2)

using the slope.

d = |x_2 - x_1|\sqrt{5}

.

Since the points lie on the line with slope

-2

, we have

y_2 - y_1 = -2(x_2 - x_1)

. The distance formula becomes

d = \sqrt{(x_2 - x_1)^2 + (-2(x_2 - x_1))^2} = \sqrt{5(x_2 - x_1)^2} = |x_2 - x_1|\sqrt{5}

.

4

Equate the distance expression to the given distance to find

(x_2 - x_1)^2

.

(x_2 - x_1)^2 = \frac{10}{9}

.

We are given

d = \frac{5\sqrt{2}}{3}

. Setting

|x_2 - x_1|\sqrt{5} = \frac{5\sqrt{2}}{3}

and squaring both sides gives

5(x_2 - x_1)^2 = \frac{50}{9}

, which simplifies to

(x_2 - x_1)^2 = \frac{10}{9}

.

5

Apply Vieta's formulas and the algebraic identity for

(x_2 - x_1)^2

.

\frac{10}{9} = \frac{4}{9} - \frac{8-4c}{3}

.

For the quadratic equation

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

, we have

x_1 + x_2 = \frac{2}{3}

and

x_1 x_2 = \frac{2-c}{3}

. We use the identity

(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2

.

6

Solve for

c

and find

2c

.

c = \frac{5}{2} \implies 2c = 5

.

Multiplying the equation by

9

gives

10 = 4 - 3(8-4c) \implies 10 = -20 + 12c \implies 12c = 30 \implies c = \frac{5}{2}

. Therefore,

2c = 5

.

Key Concept

Solving systems of linear and non-linear equations using substitution, coordinate geometry distance formula, and quadratic root relationships.

Estimated Time:3m 0s

Question 269Question →

If $x$ is a real number such that $\log_5(x) + \log_5(x - 20) = 3$ , what is the value of $x$ ?

Show answer & explanation

Answer: 25

Answer

The value of

x

is 25.

Applying the logarithmic product rule simplifies the equation to

\log_5(x^2 - 20x) = 3

. Writing this in exponential form yields

x^2 - 20x = 125

. Rearranging into standard form gives

x^2 - 20x - 125 = 0

, which factors into

(x - 25)(x + 5) = 0

. This gives potential solutions of

25

and

-5

. Because the logarithmic arguments must be strictly positive,

x = -5

is extraneous. Therefore, the only correct value is 25.

Step-by-Step Solution

1

Use the product property of logarithms to combine the terms on the left side.

\log_5(x(x - 20)) = 3

The sum of logarithms with the same base is equal to the logarithm of the product of their arguments:

\log_b(M) + \log_b(N) = \log_b(MN)

.

2

Rewrite the logarithmic equation in exponential form.

x(x - 20) = 5^3 \implies x^2 - 20x = 125

The logarithmic equation

\log_b(y) = c

is equivalent to the exponential equation

b^c = y

.

3

Rearrange the quadratic equation into standard form and solve by factoring.

x^2 - 20x - 125 = 0 \implies (x - 25)(x + 5) = 0 \implies x = 25 \text{ or } x = -5

Subtracting 125 from both sides sets the quadratic equation to 0, which can then be factored into binomials whose product is 0.

4

Verify the potential solutions in the original equation to identify any extraneous roots.

For

x = -5

, the arguments of the original logarithms are negative, which is undefined. For

x = 25

, the arguments are positive. Thus, the only valid solution is

x = 25

.

Logarithmic functions are only defined for positive real numbers. Therefore, we must have

x > 0

and

x - 20 > 0

, which requires

x > 20

.

Key Concept

Solving logarithmic equations by combining logarithmic terms and checking for extraneous solutions.

Estimated Time:1m 30s

Question 270Question →

If $9^{2x - 1} = 27^{x + 4}$ , what is the value of $x$ ?

5

-13

14

6

-4

Show answer & explanation

Answer: 14

Answer

14

Converting both bases to 3 yields

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

. Applying the exponent power rule gives

3^{4x - 2} = 3^{3x + 12}

. Equating the exponents results in

4x - 2 = 3x + 12

. Subtracting

3x

from both sides gives

x - 2 = 12

, and adding 2 to both sides results in

x = 14

. This matches the correct value of 14.

Step-by-Step Solution

1

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

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

Since

9 = 3^2

and

27 = 3^3

, rewriting the bases allows us to equate the exponents later.

2

Apply the power of a power property,

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

, to simplify the exponents on both sides.

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

which simplifies to

3^{4x - 2} = 3^{3x + 12}

To simplify an exponent raised to another power, multiply the exponents, ensuring the multiplier is distributed to both terms inside each exponent expression.

3

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

x

.

4x - 2 = 3x + 12 \implies x = 14

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

Key Concept

Solving exponential equations by expressing bases with a common base and equating the exponents.

Estimated Time:1m 30s

Question 271Question →

The circle $x^2 + y^2 = 25$ and the line $y = 2x - 5$ intersect at two points. What is the sum of the $y$ -coordinates of these two intersection points?

Show answer & explanation

Answer: -2

Answer

The sum of the

y

-coordinates of the intersection points is

-2

.

Substituting

y = 2x - 5

into the circular equation

x^2 + y^2 = 25

yields the quadratic equation

5x^2 - 20x = 0

. Factoring this equation gives

x = 0

and

x = 4

. Evaluating the linear equation at these values gives the

y

-coordinates

-5

and

3

. The sum of these coordinates is

-5 + 3 = -2

.

Step-by-Step Solution

1

Substitute

y = 2x - 5

into the circle equation

x^2 + y^2 = 25

.

x^2 + (2x - 5)^2 = 25

To find the points of intersection, we solve the system of equations by substitution.

2

Expand and simplify the resulting equation.

5x^2 - 20x = 0

Expanding

(2x - 5)^2

gives

4x^2 - 20x + 25

. Combining like terms and subtracting 25 from both sides simplifies the equation.

3

Factor the quadratic equation to solve for

x

.

x = 0

or

x = 4

Factoring out

5x

gives

5x(x - 4) = 0

, which yields the roots

x = 0

and

x = 4

.

4

Find the corresponding

y

-coordinates by substituting the

x

-values back into

y = 2x - 5

.

The intersection points are

(0, -5)

and

(4, 3)

.

For

x = 0

,

y = 2(0) - 5 = -5

. For

x = 4

,

y = 2(4) - 5 = 3

.

5

Calculate the sum of the

y

-coordinates.

-2

Adding the

y

-coordinates

-5

and

3

gives

-5 + 3 = -2

.

Key Concept

Systems of Linear and Non-Linear Equations

Question 272Question →

Consider the system of equations consisting of the quadratic function $f(x) = (x - 2)^2 - 3$ and the linear function $g(x) = 3x - 9$ . If the graphs of these functions intersect at two distinct points, $(x_1, y_1)$ and $(x_2, y_2)$ , what is the value of $x_1 y_2 + x_2 y_1$ ?

-15

-8

-3

15

24

Show answer & explanation

Answer: -3

Answer

The correct answer is

-3

. The intersection points of the two functions are

(2, -3)

and

(5, 6)

, and evaluating the expression

x_1 y_2 + x_2 y_1

gives

-3

.

To find the points where the graphs of the functions intersect, we set their expressions equal to each other:

(x - 2)^2 - 3 = 3x - 9

. Expanding the squared term gives

x^2 - 4x + 4 - 3 = 3x - 9

, which simplifies to

x^2 - 4x + 1 = 3x - 9

. Moving all terms to the left side yields

x^2 - 7x + 10 = 0

. Factoring this quadratic equation gives

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

, so the

x

-coordinates of the intersection points are

2

and

5

. Substituting these back into the linear equation gives the corresponding

y

-coordinates:

y = 3(2) - 9 = -3

and

y = 3(5) - 9 = 6

, resulting in the intersection points

(2, -3)

and

(5, 6)

. Finally, evaluating the requested expression gives

(2)(6) + (5)(-3) = 12 - 15 = -3

.

Step-by-Step Solution

1

Set the quadratic and linear functions equal to find the

x

-coordinates of their intersection points.

(x - 2)^2 - 3 = 3x - 9

At the points of intersection, the values of

f(x)

and

g(x)

must be equal.

2

Expand the binomial squared term

(x - 2)^2

.

x^2 - 4x + 4 - 3 = 3x - 9

Expanding the binomial is necessary to combine like terms and write the equation in standard quadratic form.

3

Move all terms to one side to set the equation to zero.

x^2 - 7x + 10 = 0

A quadratic equation must be set to zero before factoring or applying the quadratic formula.

4

Factor the quadratic equation.

(x - 2)(x - 5) = 0 \implies x_1 = 2 \text{ and } x_2 = 5

Factoring determines the

x

-coordinates of the intersection points.

5

Substitute the

x

-values into the linear equation

g(x) = 3x - 9

to find the corresponding

y

-coordinates.

y_1 = 3(2) - 9 = -3

y_2 = 3(5) - 9 = 6

This yields the two intersection points:

(2, -3)

and

(5, 6)

.

6

Calculate the value of the expression

x_1 y_2 + x_2 y_1

.

(2)(6) + (5)(-3) = 12 - 15 = -3

This evaluates the requested secondary value using the intersection coordinates.

Key Concept

Solving systems of linear and non-linear equations by substitution and algebraic manipulation.

Intermediate Algebra

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