Step-by-Step Guidance for MAT 2240 Section 1.2 Homework (Precalculus/Algebra)

Q1. Solve the equation: $|5x + 8| = |3 - x|$

Background

Topic: Absolute Value Equations

This question tests your understanding of how to solve equations involving absolute values, which often require considering multiple cases based on the sign of the expressions inside the absolute value.

Key Terms and Formulas:

• Absolute value: $|a|$ is the distance of $a$ from zero on the number line, always non-negative.

• To solve $|A| = |B|$, consider the cases $A = B$ and $A = -B$.

Step-by-Step Guidance

1. Set up two cases: $5x + 8 = 3 - x$ and $5x + 8 = -(3 - x)$.

2. For the first case, solve for $x$ by isolating $x$ on one side.

3. For the second case, expand and solve for $x$.

4. Check each solution in the original equation to ensure it satisfies the absolute value condition.

Try solving on your own before revealing the answer!

Final Answer:

The solutions are $x = - rac{5}{3}$ and $x = - rac{11}{6}$.

Both values satisfy the original equation when substituted back in.

Q2. Solve the inequality and write the solution using interval notation: $|6x + 5| > 3$

Background

Topic: Absolute Value Inequalities

This question tests your ability to solve inequalities involving absolute values, which require splitting the inequality into two cases.

Key Terms and Formulas:

• For $|A| > B$ (where $B > 0$), the solution is $A > B$ or $A < -B$.

• Interval notation: Expressing the solution as a range of values.

Step-by-Step Guidance

1. Set up two cases: $6x + 5 > 3$ and $6x + 5 < -3$.

2. Solve each inequality for $x$.

3. Express the solution in interval notation, combining the two intervals.

Try solving on your own before revealing the answer!

Final Answer:

The solution is $x > - rac{1}{3}$ or $x < - rac{4}{3}$, written in interval notation as $(-\infty, -\frac{4}{3}) \cup (-\frac{1}{3}, \infty)$.

Q3. Find and state the discriminant. Determine whether the solution(s) of the quadratic equation are real or complex. Then solve the equation and write the answer(s) in exact simplified form: $5x^2 + 3x - 1 = 0$

Background

Topic: Quadratic Equations and Discriminant

This question tests your ability to use the quadratic formula and discriminant to determine the nature of the roots and solve for them.

Key Terms and Formulas:

• Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Discriminant: $D = b^2 - 4ac$

• If $D > 0$, roots are real and distinct; if $D = 0$, real and equal; if $D < 0$, complex.

Step-by-Step Guidance

1. Identify $a = 5$, $b = 3$, $c = -1$.

2. Calculate the discriminant: $D = b^2 - 4ac$.

3. Determine if $D$ is positive, zero, or negative to classify the roots.

4. Set up the quadratic formula with the values for $a$, $b$, and $D$.

Try solving on your own before revealing the answer!

Final Answer:

The discriminant is $D = 3^2 - 4 \cdot 5 \cdot (-1) = 9 + 20 = 29$ (real and distinct roots).

The solutions are $x = \frac{-3 + \sqrt{29}}{10}$ and $x = \frac{-3 - \sqrt{29}}{10}$.

Q4a. Find an equation that relates weight measured in kilograms ($x$) to weight measured in pounds ($y$).

Background

Topic: Linear Equations and Unit Conversion

This question tests your ability to set up a proportional relationship between two units.

Key Terms and Formulas:

• Proportional relationship: $y = kx$

• Given: $1$kg$= 2.20$ lb

Step-by-Step Guidance

1. Set up the equation $y = kx$ where $k$ is the conversion factor.

2. Use the given information to determine $k$.

Try solving on your own before revealing the answer!

Final Answer:

The equation is $y = 2.20x$.

This means the weight in pounds is $2.20$ times the weight in kilograms.

Q4b. Use the equation found in part (a) to convert 164 pounds to kilograms rounded to two decimal places.

Background

Topic: Unit Conversion

This question tests your ability to rearrange and use a conversion equation to solve for the unknown.

Key Terms and Formulas:

• From part (a): $y = 2.20x$

• To solve for $x$ (kilograms): $x = \frac{y}{2.20}$

Step-by-Step Guidance

1. Set $y = 164$ lb in the equation.

2. Rearrange to solve for $x$: $x = \frac{164}{2.20}$.

3. Calculate the value and round to two decimal places.

Try solving on your own before revealing the answer!

Final Answer:

$x = \frac{164}{2.20} \approx 74.55$ kg

So, 164 pounds is approximately 74.55 kilograms.

Q5. Determine the equation of the line that passes through the points $(4, -3)$ and $(-8, 2)$. Write the equation in standard form.

Background

Topic: Linear Equations

This question tests your ability to find the equation of a line given two points and write it in standard form.

Key Terms and Formulas:

• Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$

• Point-slope form: $y - y_1 = m(x - x_1)$

• Standard form: $Ax + By = C$

Step-by-Step Guidance

1. Calculate the slope $m$ using the two points.

2. Use the point-slope form with one of the points.

3. Rearrange the equation to standard form $Ax + By = C$.

Try solving on your own before revealing the answer!

Final Answer:

The equation in standard form is $5x + 6y = 2$.

This line passes through both given points.

Q6. 300 cells are being monitored in a laboratory. The cells are continuously growing according to the growth model $P(t) = 300e^{0.32t}$, where $P$ is the number of cells and $t$ is time measured in days. How many days will it take for there to be 880 cells rounded to one decimal place?

Background

Topic: Exponential Growth Models

This question tests your ability to solve exponential equations for time, using logarithms.

Key Terms and Formulas:

• Exponential growth: $P(t) = P_0 e^{rt}$

• To solve for $t$: $t = \frac{1}{r} \ln\left(\frac{P}{P_0}\right)$

Step-by-Step Guidance

1. Set $P(t) = 880$ and $P_0 = 300$.

2. Write the equation: $880 = 300e^{0.32t}$.

3. Divide both sides by 300 to isolate the exponential term.

4. Take the natural logarithm of both sides to solve for $t$.

5. Set up the formula for $t$ and prepare to calculate.

Try solving on your own before revealing the answer!

Final Answer:

$t = \frac{1}{0.32} \ln\left(\frac{880}{300}\right) \approx 3.5$ days

It will take approximately 3.5 days for the cell population to reach 880.

Q7. Solve the equation for the exact answer. Then round the answer to two decimal places: $\ln(4x) - \ln(x - 10) = 7$

Background

Topic: Logarithmic Equations

This question tests your ability to use properties of logarithms to solve for $x$.

Key Terms and Formulas:

• Logarithm subtraction: $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$

• Exponentiation: $\ln(A) = B \implies A = e^B$

Step-by-Step Guidance

1. Combine the logarithms: $\ln\left(\frac{4x}{x - 10}\right) = 7$.

2. Exponentiate both sides to remove the logarithm: $\frac{4x}{x - 10} = e^7$.

3. Set up the equation and solve for $x$.

4. Prepare to calculate the value and round to two decimal places.

Try solving on your own before revealing the answer!

Final Answer:

$x = \frac{10e^7}{4 - e^7} \approx 10.01$

The exact answer is $x = \frac{10e^7}{4 - e^7}$, and rounded to two decimal places, $x \approx 10.01$.

Q8. Simplify the expression and write it in the standard form $a + bi$: $(x + 20i)(x - 20i)$

Background

Topic: Complex Numbers and Multiplication

This question tests your ability to multiply complex numbers and express the result in standard form.

Key Terms and Formulas:

• Standard form: $a + bi$

• Imaginary unit: $i^2 = -1$

• Difference of squares: $(a + b)(a - b) = a^2 - b^2$

Step-by-Step Guidance

1. Apply the difference of squares formula: $(x + 20i)(x - 20i) = x^2 - (20i)^2$.

2. Calculate $(20i)^2$ using $i^2 = -1$.

3. Write the result in the form $a + bi$.

Try solving on your own before revealing the answer!

Final Answer:

$(x + 20i)(x - 20i) = x^2 + 400$

The result is a real number, so the standard form is $x^2 + 0i$.