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Precalculus MATH F156X Summer 2026

Dr. Latrice Bowman

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Section BF-1 Linear and Quadratic Equations

Worksheet BF-1 Quick Notes

For this outcome we will be looking at one variable equations that are reducible to either a linear or a quadratic form. Specifically, for this outcome, students should be able to:
  • Determine if an equation is linear or quadratic
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  • If it is a linear equation:
    • Use correct mathematical structure to write out steps progressing to a solution.
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    • Use algebraic methods to remove grouping symbols and combine like terms.
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    • Simplify algebraic expressions to obtain the solution.
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  • If it is a quadratic equation:
    • Use correct mathematical structure to write out steps progressing to a solution.
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    • Determine the most appropriate method for solving the quadratic, if a specific method is not requested.
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    • Use algebraic methods to rewrite the quadratic expression (factoring, completing the square or setting up the quadratic formula.
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    • Simplify algebraic expressions to obtain the solution.
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Linear Equations
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An algebraic equation is a mathematical statement that relates two algebraic expressions involving at lest one variable.
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The solution set for an equation is the set of elements in the domain of the variable that make the equation true. (Each element of the solution set is a solution or root)
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An equation is called an identity if the equation is true for all elements in the domain of the variable. An equation is called conditional if it is true for certain domain values and false for others. Two equations are equivalent if the solution set is the same for both equations.
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Properties of Equality: For any real number a, b and c, we have the following properties
  1. Addition Property:\(\text{If }a=b\text{ then }a+c=b+c\)
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  2. Subtraction Property:\(\text{If }a=b\text{ then }a-c=b-c\)
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  3. Multiplication Property:\(\text{If }a=b\text{ then }ac=bc\)
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    (We assume that here \(c \ne 0\)
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  4. Division Property:\(\text{If }a=b\text{ then }\frac{a}{c}=\frac{b}{c}\)
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    (We assume that here \(c\ne 0\)
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  5. Substitution Property: \(\text{If }a=b\) then either can be used to replace the other in any statement without changing the truth value of the statement.
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Any equation that can be written in the form is considered a linear, first-degree equation in one variable.
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Absolute Value Equations
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The distance on the number line between zero and a number, \(x\) is the absolute value of x, denoted \(|x|\text{.}\)
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Note: Absolute value of a number is never negative, as it represents a distance.
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When solving absolute value equations we need to consider when the distance is both positive and negative. So if we had an equation \(|ax+b|=c\) we can rewrite this as two equations, by dropping the absolute value and setting the expression equal to the positive and negative distance.
\begin{equation*} ax+b=c \text{ and }ax+b=-c \end{equation*}
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Quadratic Equations
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Quadratic equations can be written as \(ax^2+bx+c=0\) where \(a, b,\) and \(c\) are constants and \(a\ne0\text{.}\) There are four different methods that can be used to solve a quadratic equation.
  1. Factoring:
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    Here the idea is to move all terms to one side of the equal sign and set the other side equal to \(0\text{.}\) Factoring the terms we want to try and get either \(x(ax+b)=0\) or \((ax+b)(cx+d)=0\text{.}\) These two forms will allow us to use the Zero Property of Equality to solve. This property says: If \(m(n)=0\text{ then either }m=0\text{ or }n=0\text{.}\)
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  2. Extraction of Roots:
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    This method is most useful if our quadratic is in the form \((ax+b)^2=c\text{.}\) Here we can take the square root of both sides to "extract the roots" from \(c\text{.}\) We can then solve the two resulting linear equations.
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  3. Quadratic Formula:
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    Given \(ax^2+bx+c=0\) we can use the formula \(x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\) to find the values.
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    \(b^2-4ac\) is known as the discriminant which tells the types of solutions you should expect from the equation.
    1. if \(b^2-4ac \gt 0\) then there are two distinct real roots.
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    2. if \(b^2-4ac=0\) then there one real double root.
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    3. if \(b^2-4ac \lt 0\) then there are two imaginary roots.
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  4. Completing the Square:
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    This method is most useful in rewriting the quadratic in vertex form. The idea is to use perfect square trinomial, and the properties of equality to rewrite the quadratic into the form \(a(x-h)^2+k\) where \((h,k)\) is the vertex of the quadratic.
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Section BF-1 Videos

Linear Equations in One-Variable
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Absolute Value Equations
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Radical Equations
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Rational Equations
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Solving Quadratic Equations by Factoring
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Solving Quadratic Equations by Extraction of Roots
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Solving Quadratic Equations using the Quadratic Formula
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Solving Quadratic Equations by Completing the Square
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Section BF-1 Rubric

Worksheet Worksheet

Before attempting the BF-1 outcome, you should review the following rubric. This shows the criteria you will be graded on and the expectations to earn each mastery level.
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Table 10. BF1 Rubric
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Criteria M P R N
Simplifying Algebraic Expressions All answers are fully simplified. Exact answers are given, not approximations. Radical and exponential expressions are used appropriately and are simplified in the correct manner No algebraic errors OR some expressions are not simplified. There are some algebraic errors and some expressions were not simplified. There are some algebraic errors and some expressions were not simplified.
Problem Structure and Mathematical Communication Problems have a clear beginning, middle, and end. Work progresses clearly from one step to the next. The work provided uses appropriate methods and notation and provides a clear solution. Problems have clear beginnings and ends. Work progresses from one step to the next. The work provided uses algebraic methods, correct notation, and provides a solution. Parts of the mathematical structure are missing, or steps in the progress of the solution are missing. There are errors in the mathematical notation. The work provided uses algebraic methods and provides a solution. There is not a clear structure to the solution, notation is used incorrectly, and it is unclear what methods are being used.
Solve Linear Equation Combined like terms: used distributive property, addition property of equality, and multiplication property of equality: no algebraic errors; correct answer. Combined like terms: used distributive property, addition property of equality, and multiplication property of equality.  Correct process, but ended with an incorrect answer. Errors in combining like terms, in incorrectly using distributive properties or properties of equality, or in not applying operations in the correct order resulted in an incorrect answer. No progress beyond rewriting the problem. Not enough evidence to show knowledge of how to use the properties of real numbers.
Solve Quadratic Equation by Factoring Correctly rewrote or factored the equation; correctly used the zero factor property, solved the linear equations; correct answer. Rewrote or factored the equation with minor errors; used the zero factor property, and solved the linear equation with minor errors in the solution. Errors in rewriting the equation OR incorrectly factored the equation OR did not apply zero factor property and resulted in an incorrect answer. No progress beyond rewriting the problem. Not enough evidence to show knowledge of factoring or how to use the properties of real numbers.
Solve Quadratic Equation by Extraction of Roots Rewrote and isolated the root; correctly extracted the root; correct answer. Rewrote and isolated the root with minor errors; partially extracted the root; incorrect answer. Did not rewrite or incorrectly isolated root. No progress beyond rewriting the problem. Not enough evidence to show knowledge of how to use the extraction of roots or the properties of real numbers.
Solve Quadratic Equation with the Quadratic Formula Correctly set up the formula and solved for the answer. Errors in setting up the formula or solving for the answer. Errors in setting up the formula and solving for the answer. No progress beyond rewriting the problem. Not enough evidence to show knowledge of how to use the quadratic formula.
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Worksheet BF-1 Sample Outcomes

This outcome covers solving linear and quadratic equations. You will be given 2-3 problems in this outcome; a linear equation and two quadratic equations. The linear equation will be solved using the standard algebraic techniques, while the quadratic equations will be solved using factoring, the extraction of roots method, completing the square or the quadratic formula. You should be able to write out complete steps to solve these types of problems by hand, without the use of a calculator.
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For the quiz on this outcome you would be given 30 minutes to complete 3 problems. Here are samples of the types of problems you will encounter:
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1.

Solve the following equation: \(\frac{2}{3}x+\frac{5}{18}=x-\frac{4}{9}\)
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Hint.
Start by using the LCD to clear the fractions. Then combine like terms and solve.
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Answer.
\(x=\frac{13}{6}\)
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Solution.
Start by using the LCD of 18 to clear the fractions: \(12x+5=18x-8\text{.}\) Then combine like terms: \(5+8=18x-12x\text{.}\) Next, simplify to get the solution: \(x=\frac{13}{6}\text{.}\)
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2.

Solve the following equation: \(3(2x-7)-4x+1=5x-2(x+3)\)
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Hint.
Start by distributing over the parentheses and then combining like terms.
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Answer.
\(x=-14\)
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Solution.
Start by distributing over the parentheses: \(6x-21-4x+1=5x-2x-6\text{.}\) Then combine like terms: \(2x-20=3x-6\text{.}\) Next, move all terms to one side: \(-20+6=x\text{.}\) Finally, simplify to get the solution: \(x=-14\text{.}\)
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3.

Solve \(8x^2-18x+9=0\) by factoring.
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Hint.
Look for two numbers that multiply to give the product of the coefficient of x^2 and the constant term, and add up to give the coefficient of x.
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Answer.
\(x=\frac{3}{4}\) and \(x=\frac{3}{2}\)
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Solution.
Start by multiplying the leading coefficient and the constant: \(8\cdot 9=72\text{.}\) Find the two factors of 72 that add up to -18: -12 and -6. Rewrite the middle term using these factors: \(8x^2-12x-6x+9=0\text{.}\) Factor by grouping: \(4x(2x-3)-3(2x-3)=0\text{.}\) Factor out the common binomial: \((4x-3)(2x-3)=0\text{.}\) Set each factor equal to zero and solve for x: \(x=\frac{3}{4}\) and \(x=\frac{3}{2}\text{.}\)
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4.

Solve \((6x-5)^2=49\) using the extraction of roots method.
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Hint.
Take the square root of both sides and solve the resulting linear equations.
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Answer.
\(x=2\) and \(x=\frac{-1}{3}\)
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Solution.
Start by taking the square root of both sides of the equation: \(6x-5=\pm 7\text{.}\) Then solve the resulting linear equations: \(6x-5=7\) and \(6x-5=-7\text{.}\) This gives \(x=2\) and \(x=\frac{-1}{3}\text{.}\)
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5.

Solve \(4y^2+8y-5=5\) using the quadratic formula.
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Hint.
First, move all terms to one side to get the equation in standard form, then apply the quadratic formula.
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Answer.
\(x=\frac{-1\pm\sqrt{14}}{2}\)
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Solution.
Start by setting the equation equal to zero and finding the terms: \(a=4\text{,}\) \(b=8\text{,}\) and \(c=-10\text{.}\) Then apply the quadratic formula by plugging these values into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\text{.}\) \(x=\frac{-8\pm\sqrt{64-4(4)(-10)}}{2(4)}\text{.}\) Simplify inside the root \(x=\frac{-8\pm\sqrt{64+160}}{8}\text{.}\) \(x=\frac{-8\pm\sqrt{224}}{8}\text{.}\) Now we can simplify the root \(x=\frac{-8\pm 4\sqrt{14}}{8}\text{.}\) Then simplify the resulting expression: \(x=\frac{-1\pm\sqrt{14}}{2}\text{.}\)
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6.

Solve \(3x^2+4x-5=0\) by completing the square.
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Hint.
First, divide by factor out the coefficient of x^2, then move the constant term to the other side, and complete the square.
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Answer.
\(x=\frac{-2\pm\sqrt{19}}{3}\)
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Solution.
Since the leading coefficient isn’t one, start by factoring out the coefficient from the x-terms and move the constant term to the other side. \(3(x^2+\frac{4}{3}x)=5\text{.}\) Next, complete the square by adding the square of half the coefficient of x to both sides: \(3(x^2+\frac{4}{3}x+\frac{4}{9})=5+3\left(\frac{4}{9}\right)\text{.}\) This gives us \(3(x+\frac{2}{3})^2=\frac{19}{3}\text{.}\) Now we can divide by 3 and take the square root of both sides: \(x+\frac{2}{3}=\pm\sqrt{\frac{19}{9}}\text{.}\) Finally, solve for x to get the solution: \(x=\frac{-2\pm\sqrt{19}}{3}\text{.}\)
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