(273) Algebra 1 Full Course

Variables and Algebraic Expressions

#1 Variables and Algebraic Expressions

• Variables represent unknown values, typically denoted by letters (e.g., x, y).

• Algebraic expressions combine variables, constants, and operations (e.g., 3x + 2). Example 1: If x = 4, then 2x + 3 = 2(4) + 3 = 8 + 3 = 11. Example 2: If y = -1, then 5y + 2 = 5(-1) + 2 = -5 + 2 = -3. Example 3: If z = 0, then 10z - 7 = 10(0) - 7 = 0 - 7 = -7.

#2 What is an Equation

• An equation states that two expressions are equal, represented by the equals sign (e.g., 2x + 3 = 7).

• Solving an equation means finding the value(s) of the variable(s) that make the equation true. Example 1: To solve x + 5 = 10, subtract 5 from both sides to get x = 5. Example 2: For the equation 3y - 4 = 11, add 4 to both sides to find 3y = 15, and thus y = 5. Example 3: To solve 5a + 2 = 17, subtract 2 to get 5a = 15, then divide by 5 to find a = 3.

#3 Addition Property of Equality

• States that if you add the same value to both sides of an equation, the equality remains true.

• For example, if a = b, then a + c = b + c. Example 1: If 10 = 10, then 10 + 3 = 10 + 3, resulting in 13 = 13. Example 2: If x = 3, then x + 5 = 3 + 5, giving x + 5 = 8. Example 3: If y = -2, then y + 4 = -2 + 4, resulting in y + 4 = 2.

#4 Multiplication Property of Equality

• States that if you multiply both sides of an equation by the same non-zero value, the equality remains true.

• For example, if a = b, then ac = bc. Example 1: If 7 = 7, then 7 * 2 = 7 * 2, which results in 14 = 14. Example 2: If z = 2, then z * 3 = 2 * 3, giving 3z = 6. Example 3: If m = 4, then 5m = 5 * 4, yielding 5m = 20.

#5 Multi-Step Linear Equations

• Involves solving equations that require multiple steps, such as combining like terms and using inverse operations. Example 1: For the equation 3x + 7 = 22, subtract 7 from both sides to get 3x = 15, then divide by 3 to find x = 5. Example 2: Solve 4y - 2 = 14 by adding 2 to both sides: 4y = 16, thus y = 4. Example 3: Solve 6a + 3 = 27 by first subtracting 3 to get 6a = 24, then dividing by 6, yielding a = 4.

#6 Equations with Fractions/Decimals

• To solve these equations, it may be helpful to eliminate fractions by multiplying through by the least common denominator (LCD) or converting decimals to fractions. Example 1: Solve 1/3x + 1/6 = 1 by multiplying through by 6 to eliminate fractions, leading to 2x + 1 = 6, thus x = 2. Example 2: Solve 0.5x + 2 = 5 by first subtracting 2 to get 0.5x = 3, then multiplying by 2 to yield x = 6. Example 3: Solve 3/4x - 1/2 = 1 by multiplying everything by 4 (the LCD) to clear fractions: 3x - 2 = 4, hence x = 2.

#7 Special Case Linear Equations

• These equations may have no solution (inconsistent) or infinitely many solutions (dependent). Example 1: 4x + 6 = 4x + 10 is inconsistent since the same term equals two different values, hence no solution. Example 2: 2y + 3 = 2y + 3 has infinitely many solutions; any value of y satisfies the equation. Example 3: x + 5 = x + 3 is inconsistent, as it implies 5 = 3, which is false.

#8 Solving Word Problems with Linear Equations I

• Translate the problem into an equation using known relationships, assign variables, and solve. Example 1: If a bike costs $200 and you save $50 a month, how many months until you have enough? Equation: 50m = 200, hence m = 4. Example 2: If a person walks 3 miles per hour, how long (h) will it take to walk 15 miles? Equation: 3h = 15, so h = 5. Example 3: If you buy x apples at $1 each and have $10, the equation is: x = 10.

#9 Solving Word Problems with Linear Equations II

• Continuation and advanced techniques for translating complex word problems into equations. Example 1: If Sarah has $20 and spends x dollars each week, how much does she have after 3 weeks? Equation: 20 - 3x. Example 2: If a rectangle has a length that is twice its width (w), and the perimeter is 30, then P = 2(l + w) leads to: 30 = 2(2w + w), therefore w = 6. Example 3: If a person invests $x at 5% interest and earns $50 in a year, model the equation 0.05x = 50, leading to x = 1000.

#10 Solving Proportion Equations

• Proportions are equations that state two ratios are equal. Use cross-multiplication to solve them (e.g., a/b = c/d leads to ad = bc). Example 1: Solve 1/2 = x/8 by cross-multiplying: 18 = 2x, hence x = 4. Example 2: If 4/5 = x/25, then 425 = 5x, yielding x = 20. Example 3: If 3/4 = x/12, it follows that 312 = 4x, hence x = 9.

#11 Solving Linear Inequalities I

• Linear inequalities express a relationship of comparison; similar to equations but incorporating inequality signs (e.g., x + 3 > 7). Example 1: For x + 5 < 10, subtract 5 to get x < 5. Example 2: Solve 2x - 4 ≤ 8 by adding 4: 2x ≤ 12, hence x ≤ 6. Example 3: For 5 - x > 2, subtract 5 to yield -x > -3, thus x < 3.

Solving Linear Inequalities and Equations

#12 Solving Linear Inequalities II

• Focuses on solving more complex linear inequalities that require additional algebraic manipulation. Example 1: For 4x + 1 > 9, subtract 1 to find 4x > 8, giving x > 2. Example 2: Solve 5 - 2x ≤ 3; subtracting gives -2x ≤ -2, thus x ≥ 1. Example 3: For 3x - 5 < 4, adding 5 results in 3x < 9, hence x < 3.

#13 Introduction to Linear Equations in Two Variables

• Linear equations are represented in the form y = mx + b, where m is the slope and b is the y-intercept. Example 1: For y = 3x + 2, slope m = 3, intercept b = 2. Example 2: In y = -2x + 5, the slope is -2 and y-intercept is 5. Example 3: The equation y = 1/2x - 3 has a slope of 1/2 and a y-intercept at -3.

#14 Plotting Ordered Pairs

• Ordered pairs consist of an x-coordinate and a y-coordinate, represented as (x, y). Example 1: Plot the point (2, 4); move 2 units right and 4 units up from the origin. Example 2: The pair (-3, -2) is found by moving 3 units left and 2 units down. Example 3: For (0, 5), plot directly 5 units up along the y-axis.

#15 Graphing Linear Equations in Two Variables

• Determine at least two points that satisfy the equation and connect them to form a straight line. Example 1: For y = x + 1, plot (0, 1) and (1, 2) to draw the line. Example 2: Graph y = -x + 3 using (0, 3) and (3, 0). Example 3: For y = 2x - 1, points are (0, -1) and (1, 1).

#16 How to Find the Slope of a Line

• Slope (m) is determined using m = (y2 - y1) / (x2 - x1). Example 1: For points (1, 2) and (3, 6), slope m = (6-2)/(3-1) = 4/2 = 2. Example 2: Between (2, 3) and (5, 7), m = (7-3)/(5-2) = 4/3. Example 3: From points (1, 1) to (4, 2), slope is m = (2-1)/(4-1) = 1/3.

#17 Equations of a Line | Slope-Intercept Form | Standard Form

• Slope-Intercept Form: y = mx + b, Standard Form: Ax + By = C. Example 1: The line y = 2x + 3 is in slope-intercept form. Example 2: For the standard form of 4x + 2y = 8, slope m = -2. Example 3: Convert y - 1 = 3(x + 2) into slope-intercept: y = 3x + 7.

#18 Parallel and Perpendicular Lines

• Parallel lines have the same slope, while perpendicular lines have slopes that are negative reciprocals. Example 1: y = 2x + 1 is parallel to y = 2x - 4. Example 2: y = -3x + 5 is perpendicular to y = 1/3x + 2. Example 3: The line y = 4x + 1 will not intersect with y = 4x + 7.

#19 Graphing Linear Inequalities in Two Variables

• Graph the boundary line and shade the region representing the solution set. Example 1: For y ≥ x - 3, graph the line as solid and shade above it. Example 2: In graphing y < 2x + 1, use a dashed line and shade below. Example 3: For x + y ≤ 4, shade below the solid line.

Functions and Systems of Equations

#20 Introduction to Functions

• A function is a relation where each input has exactly one output. Example 1: For f(x) = 3x + 1, if x = 2, f(2) = 3(2) + 1 = 7. Example 2: For g(x) = x², if x = 5, g(5) = 5² = 25. Example 3: h(x) = 4x - 3 gives h(0) = -3.

#21 Solving Linear Systems Using the Graphing Method

• Graph each equation to find the intersection point as the solution. Example 1: Solve y = x + 1 and y = 2x + 2 on the same graph. Example 2: Find where y = 3x - 1 intersects y = -x + 9. Example 3: For y = 2x + 5 and 3y = 9x + 6, plot both lines.

#22 Solving Linear Systems Using the Substitution Method

• Solve one equation for one variable and substitute into the other equation. Example 1: From y = 2x + 1 and 3x + y = 12, replace y in the second equation. Example 2: For y = x - 3 and 2x + y = 4, substitute y in the equation. Example 3: Solve x = 3y + 1 and y + 2x = 10 by substituting x from the first equation.

#23 Solving Linear Systems Using the Elimination Method

• Add or subtract the equations to eliminate one variable. Example 1: For x + y = 10 and 2x - y = 4, add them to solve. Example 2: In 3x + 2y = 12 and 4x - 2y = 8, add to eliminate y. Example 3: x + 3y = 15 and 2x + 3y = 18 can be solved by elimination method.

#24 Solving Word Problems with Linear Systems

• Translate the problem into a system of equations and solve. Example 1: If a video game costs $x and a magazine $y with 5x + 3y = 21, solve for costs. Example 2: If train A travels at 60 mph and train B at 80 mph, time to meet after t hours relates as 60t + 80t = D. Example 3: If 2x + 5y = 20 represents tomatoes and potatoes, find prices for x and y.

#25 Solving Systems of Linear Inequalities

• Graph inequalities and find overlapping solution region. Example 1: x + y < 10 and 2x + y ≤ 12 can show feasible region in the graph. Example 2: Shade the solution region for x > 3 and y < 2x - 1. Example 3: Solve x + y > 5 and x - y < 2 on one graph.

Exponents and Polynomials

#26 Product & Power Rules for Exponents

• The Product Rule states that when multiplying like bases, add the exponents: a^m * a^n = a^(m+n). Example 1: 2^4 * 2^3 = 2^(4+3) = 2^7 = 128. Example 2: For x² * x³ = x^(2+3) = x^5. Example 3: z^5 * z^2 = z^(5+2) = z^7.

#27 Negative Exponents & the Quotient Rule for Exponents

• Negative exponents indicate reciprocals: a^(-n) = 1/a^n.

• The Quotient Rule states: a^m / a^n = a^(m-n). Example 1: 5^(-2) = 1/5² = 1/25. Example 2: x^7 / x^3 = x^(7-3) = x^4. Example 3: For 9^(-1) = 1/9 and 2^5 / 2^3 = 2^(5-3) = 2² = 4.

#28 Scientific Notation

• A way to express very large or small numbers as N = a × 10^n, where 1 ≤ a < 10. Example 1: 0.00047 can be expressed as 4.7 × 10^(-4). Example 2: 6800000 = 6.8 × 10^6. Example 3: 0.0035 = 3.5 × 10^(-3).

#29 Adding & Subtracting Polynomials

• Combine like terms by adding or subtracting their coefficients. Example 1: (5x² + 3x) + (2x² + 4) = 7x² + 3x + 4. Example 2: (x + 2y + 3) - (3 + y) simplifies to x + y + 2. Example 3: (7x^3 - 2x^2) + (5x^3 + 4x) = 12x^3 - 2x^2 + 4x.

#30 Multiplying Polynomials

• Use distribution or the area model to multiply polynomials accurately. Example 1: (x + 1)(x + 2) = x² + 2x + x + 2 = x² + 3x + 2. Example 2: Multiplying (2x - 3)(x + 4) gives 2x² + 8x - 3x - 12 = 2x² + 5x - 12. Example 3: (x - 1)(x² + x + 1) yields x³ + x² + x - x² - x - 1 = x³ - 1.

#31 FOIL (Finding the product of two binomials)

• FOIL: First, Outer, Inner, Last helps organize the multiplication of two binomials. Example 1: (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Example 2: (2x + 1)(x + 4) = 2x² + 8x + 1x + 4 = 2x² + 9x + 4. Example 3: (x - 2)(x + 3) = x² + 3x - 2x - 6 = x² + x - 6.

#32 Special Polynomial Products

• Recognize patterns: (a + b)² = a² + 2ab + b², etc. Example 1: (2x + 3)² = 4x² + 12x + 9. Example 2: (x - 5)² = x² - 10x + 25. Example 3: (3a + 2b)² = 9a² + 12ab + 4b².

#33 Dividing Polynomials by Monomials

• Divide each term in the polynomial by the monomial. Example 1: (6x² + 12x) / 3 = 2x + 4. Example 2: (8x³ - 4x² + 16x) / 4x = 2x² - x + 4. Example 3: (10x² + 15 - 5x) / 5 = 2x² + 3 - x.

#34 Dividing Polynomials

• Use long division or synthetic division to divide polynomials. Example 1: Divide (2x³ + 3x² - x) by (x + 1) using long division. Example 2: For x² - 4 by x + 2, perform polynomial long division. Example 3: Divide (3x^3 - 2x + 1) by (x - 1) using synthetic division.

#35 Dividing Polynomials with Missing Terms

• When dividing, treat missing terms as coefficients of zero. Example 1: When dividing x³ + 3 by x - 1, write it as x³ + 0x² + 0x + 3. Example 2: For 2x^4 - x^2 by x^2 + 0x + 1, include the 0 for missing powers. Example 3: In (5x^5 + 4x^3 + 1)/(x^3), treat as 5x^5 + 4x^3 + 0x^2 + 0x + 1.

#36 GCF for a Group of Monomial Terms

• Determine the greatest common factor of a group of monomial terms. Example 1: For 8x², 12x, 16, the GCF is 4. Example 2: For 3a²b, 5a^3b², and 2a²b², the GCF is a²b. Example 3: For 10x^3y and 15xy², the GCF is 5xy.

#37 Factoring Out the GCF

• Factor out the GCF from a polynomial to simplify the expression. Example 1: 12x³ + 8x² = 4x²(3x + 2). Example 2: 15x²y + 25xy² = 5xy(3x + 5y). Example 3: 6a³ - 9a² + 3a = 3a(2a² - 3a + 1).

#38 Factoring by Grouping

• Group terms in pairs and factor out common factors to factor polynomials efficiently. Example 1: For x³ + 2x² + 3x + 6, group as (x²)(x + 2) + (3)(x + 2) = (x + 2)(x² + 3). Example 2: 2a² + 4a + 3b + 6b^2 = (2a + 3)(a + 2b). Example 3: For x²y + 3xy + x + 3, group and factor as y(x + 3) + 1(x + 3) = (x + 3)(y + 1).

Factoring and Quadratic Equations

#39 Factoring Trinomials when a is 1

• Involves finding two binomials that multiply to give the quadratic expression. Example 1: x² + 6x + 8 = (x + 4)(x + 2). Example 2: x² + 5x + 6 = (x + 2)(x + 3). Example 3: x² + 3x - 10 = (x + 5)(x - 2).

#40 Factoring Trinomials when a is not 1

• Utilize the method of splitting the middle term or by using the ac method. Example 1: 3x² + 11x + 6 = (3x + 2)(x + 3). Example 2: 2x² - 5x - 3 = (2x + 1)(x - 3). Example 3: 4x² + 12x + 9 = (2x + 3)(2x + 3) = (2x + 3)².

#41 Special Factoring Formulas

• Recognize patterns such as the difference of squares and perfect square trinomials. Example 1: x² - 16 = (x + 4)(x - 4). Example 2: x² + 10x + 25 = (x + 5)(x + 5) = (x + 5)². Example 3: a² - 49 = (a + 7)(a - 7).

#42 Solving Quadratic Equations Using Factoring

• Set the quadratic equation to zero and factor into the form (x - p)(x - q) = 0. Example 1: x² - 7x + 10 = (x - 2)(x - 5) = 0, yielding x = 2 or x = 5. Example 2: x² + 4x - 12 = (x + 6)(x - 2) = 0, so x = -6 or x = 2. Example 3: 2x² + 3x - 5 = (2x - 1)(x + 5) = 0, resulting in x = 1/2 or x = -5.

Rational Expressions

#43 Introduction to Rational Expressions

• A rational expression is a ratio of two polynomials.

• Important to identify restrictions on variables (denominators cannot be zero). Example 1: (x³ + 2x)/(x - 1), where x ≠ 1. Example 2: (x² - 4)/(x + 2), where x ≠ -2. Example 3: (3x + 1)/(x² - 1), where x ≠ 1 or x ≠ -1.

#44 Multiplying & Dividing Rational Expressions

• To multiply, multiply the numerators and denominators, then simplify.

• To divide, multiply by the reciprocal of the second expression. Example 1: (x/3) * (2/5) = 2x/15; divide: (3/x) ÷ (6/4) = (3/x) * (4/6) = 2/x. Example 2: (4x/5) * (10/x) = (40x)/(5x) = 8; divide: (2/3x) ÷ (1/4) = (2/3x) * (4/1) = 8/3x. Example 3: (a/b) * (c/d) = ac/bd; dividing: (3x/4) ÷ (6/5) = (3x/4) * (5/6) = 15x/24 = 5x/8.

#45 Finding the LCD of Rational Expressions

• Least Common Denominator (LCD) is necessary for adding and subtracting rational expressions.

• List out factors of each denominator and choose the highest power of each factor. Example 1: For 1/(x) and 1/(x²), the LCD is x². Example 2: For 2/(x²y) and 3/(xy²), the LCD is x²y². Example 3: For 1/(2x) and 3/(5), the LCD is 10x.

#46 Adding & Subtracting Rational Expressions

• Convert each expression to have the same denominator (LCD), then combine the numerators. Example 1: 1/(x) + 1/(x²) = (x + 1)/(x²). Example 2: (3/x + 2/y) = (3y + 2x)/(xy). Example 3: (5/(x + 1) + 3/(x - 1)) = ((5(x - 1) + 3(x + 1))/((x - 1)(x + 1))).

#47 Simplifying Complex Rational Expressions

• A complex rational expression has a fraction in the numerator, denominator, or both. Simplify by finding the equivalent expression. Example 1: (1/(x + 1))/(1/(x + 2)) = (x + 2)/(x + 1). Example 2: (2/(x + 3)) / (3/(x + 5)) = (2(x + 5))/(3(x + 3)). Example 3: (5/(x - 1))/(3/(y + 1)) = (5y + 5)/(3x - 3).

#48 Solving Rational Equations

• Set the rational equation to zero, clear fractions by multiplying by the LCD, then solve the resulting equation. Example 1: 1/(x-1) + 1/(x+1) = 1 leads to the equation (1+x)/(x²-1) = 1. Example 2: 2/(x+2) = 3/(x-1) can be solved by cross-multiplying. Example 3: x/(x + 3) = 1/3, clear fractions multiplying throughout by 3(x + 3).

#49 Solving Word Problems with Rational Expressions

• Translate the problem into a rational expression, solve for the variable, and interpret the solution in context. Example 1: If a car travels x miles per hour for y hours, the equation for distance d = xy. Example 2: If you complete a task in x days and it takes you y hours each day, total hours = xy. Example 3: If a worker earns $y dollars for each x hours worked, total earnings = y * x.

Variation Problems

#50 Direct Variation Problems

• In direct variation, y = kx, where k is a constant. If one variable increases, so does the other. Example 1: If y varies directly as x, and when x = 6, y = 12, then k = 2, so y = 2x. Example 2: If m varies directly with n, and when m = 10, n = 5, then k = 2, so m = 2n. Example 3: If temperature T varies directly with time t in hours, T = kt, where k is the rate of temperature rise.

#51 Inverse Variation Problems

• Inverse variation involves y = k/x, where an increase in one variable results in a decrease in another. Example 1: If xy = 12, and when y = 3, x = 4; thus x varies inversely with y. Example 2: If P = k/T, where Power varies inversely with Time. Example 3: If x is inversely proportional to y, then as y increases, x decreases; knowing x = 10 when y = 2 gives k = 20.

Square Roots and Radicals

#52 Introduction to Square Roots

• The square root of a number x is a number y such that y² = x. Example 1: √36 = 6, since 6² = 36. Example 2: √0.25 = 0.5, as (0.5)² = 0.25. Example 3: √81 = 9, since (9)² = 81.

#53 The Distance Formula

• Distance formula: d = √((x² - x¹)² + (y² - y¹)²). Example 1: Distance between points (2, 3) and (5, 7) is d = √((5-2)² + (7-3)²) = √(9 + 16) = √25 = 5. Example 2: From point (3, 4) to (1, 1), d = √((3-1)² + (4-1)²) = √(4 + 9) = √13. Example 3: Distance from (0, 0) to (4, 3) equals d = √((4-0)² + (3-0)²) = √(16 + 9) = √25 = 5.

#54 Simplifying Radicals

• Simplify square roots by factorization, removing perfect squares from under the radical. Example 1: √50 = √(25 * 2) = 5√2. Example 2: √72 = √(36 * 2) = 6√2. Example 3: √98 = √(49 * 2) = 7√2.

#55 Adding & Subtracting Radicals

• Combine like radicals (terms with the same radicand) but cannot combine unlike radicals. Example 1: 3√5 + 5√5 = 8√5. Example 2: 7√2 - 3√2 = 4√2. Example 3: 2√3 + 4√2 cannot be simplified further as they are unlike radicals.

#56 Rationalizing the Denominator

• Multiply the numerator and denominator by a suitable radical to eliminate radicals from the denominator. Example 1: To rationalize: 1/√3, multiply by √3/√3 = √3/3. Example 2: Rationalize 2/√7 by multiplying to get 2√7/7. Example 3: For 5/√2, rationalize to 5√2/2.

#57 Further Operations with Radicals

• More advanced operations, including combining, multiplying, and dividing radical expressions. Example 1: (√3)(√12) = √(3 * 12) = √36 = 6. Example 2: (4√2)/(2√2) simplifies to 4/2 = 2. Example 3: √(16/25) = √16/√25 = 4/5.

#58 Solving Radical Equations

• Isolate the radical on one side of the equation, then square both sides to eliminate the radical. Example 1: Solve √(x + 4) = 6, square: x + 4 = 36, leads to x = 32. Example 2: If √(3x - 5) = 7, then 3x - 5 = 49, solving gives x = 18. Example 3: For √(x + 2) - 3 = 0, isolate: √(x + 2) = 3, square results in x = 7.

Fractional Exponents

#59 Fractional Exponents

• A rational exponent represents both a root and a power. For example, x^(m/n) = n-th root of (x^m). Example 1: 16^(1/4) = 2, since 2⁴ = 16. Example 2: 27^(2/3) = (3³)^(2/3) = 3² = 9. Example 3: 81^(1/2) = 9, since 9² = 81.

Solving Quadratic Equations

#60 Solving Quadratic Equations with the Square Root Property

• When in the form x² = k, take the square root of both sides: x = ±√k. Example 1: If x² = 25, x = ±5 as valid solutions. Example 2: For x² = 49, x = ±7. Example 3: x² = 36 gives x = ±6.

#61 Solving Quadratic Equations by Completing the Square

• Rearrange the equation to isolate the constant, adding to both sides to complete the square. Example 1: For x² - 6x = 7, complete: (x - 3)² = 16, so x = 3 ± 4. Example 2: x² + 4x - 5 = 0, so (x + 2)² - 9 = 0, find solutions as x = -2 ± 3. Example 3: Solve x² + 6x = 8 will lead to (x + 3)² = 17, hence, x = -3 ± √17.

#62 Solving Quadratic Equations with the Quadratic Formula

• Apply: x = (-b ± √(b² - 4ac)) / (2a). Example 1: For 2x² + 4x - 6 = 0, use a = 2, b = 4, c = -6:

discriminant: 4² - 42(-6) = 16 + 48 = 64, thus roots: x = (-4 ± 8) / 4 = 1 or -3.

Example 2: If (x² - 2x - 8 = 0), then a = 1, b = -2, c = -8; calculate roots using formula gives:

discriminant: (-2)² - 4(1)(-8) = 4 + 32 = 36; thus x = (2 ± 6) / 2 gives x = 4 or -2.

Example 3: For 5x² + 4x + 1 = 0, a = 5, b = 4, c = 1; calculate:

discriminant: 4² - 4(5)(1) = 16 - 20 = -4; leads to solutions involving complex:

x = (-4 ± √-4)/(10) gives x = -2/5 ± i/5.