Part 2  Beginning Algebra Summary


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1 Part  Beginning Algebra Summary Page 1 of 4 1/1/01 1. Numbers Number Lines Interval Notation.... Inequalities Linear with 1 Variable Linear Equations The Cartesian Plane Graphing Lines Intercepts and Slope Finding the Equation of a Line Systems of Linear Equations Definitions Solving by Graphing Solving by Substitution Solving by Addition or Subtraction Word Problems Solving Polynomials Definitions Multiplication Division Factoring GCF (Greatest Common Factor) Terms Trinomials: Leading Coefficient of Trinomials: All Perfect Square Trinomials & Binomials Steps to Follow Quadratics About Graphing Solve by Factoring Solve with the Quadratic Equation Exponents Computation Rules Scientific Notation Radicals Definitions Computation Rules Rationals Simplifying Expressions Arithmetic Operations Solving Equations Summary Formulas Types of Equations Solve Any 1 Variable Equation... 4 Copyright Sally C. Zimmermann. All rights reserved.
2 1. Numbers Number Lines Real Numbers Positive Infinity (Infinity) Negative Infinity Part  Beginning Algebra Summary 1.1. Number Lines ( ) If the point is not included [ ] If the point is included Shade areas where infinite points are included Points on a number line Whole numbers, integers, rational and irrational numbers An unimaginably large positive number. (If you keep going to the right on a number line, you will never get there) An unimaginably small negative number. (If you keep going to the left on a number line, you will never get there) > 7 7, 7,, π Page of 4 1/1/01 or + Copyright Sally C. Zimmermann. All rights reserved.
3 Part  Beginning Algebra Summary Page of 4 1/1/01 Interval Notation (shortcut, instead of drawing a number line) 1.. Interval Notation 1st graph the answers on a number line, then write the interval notation by following your shading from left to right Always written: 1) Left enclosure symbol, ) smallest number, ) comma, 4) largest number, 5) right enclosure symbol Enclosure symbols ( ) Does not include the point [ ] Includes the point Infinity can never be reached, so the enclosure symbol which surrounds it is an open parenthesis 1 Ex. x 1 " x is equal to 1" Ex. x 1 " x is not equal to 1" Ex. x < 1 " x is less than 1" Ex. x 1 " x is less than or equal to 1".. Ex. x > 1 " x is greater than 1" Ex. x 1 " x is greater than or equal to 1" { } (,1) (,1] (1, ) [1, ) Copyright Sally C. Zimmermann. All rights reserved.
4 Part  Beginning Algebra Summary Page 4 of 4 1/1/01. Inequalities.1. Linear with 1 Variable Standard Form ax + b < c ax + b c ax + b > c ax + b c > x + 4 > 10 Solution A ray > x > 0 1 Multiplication When both sides of an inequality are multiplied or > 4 x Property of divided by a negative number, the direction of the 4 x Inequality inequality symbol must be reversed to form an equivalent inequality. Solving 1. Same as Solving an Equation with 1 Variable Ex 4 x (MA090), except when both sides are multiplied or 4 x divided by a negative number x x. Checking Plug solution(s) into the original equation. Should get a true inequality. Plug a number which is not a solution into the original equation. Shouldn t get a true inequality > 4 ( ) 4 6 > 4 (0) Copyright Sally C. Zimmermann. All rights reserved.
5 Part  Beginning Algebra Summary. Linear Equations.1. The Cartesian Plane Rectangular Coordinate System Two number lines intersecting at the point 0 on each number line. XAXIS  The horizontal number line YAXIS  The vertical number line ORIGIN  The point of intersection of the axes QUADRANTS  Four areas which the rectangular coordinate system is divided into ORDERED PAIR  A way of representing every point in the rectangular coordinate system (x,y) Page 5 of 4 1/1/01 Quadrant II Quadrant I Is an Ordered Pair a Solution? Yes, if the equation is a true statement when the variables are replaced by the values of the ordered pair Quadrant III Quadrant IV Ex x + y 7 (1, ) is a solution because 1 + () 7 Copyright Sally C. Zimmermann. All rights reserved.
6 Part  Beginning Algebra Summary Page 6 of 4 1/1/01 General Graphing by plotting random points Graphing linear equations by using a point and a slope.. Graphing Lines Lines which intersect the xaxis contain the variable x Lines which intersect the yaxis contain the variable y Lines which intersect both axis contain x and y 1. Solve equation for y. Pick three easy xvalues & compute the corresponding yvalues. Plot ordered pairs & draw a line through them. (If they don t line up, you made a mistake) 1. Plot the point. Starting at the plotted point, vertically move the rise of the slope and horizontally move the run of the slope. Plot the resulting point. Connect both points > x + y 7 x 7 y + x y > y x + Point 7/ Slope 1/ Copyright Sally C. Zimmermann. All rights reserved.
7 Part  Beginning Algebra Summary Page 7 of 4 1/1/01 xintercept (x, 0) yintercept (0, y) Slope of a Line Properties of Slope Standard Form SlopeIntercept Form PointSlope Form.. Intercepts and Slope WHERE THE GRAPH CROSSES THE XAXIS Ex x + y 7 Let y 0 and solve for x x + (0) 7 x 7 (7,0) WHERE THE GRAPH CROSSES THE YAXIS Ex x + y 7 Let x 0 and solve for y 0 + y 7 y.5 (0,. 5) x The slant of the line. y x y Ex Let P 1 ( 1, 1), P ( 4, 4) Let Point 1: P 1 ( x1, y1) & Point : P ( x, y) rise (change in y) m y y m (slope) x x run (change in x) y y1 x x1 POSITIVE SLOPE  Line goes up (from left to right). The greater the number, the steeper m 0 the slope NEGATIVE SLOPE  Line goes down (from left m 1/ to right). The smaller the number (more m undefined negative), the steeper the slope. HORIZONTAL LINE  Slope is 0 m 1 VERTICAL LINE  Slope is undefined m m 1/ PARALLEL LINES  Same slope m PERPENDICULAR LINES  The slope of one is the negative reciprocal of the other Ex: m ½ is perpendicular to m ax + by c > x + y 7 x and y are on the same side The equations contains no fractions and a is positive y mx + b, where m is the slope of the line, > By solving x + y 7 for y & b is the yintercept x 7 y + y equals form ; easy to graph form y y 1 m(x x 1 ), where m is the slope of the 1 line & (x 1, y 1 ) is a point on the line > Using (7, 0) and m Simplified, it can give you Standard Form or 1 SlopeIntercept Form y 0 ( x 7) Copyright Sally C. Zimmermann. All rights reserved.
8 Part  Beginning Algebra Summary Page 8 of 4 1/1/01 If you have a horizontal line.4. Finding the Equation of a Line The slope is zero y b, where b is the yintercept Ex. y If you have a vertical line The slope is undefined x c, where c is the xintercept Ex. x  If you have a slope & yintercept Plug directly into SlopeIntercept Form Ex. m 4 & yintercept ( 0, ) y 4x + 4( 0) + If you have a point & a slope If you have a point & a line that it is parallel or perpendicular to METHOD 1 1. Use PointSlope Form. Work equation into Standard Form or SlopeIntercept Form METHOD 1. Plug the point into the Slope Intercept Form and solve for b. Use values for m and b in the Slope Intercept Form 1. Determine the slope of the parallel or perpendicular line (e.g.. if it is parallel, it has the same slope). If the slope is undefined or 0, draw a picture. If the slope is a nonzero real number, go to If you have a point & a slope Ex. point (,) & m y ( x ) y x 6 y x 4 ( ) ( ) 4 Ex. point (,) & m y mx + b ( ) ( )( ) + b 6 + b b 4 y x 4 ( ) ( ) 4 Ex. point (,) & perpendicular to xaxis m undefined x Ex. point (,) & perpendicular to y x 4 m, so for perpendicular line m  1/ If you have points 1. Use the slope equation to determine the slope. Go to If you have a point & a slope Ex. ( 0, 0 ) & (, 6) 6 0 m 0 Copyright Sally C. Zimmermann. All rights reserved.
9 Part  Beginning Algebra Summary 4. Systems of Linear Equations 4.1. Definitions Type of Intersection IDENTICAL (I)  Same slope & same yintercept NO SOLUTION (N)  Same slope & different y intercept, the lines are parallel ONE POINT  Different slopes Identical Consistent Dependent Page 9 of 4 1/1/01 Terminology CONSISTENT SYSTEM  The lines intersect at a point or are identical. System has at least 1 solution INCONSISTENT SYSTEM  The lines are parallel. System has no solution DEPENDENT EQUATIONS  The lines are identical. Infinite solutions INDEPENDENT EQUATIONS  The lines are different. One solution or no solutions No solution Inconsistent Independent One point Consistent Independent 4.. Solving by Graphing 1 Graph both equations on the same Cartesian plane The intersection of the graphs gives the common solution(s). If the graphs intersect at a point, the solution is an ordered pair. Check the solution in both original equations 1 y x 1 y x 1 (0,1) 1 1 (0) 1 1 (0) 1 Copyright Sally C. Zimmermann. All rights reserved.
10 Part  Beginning Algebra Summary Page 10 of 4 1/1/ Solving by Substitution 1. Solve either equation for either variable. (pick the equation with the easiest variable to solve for). Substitute the answer from step 1 into the other equation. Solve the equation resulting from step to find the value of one variable * 4. Substitute the value form Step in any equation containing both variables to find the value of the other variable. 5. Write the answer as an ordered pair 6. Check the solution in both original equations x y 1 Solve x 4 6y 1 x 1. y 1 x. x 4 6. x + x + 4 x ( 1) y 1 5. ( 1, 1) 6. ( 1) ( 1) ( 1) 4 6( 1) 6 6 *If all variables disappear & you end up with a true statement (e.g. 5 5), then the lines are identical If all variables disappear & you end up with a false statement (e.g. 5 4), then the lines are parallel Copyright Sally C. Zimmermann. All rights reserved.
11 Part  Beginning Algebra Summary Page 11 of 4 1/1/ Solving by Addition or Subtraction 1. Rewrite each equation in standard form Ax + By C. If necessary, multiply one or both equations by a number so that the coefficients of one of the variables are opposites.. Add equations (One variable will be eliminated)* 4. Solve the equation resulting from step to find the value of one variable. 5. Substitute the value form Step 4 in any equation containing both variables to find the value of the other variable. 6. Write the answer as an ordered pair 7. Check the solution in both original equations x y 1 Solve x 4 6y 1. x y 1 x 6y 4.Multiply both sides of the first equation by x + 4 y x 6y 4. y 4. y 1 5. x ( 1) 1 x 1 6. ( 1, 1) 7. ( 1) + 4( 1) ( 1) 4 6( 1) 6 6 *If all variables disappear & you end up with a true statement (e.g. 5 5), then the lines are identical If all variables disappear & you end up with a false statement (e.g. 5 4), then the lines are parallel Copyright Sally C. Zimmermann. All rights reserved.
12 Part  Beginning Algebra Summary 5. Word Problems 1 UNDERSTAND THE PROBLEM As you use information, cross it out or underline it. DEFINE VARIABLES Create Let statement(s) The variables are usually what the problem is asking you to solve for WRITE THE EQUATION(S) You need as many equations as you have variables 4 SOLVE THE EQUATION(S) 5 ANSWER THE QUESTION Answer must include units! 6CHECK Plug answers into equation(s) 5.1. Solving 1 Variable, 1 Equation Page 1 of 4 1/1/01 Variables, Equations Method Method In a recent election for mayor 800 people voted. Mr. Smith received three times as many votes as Mr. Jones. How many votes did each candidate receive? Name what x is (Can only be one thing. When in doubt, choose the smaller thing) Define everything else in terms of x Let x Number of votes Mr. J x Number of votes Mr. S x + x 800 4x 800 x 00 Go back to your Let statement 00 Number of votes Mr. J 600 Number of votes Mr. S (00) + (00) Let x Number of votes Mr. S y Number of votes Mr. J Usually each sentence is an equation x + y 800 x y ( y) + y 800 (Substitution) 4 y 800 y 00 Go back to your Let statement 00 Number of votes Mr. J Go back to your Equations & solve for remaining variable x + (00) 800 x Number of votes Mr. S (600) + (00) (600) (00) Copyright Sally C. Zimmermann. All rights reserved.
13 6. Polynomials Term Polynomial Polynomial Name According to Number of Terms Part  Beginning Algebra Summary Degree of a Polynomial Determines number of answers (xintercepts) Polynomial Name According to Degree 6.1. Definitions Page 1 of 4 1/1/01 A constant, a variable, or a product of a constant and one or more variables raised to powers. A sum of terms which contains only whole number exponents and no variable in the denominator Number of Terms Polynomial Name Examples 1 Monomial x Binomial x + Trinomial x + x Express polynomial in simplified (expanded) form.. Sum the powers of each variable in the terms.. The degree of a polynomial is the highest degree of any of its terms Degree Polynomial Name Examples 1 Linear x Quadratic x Cubic x 4 Quartic x 4 x y Copyright Sally C. Zimmermann. All rights reserved.
14 Part  Beginning Algebra Summary Page 14 of 4 1/1/ Multiplication Multiply each term of the first polynomial by each term of the second polynomial, and then Horizontal Method Can be used for any size polynomials Vertical Method Can be used for any size polynomials. Similar to multiplying two numbers together FOIL Method 1. May only be used when multiplying two binomials. First terms, Outer terms, Inner terms, Last terms combine like terms Ex: ( x )( x + 5x 1) x( x + 5x 1) ( x + 5x 1) x x + x 5 x + x( 1) + ( ) x + ( ) 5 x + ( ) ( 1) x + 5x x x 10x + x + x 11x + Ex: ( x )( x + 5x 1 ) x 5x 1 x 1 x x 5x x 0x x + x x + 11 Ex: ( x )( x ) F O I L x x + x( ) + ( ) x + ( ) ( ) x x x + 6 x 5x + 6 Copyright Sally C. Zimmermann. All rights reserved.
15 Part  Beginning Algebra Summary Page 15 of 4 1/1/01 Write Each Numerator Term over the Denominator Method a + b + c a b c + + d d d d Factor Numerator and Cancel Method 6.. Division Dividing a Polynomial by a Monomial x + Ex : 4 x x + x + Ex: 4 ( x + 1) 4 x x + Copyright Sally C. Zimmermann. All rights reserved.
16 7. Factoring Factoring Part  Beginning Algebra Summary 7.1. GCF (Greatest Common Factor) Writing an expression as a product Numbers can be written as a product of primes. Polynomials can be written as a product of prime polynomials Useful to simplify rational expressions and to solve equations The opposite of multiplying > Factored ( x + ) > Not factored x + 4 x + factoring > x + 4 ( x + ) multiplying Page 16 of 4 1/1/01 GCF of a List of Integers GCF of a List of Variables GCF of a List of Terms Factor by taking out the GCF 1. Write each number as a product of prime numbers. Identify the common prime factors. The PRODUCT OF ALL COMMON PRIME FACTORS found in Step is the GCF. If there are no common prime factors, the GCF is 1 The variables raised to the smallest power in the list The product of the GCF of the numerical coefficients and the GCF of the variable factors 1. Find the GCF of all terms. Write the polynomial as a product by factoring out the GCF. Apply the distributive property 4. Check by multiplying > Find the GCF of 18 & GCF 6 > Find the GCF of x & x GCF x > Find the GCF of 18x& 0x GCF 6x > 1 + ( x) x + 6x ( x ) ( x ) x (1 x) x + 6x > x + 1 ( 1) ( x ) + ( 1) ( 1) 1( x 1) x + 1 Copyright Sally C. Zimmermann. All rights reserved.
17 Part  Beginning Algebra Summary Page 17 of 4 1/1/ Terms a + b + c + d (? +?)(? +?) FACTOR BY GROUPING 1. Arrange terms so the 1 st terms have a common factor and the last have a common factor. For each pair of terms, factor out the pair s GCF. If there is now a common binomial factor, factor it out 4. If there is no common binomial factor, begin again, rearranging the terms differently. If no rearrangement leads to a common binomial factor, the polynomial cannot be factored. > Factor 10ax 6xy 9y+15a 1. 10ax + 15a 6xy 9y. 5a(x + ) y(x + ) (x + )(5a y) Copyright Sally C. Zimmermann. All rights reserved.
18 Part  Beginning Algebra Summary Page 18 of 4 1/1/01 TRIAL & ERROR 7.. Trinomials: Leading Coefficient of 1 x + bx + c (x +?)(x +?) Product is c (x + one number)(x + other number) Sum is b 1. Place x as the first term in each binomial, then determine whether addition or subtraction should follow the variable x + bx + c ( x + d)( x + e) x bx + c ( x d)( x e) x ± bx c ( x + d)( x e). Find all possible pairs of integers whose product is c. For each pair, test whether the sum is b 4. Check with FOIL Ex: Factor x 1. ( x + )( x + ) x YES NO ( x + 5)( x + ) x + x + 10 Copyright Sally C. Zimmermann. All rights reserved.
19 Part  Beginning Algebra Summary Page 19 of 4 1/1/ Trinomials: All ax + bx + c (?x +?)(?x +?) METHOD 1 (trial & error) 1. Try various combinations of factors of ax and c until a middle term of bx is obtained when checking. Ex: Factor: Product is x x Product is 5 (x 1)( x + 5) + 14x 5. Check with FOIL METHOD (ac, factor by grouping) 1. Identify a, b, and c. Find magic numbers whose product is ac and whose sum is b. Factor trees can be very useful if you are having trouble finding the magic numbers (See MA090). Rewrite bx, using the magic numbers found in Step 4. Factor by grouping 5. Check with FOIL METHOD (quadratic formula) 1. Use the quadratic formula to find the x values (or roots) 15x x 14x (correct middle term) Ex: Factor: x + 14x 5 1. a b 14 c 5. ac ()( 5) 15 b 14 (15)( 1) 15 (15) + ( 1) 14 magic numbers 15, 1. x + 15x x 5 4. x(x + 5) 1(x + 5) (x + 5)(x 1) Ex: Factor: x + 14x 5 1. a b 14 c 5 14 ± 14 4( )( 5) x 6 1 x, 5. For each answer in step 1., rewrite the equation so that it is equal to zero. 1 x 1 x 0 x 1 0. Multiply the two expressions from step, and that is the expression in factored form. 4. Check with FOIL x 5 x ( x 1)( x + 5) Copyright Sally C. Zimmermann. All rights reserved.
20 Perfect Square Trinomials a ± ab + b Part  Beginning Algebra Summary 7.5. Perfect Square Trinomials & Binomials Factors into perfect squares (a binomial squared) a + ab + b ( a + b) a ab + b ( a b) Page 0 of 4 1/1/01 > 9x + 4x + 16 ( x) + ( x)(4) + (4) (x + 4) ( a x, b 4) > 9x 4x + 16 ( x) ( x)(4) + (4) (x 4) ( a x, b 4) Difference of Squares a b Sum of Squares a + b Difference of Cubes a b (MA10) Sum of Cubes a + b (MA10) Prime Polynomials (P) Factors into the sum & difference of two terms a b ( a + b)( a b) Does not factor a + b Prime a b ( a b)( a + ab + b ) a + b ( a + b)( a ab + b ) > x > 1 ( x) (1) x + ( x + 1)( x 1) 1 is prime (x )(4x 6 ( a x, b 1) > 8x 7 ( x) ( ) ( a x, b ) + > 8 7 ( ) ( (x + )(4x 6 x + 9) x + x + ) a x b Can not be factored > x + x + 1 is prime > x is prime (, ) x + 9) Copyright Sally C. Zimmermann. All rights reserved.
21 Part  Beginning Algebra Summary Page 1 of 4 1/1/ Steps to Follow 1. Put variable terms in descending order of degree with the 4 Ex. + x constant term last. 4 x. Factor out the GCF 4 ( x 16). Factor what remains inside of parenthesis ( x + 4)( x 4) TERMS see if one of the following can be applied Difference of Squares Sum of Cubes Difference of Cubes TERMS try one of the following Perfect Square Trinomial Factor Trinomials: Leading Coefficient of 1 Factoring Any Trinomial 4 TERMS try Factor by Grouping 1. If both steps & produced no results, the polynomial is prime. You re done (Skip steps 5 & 6). See if any factors can be factored further ( x + 4)( x + )( x ). Check by multiplying ( x + 4) [ ( x + )( x ) ] ( x + 8)( x 4) 4 x Copyright Sally C. Zimmermann. All rights reserved.
22 8. Quadratics Part  Beginning Algebra Summary 8.1. About Page of 4 1/1/01 Standard Form ax + bx + c 0 > x x + 0 Solutions Has n solutions, where n is the > x x + x 0 (has solutions) highest exponent Standard Form Solution Simple Form y ax Graph y ax + bx + c a, b, and c are real constants A parabola 8.. Graphing Vertex (high/low point) is (0,0) Line of symmetry is x 0 The parabola opens up if a > 0, down if a < 0 1. Plot y value at vertex. Plot y value one unit to the left of the vertex. Plot y value one unit to the right of the vertex > y x 9x + 0 > y 4x > y 4x x y Copyright Sally C. Zimmermann. All rights reserved.
23 Part  Beginning Algebra Summary Page of 4 1/1/01 Zero Factor Property Solve by Factoring 8.. Solve by Factoring 1. If a product is 0, then a factor is 0 > xy 0 (either x or y must be zero) 1. Write the equation in standard form (equal 0). Factor. Set each factor containing a variable equal to zero 4. Solve the resulting equations > x x x(x 1) (x ) 0. x 0, x 1 0, x 0. x 0, 1, Copyright Sally C. Zimmermann. All rights reserved.
24 Part  Beginning Algebra Summary Page 4 of 4 1/1/ Solve with the Quadratic Equation To solve a quadratic equation that is difficult or impossible to factor 1. Write the values for a, b, & c Ex Radicand is a perfect square ( if a term does not exist, the x x + 0 coefficient is 0) a 1, b ( ), c. Plug values into the ( ) ± ( ) 4(1)() ± 1 quadratic equation below: x (1) b ± b 4ac x, 1 a Ex Radicand breaks into perfect square and. Simplify solutions and usually leftovers leave them in their most x + 6x 1 0 exact form a 1, b 6, c ( 1) ( Negative radicand means (6) ± (6) 4(1)( 1) no real solutions) x (1) 6 ± 40 6 ± ± ± 10 Ex Radicand is just leftovers 4x x 1 0 a 4, b ( 1), c ( 1) x ( 1) ± ( 1) 4(4)( 1) 1 17 ± 8 (4) Copyright Sally C. Zimmermann. All rights reserved.
25 9. Exponents Part  Beginning Algebra Summary Exponential notation Shorthand for repeated multiplication Multiplying common bases Add powers Dividing common bases Subtract powers Raising a product to a power Raise each factor to the power Raising a quotient to a power Raise the dividend and divisor to the power 9.1. Computation Rules base x a exponent 8 a b a b x x x + m x m n x n x ( xy) x y a a a ( x y ) x y m n a ma na n x x y y Raising a power to a power a ( ) Multiply powers Raising to the zero power One Raising to a negative power Reciprocal of positive power When simplifying, eliminate negative powers x x n n b a x x b 0 n Page 5 of 4 1/1/ ( x )( y)( 4x) 4x ( ) x 9 x 4x 6 4 z ( ) 1, when x 0 1 n x z z (1) x (6) y Copyright Sally C. Zimmermann. All rights reserved.
26 Scientific Notation Standard Form Standard Form to Scientific Notation Scientific Notation to Standard Form Part  Beginning Algebra Summary 9.. Scientific Notation Shorthand for writing very small and large numbers a 10 r, where 1 a<10 & r is an integer Page 6 of 4 1/1/01 (1. 10 )(1. 10 ) Long way of writing numbers Move the decimal point in the original number to the left or right so that there is one digit before the decimal point. Count the number of decimal places the decimal point is moved in STEP 1 If the original number is 10 or greater, the count is positive If the original number is less than 1, the count is negative. Multiply the new number from STEP 1 by 10 raised to an exponent equal to the count found in STEP Multiply numbers together Copyright Sally C. Zimmermann. All rights reserved.
27 10. Radicals Roots Computation Part  Beginning Algebra Summary Undoes raising to powers 81 9 because 9 81 index Definitions 81 radical radicand If n IS AN EVEN POSITIVE INTEGER, then n n a a The radical represents only the nonnegative square root of a. The represents the negative square root of a. IF n IS AN ODD POSTIVIE INTEGER, then n n a a > > > > > > Page 7 of 4 1/1/01 (The square root of 81 is 9) (The cube root of 7 is ) 9 ( ) + ( x 1) (...09) x + 1 "Not a real number" > 9 > > > > (approximately) 7 7 ( ) Notation: Radical vs. Rational Exponent The root of a number can be expressed with a radical or a rational exponent Rational exponents The numerator indicates the power to which the base is raised. The denominator indicates the index of the radical > > 7 (7) 1 ( ) ( ) / 1/ > 7 ( 7 ) 7 7 Note, it s usually easier to compute the root before the power / 1/ Copyright Sally C. Zimmermann. All rights reserved.
28 Part  Beginning Algebra Summary Page 8 of 4 1/1/01 Operations 10.. Computation Rules Roots are powers with fractional exponents, thus power rules apply. > 8 x ( 8 x ) 1/ 1/ 1/ ( 8) ( x ) Product Rule n n n a b ab > Quotient Rule n a a n n 1 1 1,provided b 0 > n b b Simplifying 1. Separate radicand into perfect > Just perfect squares... Expressions squares and leftovers 6x 6x. Compute perfect squares > Prefect squares & leftovers.... Leftovers stay inside the radical so the answer will be exact, not rounded x 16x x 4x x > Just leftovers... x x x Copyright Sally C. Zimmermann. All rights reserved.
29 11. Rationals Rational Numbers Irrational Numbers Rational Expression Simplifying Rational Expressions (factor) Part  Beginning Algebra Summary Simplifying Expressions Can be expressed as quotient of integers (fraction) where the denominator 0 All integers are rational All terminating decimals are rational Cannot be expressed as a quotient of integers. Is a nonterminating decimal 1. An expression that can be written in the form P, where P and Q are polynomials Q. Denominator 0 1. Completely factor the numerator and denominator. Cancel factors which appear in both the numerator and denominator > 0 0/1 Page 9 of 4 1/1/01 > 4 4/1 > /4 > π > x >, Find real numbers for x + 6 which this expression is undefined: x + 6 0; x 6 4x + 0 > Simplify x 5 4( x + 5) ( x + 5) ( x 5) 4 x 5 Copyright Sally C. Zimmermann. All rights reserved.
30 Part  Beginning Algebra Summary Page 0 of 4 1/1/01 Multiplying/ Dividing Rational Expressions (multiply across) Adding/ Subtracting Rational Expressions (get common denominator) 11.. Arithmetic Operations 1. If it s a division problem, change it to a multiplication problem x x. Factor & simplify > Simplify x + 6 x x ( x +. Multiply numerators and multiply denominators 4. Write the product in simplest form x Factor & simplify each term x. Find the LCD > Simplify + x The LCD is the product of all unique LCD 6( x + 6) factors, each raised to a power equal?? to the greatest number of times that + it appears in any one factored 6( x + 6) 6( x + 6) denominator (6) x ( x + 6) +. Rewrite each rational expression as an (6)( x + 6) 6( x + 6) equivalent expression whose denominator is 6x x + 18 the LCD + 6( x + 6) 6( x + 6) 4. Add or subtract numerators and place the sum or difference over the common 9x + 18 denominator 6( x + 6) 5. Write the result in simplest form 9 ( x + ) 6 ( x + 6) x + 6 ( x + 6) 6) Copyright Sally C. Zimmermann. All rights reserved.
31 Part  Beginning Algebra Summary Page 1 of 4 1/1/01 Solving by Eliminating the Denominator Solving Proportions with the Cross Product a c b d 11.. Solving Equations 1. Factor & simplify each term. Multiply both sides (all terms) by the LCD. Remove any grouping symbols 4. Solve 5. Check answer in original equation. If it makes any of the denominators equal to 0 (undefined), it is not a solution If your rational equation is a proportion, it s easier to use this shortcut 1. Set the product of the diagonals equal to each other. Solve. Check x Solve + 1 x + 6 x LCD x( x + 6) x 1 x + 6 x x( x + 6) x x( x + 6) 1 + x x x( x) + ( x + 6) 1( x + 6 x) x + x + 18 x + 6x x 6 (6) Check + 1 (6) + 6 (6) x Solve 4 x 1 [ x( x + 6) ] + [ x( x + 6) ] ( x 1) 4x x 4x x ( ) Check 4 ( ) 1 [ x( x + 6) ] 1 Copyright Sally C. Zimmermann. All rights reserved.
32 1. Summary Geometric Part  Beginning Algebra Summary Triangle Formulas Page of 4 1/1/01 SUM OF ANGLES: Angle 1 + Angle + Angle 180 o Right Triangle PYTHAGOREAN THEOREM: a + b c c (a leg, b leg, c hypotenuse) b ~The hypotenuse is the side opposite the right angle. It is a always the longest side. Other Distance DISTANCE: d rt (r rate, t time) Copyright Sally C. Zimmermann. All rights reserved.
33 Part  Beginning Algebra Summary Page of 4 1/1/ Types of Equations 1 Variable Variables Linear Equations x 0 MA090 Solution: 1 Point y x 8 Solution: Line 0 Linear Inequalities x < 0 page 4 Solution: Ray 0 x 7 Systems of Linear Equations y 5 page 10 Solution: 1 point, infinite points or no points y > x Solution: ½ plane page y x page y x + 10 Solution: 1 point, infinite points or no points Quadratic Equations x +5x Solution: Usually points page y x Solution: Parabola page Higher Degree x + 5x + 6x 0 page y x + 5x + 6x Polynomial 4 Solution: Curve Equations (cubic, quartic, etc.) Solution: Usually x points, where x is the highest exponent Rational Equations x 1 x 1 1 page y + 1 x + 1 x Solution: Sometimes simplifies to Solution: Sometimes simplifies to a linear or quadratic equation a linear or quadratic equation * To determine the equation type, simplify the equation. Occasionally all variables cancel out. If the resulting equation is true (e.g. 5 5), then all real numbers are solutions. If the resulting equation is false (e.g. 5 4), then there are no solutions. Copyright Sally C. Zimmermann. All rights reserved.
34 Part  Beginning Algebra Summary 1.. Solve Any 1 Variable Equation Page 4 of 4 1/1/01 Is it really an equation? No It s an expression, you can t solve it. You can factor, expand & simplify it Yes Make an equivalent, simpler equation If the equation contains fractions, eliminate the fractions (multiplying both sides by the LCD) If there is a common factor in each term, divide both sides of the equation by the common factor Can the variable be isolated? Yes Solve by undoing the equation Linear equations can by undone with the addition, subtraction, multiplication & division equality properties Quadratics, of the form (x + a) b, can be undone with the square root property No Write the equation in standard form Make one side equal to zero Put variable terms in descending order of degree with the constant term last Can it easily be put in factored form? Yes Solve by Factoring No Is it a quadratic equation? Yes Solve with the Quadratic Equation or Solve by Completing the Square No Not covered in this class Check solutions in the original equation Copyright Sally C. Zimmermann. All rights reserved.
MA094 Part 2  Beginning Algebra Summary
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