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5.1 • A quadratic function f is a function of the form f(x) = ax2 + bx + c where a, b and c are real numbers and a not equal to zero. The graph of the quadratic function is called a parabola. It is a "U" shaped curve that may open up or down depending on the sign of coefficient a. Jeff White Graphing A Quadratic Function Vertex is (2,-2) Then draw the axis of symmetry which is x=2 Then plot two points on one side of the axis of symmetry. Use symmetry to plot two more points. Jeff White Graphing A Quadratic Function In Vertex Form Vertex Form First plot the vertex (H,K) = (-3,4) Then draw the axis of symmetry X=-3 and plot two points on one side of it. Use symmetry to complete the graph. Jeff White Graphing A Quadratic Function In Intercept Form Intercept Form X-intercept occur at (-2,0) and (4,0) Axis of symmetry is 1 X-coordinate of the vertex is x=1. The y-coordinate of the vertex is Jeff White 5.4 Complex Numbers • • • Imaginary unit is called i i= √ (-1) r is a positive real number √ (–r)= i √ (r) Complex number written in standard form is a+bi a & b real numbers a real part of complex number b imaginary part of complex number • b ≠0 a+bi is imaginary • a=0, b≠0 a=bi is pure imaginary number • z=a+bi is complex number Sample Problems • 1. Solve • 3. Write the (2+3i)+(7+i) as a complex number in standard form. • 5. Divide 8 1 i • 2. Write (8+5i)(1+2i) as a complex number in standard form. • 4. Multiply i(3+i). 6. Find the absolute value of 3-4i Helpful Hints • • • • • 1. Take the square root of x squared and -4. 2. Distribute the minus sign to 1 and 2i. Combine like terms. 3. Distribute the plus sign to 7 and i. Combine like terms. 4. Distribute the i to 3 and i. 5. Divide 8 by 1 and 8 by i. • 6. Consult the formula on the first page: a=3, b=-4 Answers • 1. • 2. • 3. • 4. • 5. • 6. Chapter 5.5: Completing the Square Goal 1: Solving Quadratic Equations by Completing the Square Completing the square is a process that allows you to write an expression of the form x2 + bx as the square of a binomial. To complete the square for x2 + bx, you need to add (b/2) 2. The following is a rule for completing the square: X2 + bx +(b/2)2 = (x+[b/2])2 Example 1: Completing the Square: Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial. x2 + 18x + c Write the equation out b=18 Use the formula to find b c = (b/2)2 = (18/2)2 = 92 = 81 Find the value of c that makes the expression a perfect square trinomial x2 + 18x + 81 Substitute the C value in the expression. (x+9)2 Factor to get your answer Matt Chapter 5.5: Completing the Square Example 2: Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square. X2 + 2x = 9 Write out original equation X2 + 2x + 1 = 10 Add (2/2)2 = 12 = 1 to each side (x+1)2 = 10 Write the left side as a binomial squared X + 1 = √10 Take the square roots of each side X = -1 + √10 Solve for x Example 3: Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square. 6x2 +84x +300 = 0 Write the original equation X2 +14x +50 = 0 Divide both sides by the coefficient of x2 X2 + 14x = -50 Write the left side in the form of x2 + bx X2 + 14x + 49 = -1 Add (14/2)2 = 72 = 49 to each side (X + 7)2 = -1 Write left side as a binomial squared X + 7 = √-1 Take the square roots of each side X = -7 ± √-1 Solve for x X = -7 ± i Write in terms of the imaginary unit i Chapter 5.5: Completing the Square Goal 2: Writing Quadratic Functions in Vertex Form: Given a quadratic function in standard form, y = ax2 + bx + c, you can use completing the square to write the function in vertex form, y = a(x – h)2 + k. Example 4: Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex. Y = x2 – 6x + 11 Write out the original function Y + 9 = (x2 – 6x + 9) + 11 Complete the square of x2 – 6; add (-6/2)2 = -32 = 9 Y + 9 = (x – 3)2 + 11 Write x2 – 6x + 9 as a binomial squared Y = (x – 3)2 +2 Solve for y Vertex = (3,2) Chapter 5.5: Completing the Square Practice Problems for Completing the Square: Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial. X2 – 44x + c Answer: C = 484; (x – 22)2 Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square. X2 + 20x + 104 = 0 Answer: -10 + 2i, -10 – 2i Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square. 2x2 – 12x = -14 Answer: 3 ± √2 Practice Problems for Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex. Y = x2 – 3x – 2 Answer: y = (x – [3/2])2 – (17/4) Vertex: ([3/2], [-17/4]) Algebra II Section 5.6 A presentation by: Elise Couillard; Block 5 June 9, 2010 Solving Equations With The Quadratic Formula • By completing the square once for the general equation , you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula. The Quadratic Formula • The Quadratic Formula: *Let a, b, and c be real numbers such that a does not equal 0. The solutions of the quadratic equation are: Solving a Quadratic Equation With 2 Real Solutions • Solve Number and Type of Solutions of a Quadratic Equation • Consider the quadratic equation *If >0, then the equation has two real solutions. *If =0, then the equation has one real solution. *If <0, then the equation has two imaginary solutions. Sources: Algebra II Textbook THE END! Quadratic Inequality in 2 Variables y < ax2 + bx +c y > ax2 + bx +c y < ax2 + bx +c y < ax2 + bx +c Quadratic Inequality in 1 Variable ax2 + bx +c < 0 ax2 + bx +c > 0 Example 1 1. Graph y < x2 + 8x + 16 2. Test the point (0,0) y < x2 + 8x + 16 0 < 02 + 8(0)+ 16 0 < 16 3. Shade outside region because 0 < Lesson 5.7 Jessica Semmelrock 16 ax2 + bx +c < 0 ax2 + bx +c > 0 Example 2: Graph the system of quadratic inequalities y > x2 y < x2 + 3 Only difference to this problem is when the shading overlaps that is your answer. Lesson 5.7 Jessica Semmelrock Example 3:Solve x2 + x -2 < 0 by graphing Step 1: x2 + x -2 = 0 • Find graph intercepts by replacing 0 for x. Step 2: x = -1 + 2) 1 – 4(1)(- 2(1) •Use quadratic formula to solve for x. Answer: x ≤ –1.37 or x ≥ .37 Lesson 5.7 Jessica Semmelrock Example 4:Solve x2 + 3 -18 > 0 algebraically x2 + 3 -18 > 0 x2 + 3 -18 = 0 (x-3) (x+6) = 0 Test an x-value in each interval to see if it satisfies the inequality. Test these points with the arrows. Answer x = 3 or x = -6 Lesson 5.7 Jessica Semmelrock Travis Deskus 6.1 / 7.2 - Using Properties of Exponents Evaluating Numerical Expressions • 1. Product of like bases: Example: x5 x3 = x5+3 = x8 • To multiply powers with the same base, add the exponents and keep the common base. • =32 • 2.Evaluating numerical expresions a. Simplifying Algebraic Expressions • • • • a. b. c. Scientific Notation 7.2 Properties of Rational Exponents • • • Example 1. If m= for some integer n greater than 1, the third and sixth properties can be written using radical notation as following: – – product property Quotient property – a. – b. – Using Properties of Radicals Writing Radicals in Simplest Form • • a. steps- factor out perfect cube, product property, simplify Adding and Subtracting Roots and Radicles • • Book Example The properties of rational exponent and radicals can also be applied to expressions involving variables. Because a variable can be positive, negative or zero, sometimes absolute value is needed when simplifying a variable expression. • • =x when n is odd = when n is even Simplifying expressions involving varribles Section 6.2 IDENTIFYING POLYNOMIAL FUNCTIONS • A polynomial is a function if it’s in standard form and the exponent is a whole number • ex: f(x)= 3 • If the polynomial has an exponent that is not a whole number it’s not a function • ex: f(x)= 3x1/2 – 2x2 +5 Using Synthetic Substitution • • • • • • • Write the polynomial in standard form Insert terms with coefficients of 0 for missing terms Then write the coefficients of f(x) in a row Bring down the leading coefficients and multiply them by 1 Write results in the next column and bring down your results Continue until you reach the end of the row EX: Graphing Polynomials Functions • Begin by making a table of values, including positive, negative, and zero values for x • Plot the points and connect them with a smooth curve. Then check the behavior X • • • • • -3 -2 -1 0 1 2 3 f(x) -218 is odd -18 and 6the leading 4 6 42 is positive, 262 The Degree coefficients so f(x)→ + As x → f(x) → + As x → + Examples of Graphs Work • • • • • • • • • • • Graph the polynomial function 1.) f(x)= x4+3 2.) g(x)= x3-5 3.) h(x)= 2+ x2-x4 Synthetic substitution 1.) f(x)=x3+ 5x2+4x+6, x=2 2.) f(x)= x3-x5 +3, x=-1 3.) f(x)= 5x3-4x2-2, x=0 Functions yes or no 1.) f(x)= x4+3 2.) f(x)=5x3/4-5x2+3 Adding, Subracting, and Multiplying Polynomials 6.3 By: Robert Johnson ● How to Solve To add or subtract polynomials, add or subtract the coefficients of LIKE terms. You can do this by using a vertical or horizontal format To Multiply two polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial. Then combine LIKE terms ● Adding Polynomials Add 2x3-5x2+3x-9 and x3+6x2+11 in vertical format Add 3x3+2x2-x-7 and x3-10x2+8 in horizontal format Subtracting Polynomials Subtract 3x3+2x2-x+7 from 8x3-x2-5x+2 in vertical format Subract 8x3-3x2-2x+9 from2x3+6x2-x+1 in horizontal format Multiplying Polynomials Multiply -2y2+3y-6 and y-2 in vertical format Multiply -x2+2x+4 and x-3 in horizontal format Special Product Patterns ● Sum and difference ● (a+b)(a-b)= a2 - b2 ● Square of a Binomial ● (a+b)2 = a2 + 2ab + b2 ● (a-b)2 = a2 - 2ab + b2 ● Cube of a Binomial ● (a+b)3 = a3 + 3a2b + 3ab2 + b3 ● (a-b)3 = a3 - 3a2b + 3ab2 - b3 Using Special Product Patterns (x + 2)(3x2 - x – 5) (a – 5)(a + 2)(a + 6) (xy - 4)3 Anjy Grasso x² + 8x + 12 = 0 x² + 8x = −12 Now that you have the equation you're solving, find the first factor Put the equation into Standard form In other words, make it equal zero. x² + 8x + 12 = 0 ( ) ( )=0 x² + 8x = −12 +12 +12 the only factors of x2 are x * x, you now have the first factors. x² + 8x + 12 = 0 x² + 8x + 12 = 0 (x ) (x )=0 Anjy Grasso Now that we have the first factors, the x2 goes away, and we're left with this: 8x + 12 = 0 (x ) (x )=0 Now we find the factors of 12. 1 and 12 wont work, neither will 4 and 3, so lets use 2 and 6 8x + 12 = 0 (x ) (x )=0 (x 6) (x 2) = 0 Since the entire equation was positive, Both of these should be positive too. (x + 6) (x + 2) = 0 Anjy Grasso So how exactly do we know this is right? Lets use FOIL (first, outer, inner, last) multiplication to test it out. (x + 6) (x + 2) = 0 F - First x * x = x2 O - Outer x * 2 = + 2x I - Inner 6 * x = + 6x Now Simplify… L - Last 6 * 2 = + 12 x2 + 8x + 12 x2 + 2x + 6x + 12 There's Your original equation! Anjy Grasso Polynomial Long Division – when dividing a polynomial f(x) by a divisor d(x), you get a quotient polynomial -q(x) and a remainder polynomial r(x) f(x) = q(x) + r(x) d(x) d(x) Remainder Theorem – If polynomial f(x) is divided by x-k, then the remainder is r = f(x) Synthetic Division – only use the Coefficients of the polynomial and the x – k must be in the form of a divisor. Factor Theorem – A polynomial f(x) has a factor x-k if and only if f(k) = 0 Rachael SKinner Long Division f(x) = 3x² + 4x -3 by x² - 3x + 5 3x² + 4x -3 x² - 3x + 5 3x⁴ - 5x³ 0x² + 4x – 6 -3x⁴- 9x³+ 15x² 4x² - 15x² + 4x -4x² - 12x² - 20x -3x² - 16x-6 -3x² + 9x -15 25x+9 3x² + 4x -3 25x+9 x² - 3x + 5 Remainder Rachael SKinner * Don't forget to add in exponents if needed exponents must go in numerical order * Remember to subtract – which means the signs will change Synthetic Division Factoring Completely f(x) = 2x³ +x² - 8x +5 by x + 3 x+3=0 x–3=-3 x=-3 f(x) = 3x³ - 4x² - 28x – 16 x +2 is a factor x=2 3 2 2 1 -8 5 -6 2 -5 -28 -16 -6 20 16 *Use only coefficients 3 -3 -4 -10 -8 0 no remainder 15 -21 f(x) (2 + x) (3x² - 10x -8) 7 -16 Remainder f(x) (x+2) (3x+2) (x – 4) finding the 0's of f(x) (x+2) = -2 (3x+2) = -2/3 (x – 4) = 4 2x³ +x² - 8x +5 -16 x+3 Rachael SKinner Factor Solve 1.) Divide 2x⁴ + 3x³ + 5x – 1 by x² – 2x +2 Using Long Division 2.) Divide x³ + 2x² - 6x – 9 by x-2 Using Synthetic Division 3.) Factor Completely given that x-4 is a factor f(x) = x³ + 6x² + 5x +12 Rachael SKinner 1.) 2x² + 7x +10 + 11x-21 x² -2x +2 2.) x² + 4x + 2 + -5 x- 2 3.) f(x) = (x-3) (x-4) (x+1) Rachael SKinner How to find the rational zeros of a polynomial function Lesson 6.6 Nate Rational Zero Theorem • If f(x)=anxn+…..+a1x+a0 has integer coefficients, then every rational zero of f has the form: P/Q = factor of constant term a0 / factor of leading coefficient an Nate Steps to Solve • List all possible rational zeros, both positive and negative. • Change the number that is outside until one of these numbers makes your answer zero. • When your answer is zero, insert the numbers you got in order to find the value of x. Nate Example Solve: f(x)=3x3-4x2-17x+6 Nate Answer 3 -4 -17 6 -2 -6 20 -6 Synthetic Division 3 -10 3 0 x= 1,2,3,6 (all #’s are +or -) / 1,3 (all #’s are + or -) f(x)=(x+2)(3x2-10x+3) f(x)=(x+2)(3x-1)(x-3) x=-2,3,1/3 Factor the trinomial and use the factor theorem Nate 7.1 nth roots • n is an integer greater than 1 and a is a real number: • n is odd, a has one nth root: =a • n is even and a>0 a has 2 nth roots: = a • n is even and a=0 a has 1 nth root: =0 =0 • n is even and a<0 a has 0 nth roots Loren 7.1 nth roots • a is the nth root of a and m is a positive integer: • a =(a ) m =( ) m • a = = m= m,a 0 Loren Practice Problems • Rewrite using rational exponent notation. • Fine the indicated real nth root(s) of a n=4, a=0 • Evaluate the expression Loren Section 7.3 Power Functions and Function Operations Andy Examples 1 2 1 2 Let f ( x) 4 x and g ( x) 9 x . Find the following. a.) f ( x) g ( x) = 4x1/2 + (–9x1/2) b.) f ( x) g ( x) = 4x1/2 – (–9x1/2) 1. Combine the coefficients1. Combine the = [4 + (–9)]x1/2 = [4 – (–9)]x1/2 coefficients 2. Simplify 2. Simplify = –5x1/2 = 13x1/2 Andy Practice Problems 2 3 2 3 Let f ( x) 2 x and g ( x) 4 x . Find the following. 1. f (x) + g(x) 2. f (x) – g(x) Andy Examples 3 4 Let f ( x) 6 x and g ( x) x . Find the following. a. f (x) f (x) g(x) b. Plug in equation = (6x)(x3/4) Combine powers by adding = 6x(1 + 3/4) Simplify = 6x7/4 g(x) Plug in equation 6x = 3/4 x Combine powers by subtraction = 6x(1 – 3/4) Simplify = 6x1/4 Andy Practice Problems 1 5 Let f ( x) 4 x and g ( x) x . Find the following. 4. f (x) g(x) 5. f (x) g(x) Andy Examples Let f ( x) 4 x a. 1 and g ( x) 5x 2. Find the following. f(g(x)) Plug in equation = f(5x – 2) Plug g(x) into the equation = 4(5x – 2)–1 g(f(x)) b. Plug in equation = g(4x–1) Plug g(x) into the equation = 5(4x–1) – 2 Distribute the 5 Plug in the coefficient and power of f(x) = –20x–1 – 2 Simplify 20 = x –2 Andy Examples Let f ( x) 4 x c. 1 and g ( x) 5x 2. Find the following. f(f(x)) Plug in equation = f(4x–1) Plug f(x) into the equation = 4(4x–1)–1 Distribute the exponent = 4(4–1 x) Distribute the 4 = 40x OR =x Andy Practice Problems Let f ( x) 2 x 6 and g ( x) 3x 2 . Find the following. a. f(g(x)) b. g(f(x)) c. f(f(x)) Andy 7.4 Inverse Functions Concepts Inverse relations – maps the output values back to their original input values •Domain – of inverse relation is the range of the original relation. •Range – of inverse relation is the domain of the original relation. •Inverse of linear functions: reflect original relation over y=x to obtain the inverse relation. •Which means switch the roles of x & y and solve for y (if possible). Ileishka Ortiz 7.4 Original relation Inverse relation DOMAIN DOMAIN RANGE RANGE Ileishka Ortiz 1.) Find an equation for the inverse of the relation y =2x–4 SOLUTION y=2x–4 Write original relation. x Switch x and y =2y – 4 x + 4 = 2y Add 4 to each side. 1 x+2=y 2 Divide each side by 2. Ileishka Ortiz 7.4 Functions f and g are inverses of each other provided: f (g (x)) = x and The function g is denoted by f g ( f (x)) = x –1 , read as “f inverse.” Given any function, you can always find its inverse relation by switching x and y. For a linear function f (x ) = mx + b where m 0, the inverse is itself a linear function. Ileishka Ortiz 7.4 1.) Verify that f (x) = 2 x – 4 and g (x) = 1/2 x + 2 are inverses. SOLUTION Show that f (g (x)) = x and g (f (x)) = x. f (g (x)) = f ( 1/2 x) + 2 = 2 (1/2 x ) + 2 – 4 = x+4 – 4 = x Ileishka Ortiz g (f (x)) = g (2x – 4) = 1 (2x – 4) + 2 2 = x–2+2 = x 7.4 PRACTICE Ileishka Ortiz 7.5 + 7.6 With Da Gawd • By: Jon Reyyashi Its Easy To graph y=a√x-h + k or y=a^3√x-h + k y=a√x 0r y=a^3√x H+K H means shift the graph left our right and K means move the point either up or down. 7.5 LET US START WITH 7.5 SHALL WE 1. √X+1-3 SO LETS SAY IT IS √X-(-1)+(-3). H=-1 AND K= 3’ TO OBTAIN THE GRAPH OF Y=√X+1-3, SHIFT THE GRAPH OF Y=√X LEFT 1 UNIT AND DOWN 3 UNITS GRAPHING IT EXAMPLE - Y=-3√X-2 + 1 DRAW THE GRAPH OF Y=-3√X DASHED LINE IN GRAPH IT BEGINS AT THE ORIGIN AND PASSES THROUGH (1, -3) YOU THEN SHIFT THE GRAPH 2 UNITS UP AND UP 1 UNIT. THE GRAPH STARTS (2,1) AND PASSES THROUGH POINT (3, -2) Domain and Range State the domain and range of the functions From the graph of y=-3√x-2 + 1 the domain is x≥2 and the range is y≤1 Another Example ANOTHER EXAMPLE 3^3√X+2 - 1 1. Sketch the graph of y=3^3√x+2 - 1 This means h=-2 and k=-1 Sketch the graph of y=3^3√x It passes the origin and points (-1, -3) and (1, 3) You then shift the graph to the left two and down one You repeat these steps for each point. Example - (0,0) becomes (-2, -1) because you basically move the point over -2 and down one so it becomes (-2, -1) Domain and Range Domain and range are both all real numbers 7.6 Powers Property of Equality: If a=B then A^n = B^n This means that you can raise each side of an equation to the same power An extraneous solution is a trial solution that does not satisfy the original equation Example √x + 5 = 9 Isolate the radical by subtracting 5 by each side (√x)^2 = 4^2 Simplify x = 16 Radical Exponents Example 3x^3/4 = 192 Isolate the power by dividing each side by 3 (x^3/4) = 64 Raise each side by 4/3 power, and cancel the original 4/3 power by performing its reciprocal x = (64^1/3)4 Apply Properties of roots x = 4^ 4 = 256 Simplify (The solution is 256) Check answer by substituting. With Only One Radical 3√8x+3 - 5 = -2 Isolate the radical by adding 5 to each side 3√8x+3 = 3^3 Cube each side 8x +3 = 27 Subtract each side of by 3 8x=24 Simplify x=3 Two Radicals 3√2x+4 = 2^3√3-x Cube each side (3√2x+4)^3 = (2^3√3-x)^3 Simplify 2x + 4 = 8(3-x) Distribute 2x + 4 = 24 - 8x ( Add 8 x to each side 10x + 4 = 24 x=2 Follow These Rules and You will be GAWDLY like Me and get a 100 on the Final. Thank You :) Section 9.3 Y = p (x) / q (x) = amxm+am-1xm-1+…+ax+a0 bnxn+bn-1xn-1+…+bx+b0 Sydney Section 9.3 - Concepts • P (x) and q (x) are polynomials with no common factors other than 1. • The graph of the function p (x) /q (x) has the following characteristics -the x intercepts are the real zeros of p (x) *you set the polynomial p (x) equal to zero and solve* -there is a vertical asymptote at each real zero of q (x) *you set the polynomial q (x) equal to zero and solve* -at most there is one horizontal asymptote Sydney Section 9.3 – Concepts cont. • If m < n, the line y=0 is the horizontal asymptote • If m = n, the line y = Am / Bn is the horizontal asymptote • If m > n, the graph has no horizontal asymptote. Sydney Section 9.3 – Example 1 (m<n) • Graph: y=4 x2 + 1 • Answer: P (x) has no real zeros, so no xintercepts The denominator has no real zeros, so no vertical asymptote M<n, so the horizontal asymptote is y = 0 Sydney Section 9.3 – Example 2 (m=n) • Graph: y = 3x2 x2-4 • Answer: The numerator’s only zero is zero. X-intercept is (0,0) Vertical asymptote’s at 2 and 2 M=N so horizontal asymptote is at 3 Sydney Section 9.3 – Example 3 (M>N) • Graph: y = x2-2x-3 x+4 • Answer: X-intercepts are 3 and -1 Vertical asymptote is -4 M>n, so no horizontal asymptote Sydney 9.4 Multiplying and Dividing Rational Expressions Brett Robinson • Let a, b, and c be nonzero real numbers or variable expressions. Then the following property applies: • Divide out common Factor c Brett Robinson Multiplying Factor Numerator and Denominator Divide out common factors before multiplying if possible Simplified Form Brett Robinson Dividing Multiply by the reciprocal Factor Divide out common Factors Simplified Form Brett Robinson Section 9.5 Addition, Subtraction, and Complex Fractions Adding and Subtracting with Like Denominators: Solve the following problems: Step I: Add numerators and simplify the expressions 4 5 4 5 9 3 3x 3x 3x 3x x QuickTime™ and a decompressor are needed to see this picture. Step I: Subtract numerators 2x 4 2x 4 x 3 x 3 x 3 Jorge Verde Adding with Unlike Denominators QuickTime™ and a decompressor are needed to see this picture. Solution: 5 Step I: First find the LCD of 6 x 2 and 4 x The LCD is 12 x 2 2 x 12 x ( x 3) each expression. Step 2: Use this to rewrite 5 x 5 x 5[2(x 3)] x(3x) 2 2 2 2 6x 4x 12x 6x 4x(x 3) 6x [2(x 3)] 4x(x 3)(3x) 10x 30 3x 2 2 12x (x 3) 12x 2 (x 3) 3x 2 10x 30 12x 2 (x 3) Jorge Verde Subtracting with Unlike Denominators QuickTime™ and a decompressor are needed to see this picture. Solution x 1 2 x 1 2 2 2 2 x 4x 4 x 4 ( x 2) ( x 2)( x 2) ( x 1)( x 2) 2(x 2) 2 (x 2) ( x 2) (x 2)( x 2)( x 2) x 2 x 2 (2 x 4) ( x 2) 2 (x 2) x 2 3x 6 (x 2) 2 ( x 2) Jorge Verde Simplifying a Complex Fraction QuickTime™ and a decompressor are needed to see this picture. Solution Step I: Add fractions in denominator. Step 2: Multiply by reciprocal. QuickTime™ and a decompressor are needed to see this pictur e. Step 3: Divide out common factor. Step 4: Write in simplified form. Quick Time™a nd a dec ompr esso r ar e nee ded to see this pictur e. QuickTime™ and a decompressor are needed to see this picture. Qu ic kT i me™ a nd a de co mpre ss or are n ee de d t o s ee t h is pi ct u r e. Jorge Verde Section 9.6 Solving Rational Equations Morgan Hillhouse Different ways to Solve Rational Expressions Cross Multiplying • Set up your equation. • Multiply together the top expression on the left with the bottom on the right and the top expression on the right with the bottom of the left. • Set the result equal to zero. • Solve for each part. Morgan Hillhouse Multiple Rational Expressions • Set up your equation. • Multiply each side if the equation by the LCD of both terms. • Simplify each side. • Set the simplified form to zero and solve for each result. • Check for extraneous solutions. Cross Multiplying X ― X-4 = 3 ― 4 4X=3X-12 -3X 4X=-12 4 X=-3 Morgan Hillhouse Continued Problem #2 X+6 ―― X = Morgan Hillhouse 7 ― 8 8X+48=7X -48 -7X X=-48 Multiple Rational Expressions 2X ―― X-2 - (3X+2) (X-2) Morgan Hillhouse 4X 17X+4 ―― = ―――― 3X+2 3X₂-4X-4 2X ―― X-2 (3X+2) (X-2) - 4X 17X+4 ―― = ―――― 3X+2 (3X+2)(X-2) CONTINUED ON NEXT SLIDE Continued 2X(3X+2)-(4X-1)(X-2)=17X+4 6X₂+4X-4X₂+9X-2=17+4 -17-4 2X₂-4X-6=0 6X₂+4X-(4X₂-9X+2)=17X+4 Morgan Hillhouse 2(X₂-2X-3)=0 2(X-3)(X+1)=0 X-3=0 X+1=0 X=3 or -1