P.5 Part I Solving Equations Objective: Solve equations with a linear nature. Example 1 (Example 2, p. 54) Solve x 3x 2 3 4 Example 2 (#30, p. 62) Solve 1 3 4 2 x2 x3 x x6 Example 3 (#36, p. 62) Solve (x + 1)2 + 2(x – 2) = (x + 1)(x – 2) Text p. 62 #15-37 odd P.5 Part II Solving Equations Objective: Solve quadratic equations. Quadratic Equation – an equation in the form ax2 + bx + c = 0, a ≠ 1 Four methods for solving quadratic equations: Factoring Root Extraction Completing the Square Quadratic Formula Factoring 1. Write equation in standard form with one side of the equation = 0 2. Factor the quadratic expression 3. Let each factor equal 0, and solve each equation. Example 1 Solve 2x2 – 3x – 2 = 7 2x2 – 3x – 9 = 0 (2x + 3)(x – 3) = 0 2x + 3 = 0 x–3=0 x= 3 2 x=3 Set one side equal to zero Factor the quadratic expression Set each factor equal to zero Root Extraction To perform root extraction, b = 0, and the equation must be in standard form ax2 + c = 0 Example 2 (Similar to example 5a, p. 56) 4x2 – 12 = 0 Notice that the “x” term is absent, meaning b = 0 + 12 + 12 Move constant to other side 4x2 = 12 4 4 Divide both sides by 4 to isolate the “x2” x2 = 3 x= 3 Extract the square roots In the case that we must take the square root of a negative number, the solutions to the equation will be imaginary. Completing the square To complete the square, we try to rewrite an equation so that one side of the equation is a perfect square trinomial. While this method can be done when “a” is any value, it is recommended that a = 1 when using this method. Example 3 (#68, p. 63) Solve x2 + 8x + 14 = 0 - 14 -14 Remove constant term from left side. x2 + 8x + ____ = -14 + ____ Take half of b and square the result. 2 2 b 8 16 2 2 Place this in each of the blanks. x2 + 8x + 16 = -14 + 16 (x + 4)(x + 4) = 2 (x + 4)2 = 2 Perform root extraction on both sides x+4= 2 -4 -4 x = -4 2 If it is necessary to take the square root of a negative number, the solutions will be complex Quadratic Formula To solve an equation in the form ax2 + bx + c = 0, use the formula b b2 4ac x 2a This formula appears on p. 57 of the text. It may be used for any quadratic equation. Example 4 Solve 2x2 – 3x – 9 = -1 +1 + 1 2x2 – 3x – 8 = 0 Employ the quadratic formula: x x x x b b2 4ac 2a ( 3) ( 3)2 4(2)( 8) 2(2) 3 9 64 4 3 73 4 x = 2.886 or x = -1.386 Set one side equal to zero Write the quadratic formula Make substitution Four types of solutions natures: If the discriminant (b2 – 4ac) is a positive perfect square, then there are two rational solutions. If the discriminant (b2 – 4ac) is a positive number but not a perfect square, then there are two irrational solutions. If the discriminant (b2 – 4ac) is zero, there is one rational solution, which is a double root. If the discriminant (b2 – 4ac) is negative, there are two complex solutions. Text p. 63 #43-81 odd P.5 Part III Solving Equations Objective: Solve radical and absolute value equations. With these two types of equations, a key component of solving them is to take steps to isolate the special type of expression on one side of the equation. Example 1 (Example 11, p. 59) 2x 7 x 2 Isolate radical expression on one side. +x +x 2x 7 x 2 2 2 Square both sides 2x 7 x 2 2x + 7 = (x+2)(x+2) 2x + 7 = x2 + 4x + 4 0 = x2 + 2x – 3 Solve as a quadratic 0 = (x + 3)(x – 1) Factor to solve x + 3 = 0x – 1 = 0 x = -3 x=1 Go back and check for extraneous solutions. 2(3) 7 (3) 2 2(1) 7 (1) 2 1 3 2 9 1 2 -3 is not a solution 1 is a solution Therefore, { 1 } is the solution set. Example 2 x 1 x 10 1 When there are two radical expressions in the equation, isolate one of them first. x 1 x 10 1 Now square both sides of the equation, remembering that the one side is in binomial form. ( x 1)2 ( x 10 1)2 x 1 x 10 1 x 10 1 x 10 1 Notice the use of the “FOIL” method x 1 x 9 2 x 10 Combine like terms -x +9 -x + 9 Isolate the remaining radical 8 2 x 10 2 2 4 = x 10 (4)2 = ( x 10 )2 Square both sides again 16 = x – 10 26 = x Check solution Substitute (26) 1 26 10 1 25 16 1 Evaluate both sides 5-4=1 Therefore {26} is the solution set. Absolute Value Equations Solve |2x – 7| + 3 = 16 - 3 -3 First, isolate the absolute value expression |2x – 7| = 13 Once the absolute value expression is isolated, split the equation into two parts to remove the absolute value. 2x – 7 = 13 or 2x – 7 = -13 + 7 +7 +7 +7 2x = 20 or 2x = -6 x = 10 or x = -3 The solution set is {-3, 10} If the absolute value expression is equal to zero, there will only be one equation and only one solution. If the absolute value expression is equal to a number less that zero, there will be no solution. Text p. 64, #95-104 (skip 102), 115-118

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# 1-6 Solving Compound and Absolute Value Inequalities

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