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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §8.2 Quadratic Equation

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review § Any QUESTIONS About §8.1 → Complete the Square Any QUESTIONS About HomeWork §8.1 → HW-37 8.1 MTH 55

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 3 Bruce Mayer, PE Chabot College Mathematics The Quadratic Formula The solutions of ax 2 + bx + c = 0 are given by This is one of the MOST FAMOUS Formulas in all of Mathematics

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 4 Bruce Mayer, PE Chabot College Mathematics §8.2 Quadratic Formula The Quadratic Formula Problem Solving with the Quadratic Formula

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 5 Bruce Mayer, PE Chabot College Mathematics Derive Quadratic Formula - 1 Consider the General Quadratic Equation Where a, b, c are CONSTANTS Solve This Eqn for x by Completing the Square First; isolate the Terms involving x Next, Divide by “a” to give the second degree term the coefficient of 1 Now add to both Sides of the eqn a “quadratic supplement” of (b/2a) 2

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 6 Bruce Mayer, PE Chabot College Mathematics Derive Quadratic Formula - 2 Now the Left-Hand-Side (LHS) is a PERFECT Square Take the Square Root of Both Sides Combine Terms inside the Radical over a Common Denom

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 7 Bruce Mayer, PE Chabot College Mathematics Derive Quadratic Formula - 4 Note that Denom is, itself, a PERFECT SQ Next, Isolate x But this the Renowned QUADRATIC FORMULA Note That it was DERIVED by COMPLETING the SQUARE Now Combine over Common Denom

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example a) 2x 2 + 9x − 5 = 0 Solve using the Quadratic Formula: 2x 2 + 9x − 5 = 0 Soln a) Identify a, b, and c and substitute into the quadratic formula: 2x 2 + 9x − 5 = 0 Now Know a, b, and c a b c

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 9 Bruce Mayer, PE Chabot College Mathematics Solution a) 2x 2 + 9x − 5 = 0 Using a = 2, b = 9, c = −5 Be sure to write the fraction bar ALL the way across. Recall the Quadratic Formula → Sub for a, b, and c

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 10 Bruce Mayer, PE Chabot College Mathematics Solution a) 2x 2 + 9x − 5 = 0 From Last Slide: So: The Solns:

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example b) x 2 = −12x + 4 Soln b) write x 2 = −12x + 4 in standard form, identify a, b, & c, and solve using the quadratic formula: 1x 2 + 12x – 4 = 0 a b c

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example c) 5x 2 − x + 3 = 0 Soln c) Recognize a = 5, b = −1, c = 3 → Sub into Quadratic Formula Since the radicand, – 59, is negative, there are NO real-number solutions. The COMPLEX No. Soln

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 13 Bruce Mayer, PE Chabot College Mathematics Quadratic Equation Graph The graph of a quadratic eqn describes a “parabola” which has one of a: Bowl shape Dome shape The graph, depending on the “Vertex” Location, may have different numbers of of x-intercepts: 2 (shown), 1, or NONE x intercepts vertex

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 14 Bruce Mayer, PE Chabot College Mathematics The Discriminant It is sometimes enough to know what type of number (Real or Complex) a solution will be, without actually solving the equation. From the quadratic formula, b 2 – 4ac, is known as the discriminant. The discriminant determines what type of number the solutions of a quadratic equation are. The cases are summarized on the next sld

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 15 Bruce Mayer, PE Chabot College Mathematics Soln Type by Discriminant Discriminant b 2 – 4ac Nature of Solutions x- Intercepts 0 Only one solution; it is a real number Only one Positive Two different real-number solutions Two different Negative Two different NONreal complex-number solutions (complex conjugates) None

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example Discriminant Determine the nature of the solutions of: 5x 2 − 10x + 5 = 0 SOLUTION Recognize a = 5, b = −10, c = 5 Calculate the Discriminant b 2 − 4ac = (−10) 2 − 4(5)(5) = 100 − 100 = 0 There is exactly one, real solution. This indicates that 5x 2 − 10x + 5 = 0 can be solved by factoring 5(x − 1) 2 = 0

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example Discriminant Determine the nature of the solutions of: 5x 2 − 10x + 5 = 0 SOLUTION Examine Graph Notice that the Graph crosses the x-axis (where y = 0) at exactly ONE point as predicted by the discriminant

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example Discriminant Determine the nature of the solutions of: 4x 2 − x + 1 = 0 SOLUTION Recognize a = 4, b = −1, c = 1 Calculate the Discriminant b 2 – 4ac = (−1) 2 − 4(4)(1) =1 − 16 = −15 Since the discriminant is negative, there are two NONreal complex-number solutions

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example Discriminant Determine the nature of the solutions of: 4x 2 − 1x + 1 = 0 SOLUTION Examine Graph Notice that the Graph does NOT cross the x-axis (where y = 0) indicating that there are NO real values for x that satisfy this Quadratic Eqn

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example Discriminant Determine the nature of the solutions of: 2x 2 + 5x = −1 SOLUTION: First write the eqn in Std form of ax 2 + bx + c = 0 → 2x 2 + 5x + 1 = 0 Recognize a = 2, b = 5, c = 1 Calculate the Discriminant b 2 – 4ac = (5) 2 – 4(2)(1) = 25 – 8 = 17 There are two, real solutions

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example Discriminant Determine the nature of the solutions of: 0.3x 2 − 0.4x + 0.8 = 0 SOLUTION Recognize a = 0.3, b = −0.4, c = 0.8 Calculate the Discriminant b 2 − 4ac = (−0.4) 2 − 4(0.3)(0.8) =0.16–0.96 = −0.8 Since the discriminant is negative, there are two NONreal complex-number solutions

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 22 Bruce Mayer, PE Chabot College Mathematics Writing Equations from Solns The principle of zero products informs that this factored equation (x − 1)(x + 4) = 0 has solutions 1 and −4. If we know the solutions of an equation, we can write an equation, using the principle of Zero Products in REVERSE.

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example Write Eqn from solns Find an eqn for which 5 & −4/3 are solns SOLUTION x = 5 or x = –4/3 x – 5 = 0 or x + 4/3 = 0 (x – 5)(x + 4/3) = 0 x 2 – 5x + 4/3x – 20/3 = 0 3x 2 – 11x – 20 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms and clearing fractions

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example Write Eqn from solns Find an eqn for which 3i & −3i are solns SOLUTION x = 3i or x = –3i x – 3i = 0 or x + 3i = 0 (x – 3i)(x + 3i) = 0 x 2 – 3ix + 3ix – 9i 2 = 0 x 2 + 9 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 25 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work Problems From §8.2 Exercise Set 18, 30, 44, 58 Solving Quadratic Equations 1. Check to see if it is in the form ax 2 = p or (x + c) 2 = d. If it is, use the square root property 2. If it is not in the form of (1), write it in standard form: ax 2 + bx + c = 0 with a and b nonzero. 3. Then try factoring. 4. If it is not possible to factor or if factoring seems difficult, use the quadratic formula. The solns of a quadratic eqn cannot always be found by factoring. They can always be found using the quadratic formula.

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 26 Bruce Mayer, PE Chabot College Mathematics All Done for Today The Quadratic Formula

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 27 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 28 Bruce Mayer, PE Chabot College Mathematics Graph y = |x| Make T-table

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 29 Bruce Mayer, PE Chabot College Mathematics

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BMayer@ChabotCollege.edu MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 30 Bruce Mayer, PE Chabot College Mathematics Quadratic Equation Graph The graph of a quadratic eqn describes a “parabola” which has one of a: Bowl shape Dome shape The graph, depending on the “Vertex” Location may have different numbers of x-intercepts: 2 (shown), 1, or NONE

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