 MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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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

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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.

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

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 –

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

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

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