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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Chabot Mathematics. §8.2 Quadratic Equation. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. MTH 55. 8.1. Review §. Any QUESTIONS About §8.1 → Complete the Square Any QUESTIONS About HomeWork §8.1 → HW-37. The Quadratic Formula.

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Chabot Mathematics §8.2 QuadraticEquation Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. MTH 55 8.1 Review § • Any QUESTIONS About • §8.1 → Complete the Square • Any QUESTIONS About HomeWork • §8.1 → HW-37

  3. The Quadratic Formula • The solutions of ax2 + bx + c = 0 are given by This is one of theMOST FAMOUSFormulas in allof Mathematics

  4. §8.2 Quadratic Formula • The Quadratic Formula • Problem Solving with the Quadratic Formula

  5. Consider the General Quadratic Equation Derive Quadratic Formula - 1 • Next, Divide by “a” to give the second degree term the coefficient of 1 • Where a, b, c are CONSTANTS • Solve This Eqn for x by Completing the Square • First; isolate the Terms involving x • Now add to both Sides of the eqn a “quadratic supplement” of (b/2a)2

  6. Now the Left-Hand-Side (LHS) is a PERFECT Square Derive Quadratic Formula - 2 • Combine Terms inside the Radical over a Common Denom • Take the Square Root of Both Sides

  7. Note that Denom is, itself, a PERFECT SQ Derive Quadratic Formula - 4 • Now Combine over Common Denom • But this the Renowned QUADRATIC FORMULA • Note That it was DERIVED by COMPLETING theSQUARE • Next, Isolate x

  8. Example a) 2x2 + 9x− 5 = 0 • Solve using the Quadratic Formula: 2x2 + 9x− 5 = 0 • Soln a) Identify a, b, and c and substitute into the quadratic formula: 2x2 + 9x−5 = 0 a bc • Now Know a, b, and c

  9. Solution a) 2x2 + 9x− 5 = 0 • Using a = 2, b = 9, c = −5 Recall the Quadratic Formula→ Sub for a, b, and c Be sure to write the fraction bar ALL the way across.

  10. Solution a) 2x2 + 9x− 5 = 0 • From Last Slide: • So: • The Solns:

  11. Example b) x2 = −12x + 4 • Soln b) write x2 = −12x + 4 in standard form, identify a, b, & c, and solve using the quadratic formula: 1x2 + 12x–4 = 0 a bc

  12. Example c) 5x2−x + 3 = 0 • Soln c) Recognize a = 5, b = −1, c = 3 → Sub into Quadratic Formula • The COMPLEX No. Soln Since the radicand, –59, is negative, there are NO real-number solutions.

  13. vertex Quadratic Equation Graph • The graph of a quadratic eqn describes a “parabola” which has one of a: • Bowl shape • Dome shape x intercepts • The graph, dependingon the “Vertex” Location,may have different numbers of of x-intercepts: 2 (shown), 1, or NONE

  14. 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, b2 – 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

  15. Soln Type by Discriminant

  16. Example  Discriminant • Determine the nature of the solutions of: 5x2− 10x + 5 = 0 • SOLUTION • Recognize a = 5, b = −10, c = 5 • Calculate the Discriminant b2− 4ac = (−10)2− 4(5)(5) = 100 − 100 = 0 • There is exactly one, real solution. • This indicates that 5x2− 10x + 5 = 0 can be solved by factoring  5(x− 1)2 = 0

  17. Example  Discriminant • Determine the nature of the solutions of: 5x2− 10x + 5 = 0 • SOLUTION Examine Graph • Notice that the Graphcrosses the x-axis (where y = 0) atexactly ONE point aspredicted by the discriminant

  18. Example  Discriminant • Determine the nature of the solutions of: 4x2−x + 1 = 0 • SOLUTION • Recognize a = 4, b = −1, c = 1 • Calculate the Discriminant b2 – 4ac = (−1)2− 4(4)(1) =1 − 16 = −15 • Since the discriminant is negative, there are two NONreal complex-number solutions

  19. Example  Discriminant • Determine the nature of the solutions of: 4x2− 1x + 1 = 0 • SOLUTION Examine Graph • Notice that the Graphdoes NOT cross the x-axis (where y = 0) indicating that there are NO real values for x that satisfy this Quadratic Eqn

  20. Example  Discriminant • Determine the nature of the solutions of: 2x2 + 5x = −1 • SOLUTION: First write the eqn in Std form of ax2 + bx + c = 0 → 2x2 + 5x + 1 = 0 • Recognize a = 2, b = 5, c = 1 • Calculate the Discriminant b2 – 4ac = (5)2 – 4(2)(1) = 25 – 8 = 17 • There are two, real solutions

  21. Example  Discriminant • Determine the nature of the solutions of: 0.3x2− 0.4x + 0.8 = 0 • SOLUTION • Recognize a = 0.3, b = −0.4, c = 0.8 • Calculate the Discriminant b2− 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

  22. Writing Equations from Solns • The principle of zero products informs that this factored equation (x − 1)(x + 4) = 0 has solutions1 and −4. • If we know the solutions of an equation, we can write an equation, using the principle of Zero Products in REVERSE.

  23. Example  Write Eqn from solns • Find an eqn for which 5 & −4/3 are solns • SOLUTION x = 5 orx = –4/3 x – 5 = 0 orx + 4/3 = 0 Get 0’s on one side Using the principle of zero products (x – 5)(x + 4/3) = 0 x2 – 5x + 4/3x – 20/3 = 0 Multiplying 3x2 – 11x – 20 = 0 Combining like terms and clearing fractions

  24. Example  Write Eqn from solns • Find an eqn for which 3i & −3i are solns • SOLUTION x = 3iorx = –3i x – 3i = 0 orx + 3i = 0 Get 0’s on one side Using the principle of zero products (x – 3i)(x + 3i) = 0 x2 – 3ix + 3ix – 9i2= 0 Multiplying x2+ 9 = 0 Combining like terms

  25. WhiteBoard Work • Problems From §8.2 Exercise Set • 18, 30, 44, 58 Solving Quadratic Equations 1. Check to see if it is in the formax2 = 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: • ax2 + 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.

  26. All Done for Today TheQuadraticFormula

  27. Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –

  28. Graph y = |x| • Make T-table

  29. Quadratic Equation Graph • The graph of a quadratic eqn describes a “parabola” which has one of a: • Bowl shape • Dome shape • The graph, dependingon the “Vertex” Locationmay have different numbers of x-intercepts: 2 (shown), 1, or NONE

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