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# Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege - PowerPoint PPT Presentation

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

8.1

Review §

• §8.1 → Complete the Square

• §8.1 → HW-37

• The solutions of ax2 + bx + c = 0 are given by

This is one of theMOST FAMOUSFormulas in allof Mathematics

• Problem Solving with the Quadratic Formula

• 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

• Combine Terms inside the Radical over a Common Denom

• Take the Square Root of Both Sides

• Now Combine over Common Denom

• But this the Renowned QUADRATIC FORMULA

• Note That it was DERIVED by COMPLETING theSQUARE

• Next, Isolate x

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

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.

Solution a) 2x2 + 9x− 5 = 0

• From Last Slide:

• So:

• The Solns:

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

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.

• 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

• 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

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

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

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

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

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

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

• 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.

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

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

• Problems From §8.2 Exercise Set

• 18, 30, 44, 58

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.

Appendix

Bruce Mayer, PE

Graph y = |x|

• Make T-table

• 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