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[email protected] • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §8.2 Quadratic §8.2 Quadratic Equation Equation
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Chabot Mathematics. §8.2 Quadratic Equation. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. MTH 55. 8.1. Review §. Any QUESTIONS About §8.1 → Complete the Square Any QUESTIONS About HomeWork §8.1 → HW-37. The Quadratic Formula. - PowerPoint PPT Presentation

### Transcript of Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics Bruce Mayer, PE Chabot College Mathematics

Review §Review §

Any QUESTIONS About• §8.1 → Complete the Square

Any QUESTIONS About HomeWork• §8.1 → HW-37

8.1 MTH 55 Bruce Mayer, PE Chabot College Mathematics

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

a

acbbx

2

42

This is one of theMOST FAMOUSFormulas in allof Mathematics Bruce Mayer, PE Chabot College Mathematics

Problem Solving with the Quadratic Formula Bruce Mayer, PE Chabot College Mathematics

• 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

02 cbxax

cbxax 2

a

cx

a

bx 2

a

cababx

a

bx

222

22 Bruce Mayer, PE Chabot College Mathematics

Derive Quadratic Formula - 2Derive Quadratic Formula - 2 Now the Left-Hand-Side

(LHS) is a PERFECT Square

a

c

a

b

a

bx

a

cababx

a

bx

22

222

22

22

Take the Square Root of Both Sides

a

c

a

b

a

bx

2

22

Combine Terms inside the Radical over a Common Denom

2

2

2

2

2

2

4

4

2

4

4

42

42

a

acb

a

bx

aa

ac

a

b

a

bx

a

c

a

b

a

bx Bruce Mayer, PE Chabot College Mathematics

Derive Quadratic Formula - 4Derive 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 theSQUARE

a

acb

a

bx

a

acb

a

bx

2

4

2

4

4

22

2

2

a

acb

a

bx

2

4

2

2

a

acbbx

2

42

Now Combine over Common Denom Bruce Mayer, PE Chabot College Mathematics

Example a) 2Example a) 2xx22 + 9 + 9xx −− 5 = 0 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

Now Know a, b, and c

a b c Bruce Mayer, PE Chabot College Mathematics

Solution a) 2Solution a) 2xx22 + 9 + 9xx −− 5 = 0 5 = 0

Using a = 2, b = 9, c = −52 4

2

b b acx

a

2( ) 49 9 2( )( )

2 )2(

5x

9 81 ( 40)

4x

Be sure to write the fraction bar ALL the way across.

9 121

4x

Recall the Quadratic Formula→ Sub for a, b, and c Bruce Mayer, PE Chabot College Mathematics

Solution a) 2Solution a) 2xx22 + 9 + 9xx −− 5 = 0 5 = 0

From Last Slide: 9 121

4x

9 11

4x

9 11 9 11 or

4 4x x

2 20 or

4 4x x

1 or 5

2x x

So:

The Solns: Bruce Mayer, PE Chabot College Mathematics

Example b) Example b) xx22 = = −−1212xx + 4 + 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

acbbx

2

42

a b c

2( ) 412 12 1( )( 4)

)12(x

12 144 16 12 160

2 2x

12 16 10

2 2x

4 106

2x 6 2 10 Bruce Mayer, PE Chabot College Mathematics

Example c) 5Example c) 5xx22 −− xx + 3 = 0 + 3 = 0

Soln c) Recognize a = 5, b = −1, c = 3 → Sub into Quadratic Formula

a

acbbx

2

42

2( ) ( ) 4( )( )

2(

1 1 5

5)

3x

1 1 60

10x

1 59

10x

Since the radicand, –59, is negative, there are NO real-number solutions.

The COMPLEX No. Soln

10

59

10

1ix Bruce Mayer, PE Chabot College Mathematics

The graph of a quadratic eqn describes a “parabola” which has one of a:• Bowl shape

• Dome shape

The graph, dependingon the “Vertex” Location,may have different numbers of of x-intercepts: 2 (shown), 1, or NONE

1282 xxy

163 2 xxy

x intercepts

vertex Bruce Mayer, PE Chabot College Mathematics

The DiscriminantThe 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 Bruce Mayer, PE Chabot College Mathematics

Soln Type by DiscriminantSoln Type by Discriminant

Discriminant b2 – 4ac

Nature of Solutionsx-

Intercepts

0Only one solution; it is a real number

Only one

Positive

Two different real-number solutions

Two different

NegativeTwo different NONreal complex-number solutions (complex conjugates)

None Bruce Mayer, PE Chabot College Mathematics

Example Example Discriminant 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 Bruce Mayer, PE Chabot College Mathematics

Example Example Discriminant 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 Bruce Mayer, PE Chabot College Mathematics

Example Example Discriminant 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 Bruce Mayer, PE Chabot College Mathematics

Example Example Discriminant 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 Bruce Mayer, PE Chabot College Mathematics

Example Example Discriminant 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 Bruce Mayer, PE Chabot College Mathematics

Example Example Discriminant 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 Bruce Mayer, PE Chabot College Mathematics

Writing Equations from Solns 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. Bruce Mayer, PE Chabot College Mathematics

Example Example Write Eqn from solns 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

x2 – 5x + 4/3x – 20/3 = 0

3x2 – 11x – 20 = 0

Get 0’s on one side

Using the principle of zero products

Multiplying

Combining like terms and clearing fractions Bruce Mayer, PE Chabot College Mathematics

Example Example Write Eqn from solns 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

x2 – 3ix + 3ix – 9i2 = 0

x2 + 9 = 0

Get 0’s on one side

Using the principle of zero products

Multiplying

Combining like terms Bruce Mayer, PE Chabot College Mathematics

WhiteBoard WorkWhiteBoard Work

Problems From §8.2 Exercise Set• 18, 30, 44, 58

1. Check to see if it is in the form ax2 = 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. Bruce Mayer, PE Chabot College Mathematics

All Done for TodayAll Done for Today Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

srsrsr 22 Bruce Mayer, PE Chabot College Mathematics

Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

x

y

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls Bruce Mayer, PE Chabot College Mathematics

-3

-2

-1

0

1

2

3

4

5

-3 -2 -1 0 1 2 3 4 5

M55_§JBerland_Graphs_0806.xls -5

-4

-3

-2

-1

0

1

2

3

4

5

-10 -8 -6 -4 -2 0 2 4 6 8 10

M55_§JBerland_Graphs_0806.xls

x

y 