Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§8.2 Quadratic§8.2 QuadraticEquationEquation



Review §Review §

Any QUESTIONS About• §8.1 → Complete the Square

Any QUESTIONS About HomeWork• §8.1 → HW-37

8.1 MTH 55



The Quadratic FormulaThe Quadratic Formula

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

a

acbbx

2

42

This is one of theMOST FAMOUSFormulas in allof Mathematics



§8.2 Quadratic Formula§8.2 Quadratic Formula

The Quadratic Formula

Problem Solving with the Quadratic Formula



Derive Quadratic Formula - 1Derive 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

02 cbxax

cbxax 2

a

cx

a

bx 2

a

cababx

a

bx

222

22



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



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



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



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



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:



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



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



Quadratic Equation GraphQuadratic 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” Location,may have different numbers of of x-intercepts: 2 (shown), 1, or NONE

1282 xxy

163 2 xxy

x intercepts

vertex



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



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



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





5x2 − 10x + 5 = 0 SOLUTION

Examine Graph

• Notice that the Graphcrosses the x-axis (where y = 0) atexactly ONE point aspredicted by the discriminant





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





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 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





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



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.



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



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



WhiteBoard WorkWhiteBoard Work

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

Solving Quadratic Equations

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.



All Done for TodayAll Done for Today

TheQuadraticFormula



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



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



-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



Quadratic Equation GraphQuadratic Equation Graph

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

• Dome shape

542 xxy

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

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]