Section1.4 QUADRATIC EQUATIONS

Section1.4

QUADRATIC EQUATIONSThis presentation is base on Power Point slides found at http://cwx.prenhall.com/bookbind/pubbooks/sullivan13/

with modifications by Jeffrey Linek Ed. D.

http://cwx.prenhall.com/bookbind/pubbooks/sullivan13/

A quadratic equation is anequation equivalent to one ofthe form

ax bx c2 0

where a, b, and c are realnumbers and a 0.

Quadratic Equations

• We can solve quadratic equations– Graphically– Algebraically by factoring– Using the Completing the Square Method– Extracting a Zero or Root– Using the Quadratic Formula

2 5 3 02x x

Solve: 2 3 52x x

2 1 3 0x x

2 1 0x or x 3 0

x 12

or x3

Solution Set:

12

3,

Solve: x 2 252

x 2 25

x 2 5 or x 2 5

x 3 o r x 7

Solution set: {-7, 3}

Solve by completing the square:

3 2 1 02x x

3 2 12x x

x x2 23

13

1

2

2

3

1

9

2

x x2 23

19

13

19

x x2 23

19

49

x

1

349

2

x 13

23

x 13

23

13

x 13

23

1

Solution set:

13

1,

C o n s i d e r t h e q u a d r a t i c e q u a t i o n

a x b x c a2 0 0

I f b a c2 4 0 , t h i s e q u a t i o n h a s n o r e a ls o l u t i o n .

I f b a c2 4 0 , t h e r e a l s o l u t i o n ( s ) o f t h i se q u a t i o n i s ( a r e ) g i v e n b y t h e q u a d r a t i cf o r m u l a :

Theorem Quadratic Formula

xb b ac

a 2 4

2

F o r a q u a d r a t i c e q u a t i o n a x b x c2 0 :

1 . I f b a c2 4 0 , t h e r e a r e t w o u n e q u a l r e a ls o l u t i o n s .

2 . I f b a c2 4 0 , t h e r e i s a r e p e a t e d r e a ls o l u t i o n , a r o o t o f m u l t i p l i c i t y 2 .

3 . I f b a c2 4 0 , t h e r e i s n o r e a l s o l u t i o n .

Discriminant of a Quadratic Equation

4 2 02x x

b ac2 4 ( ) ( )( ) 1 4 4 221 32 33

xb b ac

a 2 4

2 ( )

( )1 332 4

1 338

Find the real solutions, if any, of theequation 4 2 02x x .

Quadratic Equations

Solve the following equation graphically.

x2 - 4 = x + 6

Solve the following equation graphically.x2 - 4 = x + 6

On the TI-83 press the Y= key and enter the equations

Y1 = x2 - 4 and

Y2 = x + 6


Press the GRAPH key.

The points of intersection are the solutions to the problem.

We will use the intersect feature to find their values.


TI-83: The Intersect Feature

Press the 2nd key, then the TRACE key

Select 5: intersect

Then, press the ENTER key


TI-83: The Intersect Feature• The graph of the two

equations will be displayed. The TI-83 will ask if the cursor is on the first graph. Press the ENTER key.

• Next, the TI-83 will ask if the cursor is on the second graph. Press the ENTER key.



Notice that the calculator asks you to guess at the answer, and that the cursor is midway between the two points of intersection.

We will just press the ENTER key to see what happens.



• Notice that the TI-83 solved for one of the points of intersection, namely,

(-2.702, 3.298).

We need to find the other point as well.


TI-83: The Intersect Featureo As we did earlier press the

2nd Key, the Trace Key, then the number 5 Key for 5: intersect.

o Once again the calculator will ask if the cursor is on the first graph. This time, use the key to move the cursor closed to the point of intersection. Then, press the ENTER key

>



The TI-83 will ask if the cursor is on the second graph. Press the ENTER key.



The calculator asks you to guess at the answer.

We will just press the ENTER key to get the answer.



The answer (3.702, 9.702)

appears.

However, we need to interpret the answer because our original equation did not contain y.

Solve the following equation graphically.x2 - 4 = x + 6The answers.

Remember that original we set

Y1 = x2 - 4 and Y2 = x + 6.

Therefore, Y1 = Y2 since,

x2 - 4 = x + 6

Y1 and Y2 are result of placing a value for x into the equation x2 - 4 = x + 6

So, our answers are x = -2.702 or x = 3.702

Now, solve the equation algebraically

x2 - 4 = x + 6

Section1.4 QUADRATIC EQUATIONS

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