On Page 234, complete the Prerequisite skills #1-14

ON PAGE 234, COMPLETE THE PREREQUISITE SKILLS #1-14.

CHAPTER 4:QUADRATIC FUNCTION AND

FACTORING

BIG IDEAS:1. Graphing and writing quadratic

functions in several forms2. Solving quadratic equations using a

variety of methods3. Performing operations with square

roots and complex numbers

LESSON 1: GRAPH QUADRATIC FUNCTIONS IN

STANDARD FORM

How are the values of a, b, and c in the equation y =

ax2 + bx + c related to the graph of a quadratic

function?

Essential question

VOCABULARY Quadratic Function – y = ax2 + bx + c

Parabola – The set of all points equidistant from a point called the focus and a line called the directrix.

Vertex – The point on a parabola that lies on the axis of symmetry

Axis of Symmetry – the line perpendicular to the parabola’s directrix and passing through its focus and vertex

EXAMPLE 1 Graph a function of the form y = ax2

STEP 4 Compare the graphs of y = 2x2 and y = x2.Both open up and have the same vertex andaxis of symmetry. The graph of y = 2x2 isnarrower than the graph of y = x2.

EXAMPLE 2 Graph a function of the form y = ax2 + c

Graph y = – Compare the graph with thex2 + 3 12

graph of y = x2

SOLUTION

STEP 1 Make a table of values for y = – x2 + 3 12

STEP 2 Plot the points from the table.

STEP 3 Draw a smooth curve through the points.

EXAMPLE 2 Graph a function of the form y = ax2

STEP 4 Compare the graphs of y = – and y = x2. Both graphs have the same axis of symmetry. However, the graph of y = – opens down and is wider than the graph of y = x2. Also, its vertex is 3 units higher.

x2 + 3 12

x2 + 3 12

GUIDED PRACTICE for Examples 1 and 2

Graph the function. Compare the graph with the graph of y = x2.

1. y = – 4x2

Same axis of symmetry and vertex, opens down, and is narrower

ANSWER


2. y = – x2 – 5

ANSWER

Same axis of symmetry, vertex is shifted down 5 units, and opens down


3. f(x) = x2 + 2 14

ANSWER

Same axis of symmetry, vertex is shifted up 2 units, opens up, and is wider

EXAMPLE 3 Graph a function of the form y = ax2 + bx + c

Graph y = 2x2 – 8x + 6.

SOLUTION

Identify the coefficients of the function. The coefficients are a = 2, b = –8, and c = 6. Because a > 0, the parabola opens up.

STEP 1

STEP 2 Find the vertex. Calculate the x-coordinate.

x = b 2a

=(–8) 2(2)

– – = 2

Then find the y-coordinate of the vertex.

y = 2(2)2 – 8(2) + 6 = –2

So, the vertex is (2, –2). Plot this point.


STEP 3 Draw the axis of symmetry x = 2.

STEP 4 Identify the y-intercept c, which is 6. Plot the point (0, 6). Then reflect this point in the axis of symmetry to plot another point, (4, 6).

STEP 5 Evaluate the function for another value of x, such as x = 1.

y = 2(1)2 – 8(1) + 6 = 0

Plot the point (1,0) and its reflection (3,0) in the axis of symmetry.


STEP 6 Draw a parabola through the plotted points.

GUIDED PRACTICE for Example 3

Graph the function. Label the vertex and axis of symmetry.

4. y = x2 – 2x – 1 5. y = 2x2 + 6x + 3


6. f (x) = x2 – 5x + 2 1 3

–

How are the values of a, b, and c in the equation y = ax2 + bx + c

related to the graph of a quadratic function?

A – tell whether the parabola opens up or down

A/b – used to find the equation of the axis of symmetry and x-

coordinate of the vertexC – y-intercept

Essential question

FIND THE PRODUCTA. (X + 6) (X + 3)B. (X-5)2

C. 4(X+5)(X-5)

LESSON 2: GRAPH QUADRATIC FUNCTIONS IN VERTEX

OR INTERCEPT FORM

Why do we write the graph of quadratic functions in vertex form or intercept

form?

Essential question

VOCABULARY Vertex form – y = a(x – h)2 + k

Intercept form – y = a(x-p)(x-q)

EXAMPLE 1 Graph a quadratic function in vertex form

Graph y = – (x + 2)2 + 5.14

SOLUTION

STEP 1 Identify the constants a = – , h = – 2, and k = 5.

Because a < 0, the parabola opens down.

14

STEP 2 Plot the vertex (h, k) = (– 2, 5) and draw the axis of symmetry x = – 2.

EXAMPLE 1 Graph a quadratic function in vertex form

STEP 3 Evaluate the function for two values of x.

x = 0: y = (0 + 2)2 + 5 = 414

–

x = 2: y = (2 + 2)2 + 5 = 114

–

Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry.

STEP 4 Draw a parabola through the plotted points.

EXAMPLE 2 Use a quadratic model in vertex form

Civil Engineering

The Tacoma Narrows Bridge in Washington has two towers that each rise 307 feet above the roadway and are connected by suspension cables as shown. Each cable can be modeled by the function.

y = (x – 1400)2 + 27 1 7000

where x and y are measured in feet. What is the distance d between the two towers ?

EXAMPLE 2 Use a quadratic model in vertex form

SOLUTION

The vertex of the parabola is (1400, 27). So, a cable’s lowest point is 1400 feet from the left tower shown above. Because the heights of the two towers are the same, the symmetry of the parabola implies that the vertex is also 1400 feet from the right tower. So, the distance between the two towers is d = 2 (1400) = 2800 feet.


Graph the function. Label the vertex and axis of symmetry.

1. y = (x + 2)2 – 3 2. y = –(x + 1)2 + 5


3. f(x) = (x – 3)2 – 412


4. WHAT IF? Suppose an architect designs a bridge

y =1

6500(x – 1400)2 + 27

where x and y are measured in feet. Compare this function’s graph to the graph of the function in Example 2.

with cables that can be modeled by

ANSWER This graph is slightly steeper than the graph in Example 2. They both have the same vertex and axis of symmetry, and both open up.

EXAMPLE 3 Graph a quadratic function in intercept form

Graph y = 2(x + 3)(x – 1).

SOLUTION

STEP 1 Identify the x-intercepts. Because p = –3 and q = 1, the x-intercepts occur at the points (–3, 0) and (1, 0).

STEP 2 Find the coordinates of the vertex.

x = p + q

2

–3 + 1

2= –1=

y = 2(–1 + 3)(–1 – 1) = –8

So, the vertex is (–1, –8)

EXAMPLE 3 Graph a quadratic function in intercept form

STEP 3 Draw a parabola through the vertex and the points where the x-intercepts occur.

EXAMPLE 4 Use a quadratic function in intercept form

Football

The path of a placekicked football can be modeled by the function y = –0.026x(x – 46) where x is the horizontal distance (in yards) and y is the corresponding height (in yards).

a. How far is the football kicked ?

b. What is the football’s maximum height ?

EXAMPLE 4 Use a quadratic function in intercept form

SOLUTION

a. Rewrite the function as y = –0.026(x – 0)(x – 46). Because p = 0 and q = 46, you know the x-intercepts are 0 and 46. So, you can conclude that the football is kicked a distance of 46 yards.b. To find the football’s maximum height, calculate the coordinates of the vertex.

x = p + q

2

0 + 46

2= 23=

y = –0.026(23)(23 – 46) 13.8

The maximum height is the y-coordinate of the vertex, or about 13.8 yards.


Graph the function. Label the vertex, axis of symmetry, and x-intercepts.

5. y = (x – 3)(x – 7) 6. f (x) = 2(x – 4)(x + 1)


7. y = –(x + 1)(x – 5)

8. WHAT IF? In Example 4, what is the maximum height of the football if the football’s path can be modeled by the function y = –0.025x(x – 50)?

15.625 yardsANSWER

EXAMPLE 5 Change from intercept form to standard form

Write y = –2(x + 5)(x – 8) in standard form.

y = –2(x + 5)(x – 8) Write original function.= –2(x2 – 8x + 5x – 40) Multiply using FOIL.

= –2(x2 – 3x – 40) Combine like terms.

= –2x2 + 6x + 80 Distributive property

EXAMPLE 6 Change from vertex form to standard form

Write f (x) = 4(x – 1)2 + 9 in standard form.

f (x) = 4(x – 1)2 + 9 Write original function.= 4(x – 1) (x – 1) + 9

= 4(x2 – x – x + 1) + 9 Multiply using FOIL.

Rewrite (x – 1)2.

= 4(x2 – 2x + 1) + 9 Combine like terms.

= 4x2 – 8x + 4 + 9 Distributive property

= 4x2 – 8x + 13 Combine like terms.


Write the quadratic function in standard form.

9. y = –(x – 2)(x – 7)

–x2 + 9x – 14

10. y = – 4(x – 1)(x + 3)

–4x2 – 8x + 12

11. f(x) = 2(x + 5)(x + 4)

2x2 + 18x + 40

ANSWER

ANSWER ANSWER

ANSWER

12. y = –7(x – 6)(x + 1)

–7x2 + 35x + 42


13. y = –3(x + 5)2 – 1

–3x2 – 30x – 76

14. g(x) = 6(x – 4)2 – 10

6x2 – 48x + 86

15. f(x) = –(x + 2)2 + 4

–x2 – 4x

16. y = 2(x – 3)2 + 9

2x2 – 12x + 27

ANSWER

ANSWER ANSWER

ANSWER

Essential question

Why do we write the graph of quadratic functions in vertex form or

intercept form?

When a quadratic function is written in vertex form, you can read the

coordinates of the vertex directly from the equation. When a quadratic

function is written in intercept form, you can read the x-intercepts directly

from the equation.

FIND THE PRODUCT:

A. (M-8) (M-9)B. (D+9)2

C. (Y+20) (Y-20)

LESSON 3:

SOLVE X2 + BX + C = 0 BY FACTORING

How can factoring be used to solve quadratic

equations when A = 1?

Essential question

VOCABULARY Monomial – An expression that is either a

number, variable or the product of a number and one or more variables with whole number exponents

Binomial – The sum of two monomials

Trinomial – The sum of three monomials

Quadratic Equation – ax2 + bx + c = 0, a≠ 0

Root of an equation – The solution of a quadratic equation

EXAMPLE 1 Factor trinomials of the form x2 + bx + c

Factor the expression.

a. x2 – 9x + 20

b. x2 + 3x – 12

SOLUTION

a. You want x2 – 9x + 20 = (x + m)(x + n) where mn = 20 and m + n = –9.

ANSWER

Notice that m = –4 and n = –5. So, x2 – 9x + 20 = (x – 4)(x – 5).

EXAMPLE 1 Factor trinomials of the form x2 + bx + c

b. You want x2 + 3x – 12 = (x + m)(x + n) where mn = – 12 and m + n = 3.

ANSWER

Notice that there are no factors m and n such that m + n = 3. So, x2 + 3x – 12 cannot be factored.


Factor the expression. If the expression cannot be factored, say so.

1. x2 – 3x – 18

ANSWER

(x – 6)(x + 3)

2. n2 – 3n + 9

cannot be factored

ANSWER

3. r2 + 2r – 63

(r + 9)(r –7)

ANSWER

EXAMPLE 2 Factor with special patterns


a. x2 – 49

= (x + 7)(x – 7)

Difference of two squares

b. d 2 + 12d + 36

= (d + 6)2

Perfect square trinomial

c. z2 – 26z + 169

= (z – 13)2


= x2 – 72

= d 2 + 2(d)(6) + 62

= z2 – 2(z) (13) + 132


4. x2 – 9

(x – 3)(x + 3)

5. q2 – 100

(q – 10)(q + 10)

6. y2 + 16y + 64

(y + 8)2


ANSWER

ANSWER

ANSWER


7. w2 – 18w + 81

(w – 9)2

Essential question

How can factoring be used to solve quadratic equations when a = 1?

The left side of ax2 + bx + c = 0 can be factored, factor the trinomial, use the zero product property to set each factor equal to 0 and solve resulting

linear equations.

FIND THE PRODUCT:

A. (4Y-3)(3Y+8)

B. (5M+6)(5M-6)

LESSON 4:

SOLVE AX2 + BX + C = 0 BY FACTORING

How can factoring be used to solve quadratic equations when

a ≠ 1?

Essential question

VOCABULARY There is no new vocabulary for this

lesson.

EXAMPLE 3 Factor with special patterns


a. 9x2 – 64

= (3x + 8)(3x – 8)Difference of two squares

b. 4y2 + 20y + 25

= (2y + 5)2


c. 36w2 – 12w + 1

= (6w – 1)2


= (3x)2 – 82

= (2y)2 + 2(2y)(5) + 52

= (6w)2 – 2(6w)(1) + (1)2

GUIDED PRACTICEGUIDED PRACTICE for Example 3


7. 16x2 – 1

(4x + 1)(4x – 1)

8. 9y2 + 12y + 4

(3y + 2)2

9. 4r2 – 28r + 49

(2r – 7)2

10. 25s2 – 80s + 64

(5s – 8)2ANSWER

ANSWER

ANSWER

ANSWER


11. 49z2 + 4z + 9

(7z + 3)2

12. 36n2 – 9 = (3y)2

(6n – 3)(6n +3)

ANSWER

ANSWER

EXAMPLE 4 Factor out monomials first


a. 5x2 – 45

= 5(x + 3)(x – 3)

b. 6q2 – 14q + 8

= 2(3q – 4)(q – 1)

c. –5z2 + 20z

d. 12p2 – 21p + 3

= 5(x2 – 9)

= 2(3q2 – 7q + 4)

= –5z(z – 4)

= 3(4p2 – 7p + 1)



13. 3s2 – 24

14. 8t2 + 38t – 10

2(4t – 1) (t + 5)

3(s2 – 8)

15. 6x2 + 24x + 15

3(2x2 + 8x + 5)

16. 12x2 – 28x – 24

4(3x + 2)(x – 3)

17. –16n2 + 12n

–4n(4n – 3)ANSWER

ANSWER

ANSWER

ANSWER

ANSWER


18. 6z2 + 33z + 36

3(2z + 3)(z + 4)

ANSWER

EXAMPLE 5 Solve quadratic equations

Solve (a) 3x2 + 10x – 8 = 0 and (b) 5p2 – 16p + 15 = 4p – 5.

a. 3x2 + 10x – 8 = 0

(3x – 2)(x + 4) = 0

3x – 2 = 0 or x + 4 = 0

Write original equation.

Factor.

Zero product property

Solve for x.or x = –4x = 23

EXAMPLE 5 Solve quadratic equations

b. 5p2 – 16p + 15 = 4p – 5. Write original equation.5p2 – 20p + 20 = 0

p2 – 4p + 4 = 0

(p – 2)2 = 0

p – 2 = 0

p = 2

Write in standard form.

Divide each side by 5.

Factor.

Zero product property

Solve for p.

Essential question

How can factoring be used to solve quadratic equations when a ≠ 1?

First, write the equation in standard form.

Then factor any common monomial.Next, factor the expression.

Use the zero product property to solve the equation.

FIND THE EXACT VALUE:

A.

B. -

LESSON 5:

SOLVE QUADRATIC EQUATIONS BY FINDING SQUARE ROOTS

How can you use square roots to solve a quadratic

equation?

Essential question

VOCABULARY Square root – A number that is

multiplied by itself to solve a square

Radical – An expression of the form or where x is a number or an expression

Radicand – The number or expression beneath a radical sign

Conjugate – The expression a + and a - where and a b are rational numbers.

EXAMPLE 1 Use properties of square roots

Simplify the expression.

5= 4a.

80 516=

= 3 14b.

6 21 126= 9 14=

c.

4

81=

4

81=

2 9

d.

7

16=

7

16=

47


271.

3 3

982.

2 7ANSWER

ANSWER


3.

6 5

10 15

4. 8 28

14 4ANSWER

ANSWER


5.

3 8

9

64

6. 15

4

215ANSWER

ANSWER


7. 11

25

8. 36

49

511

7

6 ANSWER

ANSWER

EXAMPLE 2 Rationalize denominators of fractions.

Simplify (a) 5

2and 3

7 + 2

SOLUTION

(a) 5

2

210

=

=5

2

=5

2

2

2

(b)

EXAMPLE 2 Rationalize denominators of fractions.

SOLUTION

(b)3

7 + 2=

3

7 + 2 7 – 2

7 – 2

=21 – 3 2

49 – 7 + 7 – 2 2 2

=21 – 3 2

47

EXAMPLE 3 Solve a quadratic equation

Solve 3x2 + 5 = 41.

3x2 + 5 = 41 Write original equation.

3x2 = 36 Subtract 5 from each side.

x2 = 12 Divide each side by 3.

x = + 12 Take square roots of each side.

x = + 4 3

x = + 2 3

Product property

Simplify.


ANSWER

The solutions are and 2 3 2 3–

Check the solutions by substituting them into the original equation.

3x2 + 5 = 41

3( )2 + 5 = 412 3 ?

41 = 41

3(12) + 5 = 41?

3x2 + 5 = 41

3( )2 + 5 = 41 – 2 3 ?

41 = 41

3(12) + 5 = 41?

EXAMPLE 4 Standardized Test Practice

SOLUTION

15 (z + 3)2 = 7 Write original equation.

(z + 3)2 = 35 Multiply each side by 5.

z + 3 = + 35 Take square roots of each side.

z = –3 + 35 Subtract 3 from each side.

The solutions are –3 + and –3 – 35 35


ANSWER

The correct answer is C.

GUIDED PRACTICE for Examples 2, 3, and 4GUIDED PRACTICE

Simplify the expression.

9.

530

6

5

10. 9

8

2

4

3

11. 17

12

51

6ANSWER

ANSWER

ANSWER

12.

399

21

19

21

– 21 – 3 5

22

13. – 6

7 – 5

8 – 2 11

5

14. 2

4 + 11

ANSWER

ANSWER

ANSWER

– 9 + 7

74

for Examples 2, 3, and 4GUIDED PRACTICE

15. – 1

9 + 7

32 + 4 3

61

16. 4

8 – 3

ANSWER

ANSWER

for Examples 2, 3, and 4GUIDED PRACTICE

17.

Solve the equation.

5x2 = 80

+ 4

18. z2 – 7 = 29

+ 6

19. 3(x – 2)2 = 40

120

32 +

ANSWER

ANSWER

ANSWER

Essential question

How can you use square roots to solve a quadratic equation?

If a quadratic equation is written in the form x2 = s where s > 0, then you can solve it by taking the square roots

of both sides and simplifying the results.

SOLVE THE EQUATION:

A. 3X2 + 8 = 23

B. 2(X+7)2 = 16

LESSON 6:

PERFORM OPERATIONS WITH COMPLEX NUMBERS

How do you perform opations on complex

numbers?

Essential question

VOCABULARY Imaginary unit (i) – i = so i2

Complex number – A number a + bi where a and b are real numbers and i is the imaginary unit


Solve 2x2 + 11 = –37.

2x2 + 11 = –37 Write original equation.

2x2 = –48 Subtract 11 from each side.

x2 = –24 Divide each side by 2.

Take square roots of each side.x = + –24

Write in terms of i.x = + i 24

x = + 2i 6 Simplify radical.

ANSWER

The solutions are 2i 6 and –2i 6 .


Solve the equation.

x2 = –13.1.

+ i 13ANSWER

x2 = –38.2.

+ i 38ANSWER

x2 + 11= 3.3.

+ 2i 2ANSWER

x2 – 8 = –36 .4.

+ 2i 7ANSWER

3x2 – 7 = –31 .5.

+ 2i 2

5x2 + 33 = 3 .6.

+ i 6

ANSWER

ANSWER

EXAMPLE 2 Add and subtract complex numbers

Write the expression as a complex number in standard form.

a. (8 – i) + (5 + 4i) b. (7 – 6i) – (3 – 6i) c. 10 – (6 + 7i) + 4i

SOLUTION

a. (8 – i) + (5 + 4i) =

(8 + 5) + (–1 + 4)i

Definition of complex addition

= 13 + 3i Write in standard form.

b. (7 – 6i) – (3 – 6i) =

(7 – 3) + (–6 + 6)i

Definition of complex subtraction

= 4 + 0i Simplify.

= 4 Write in standard form.

EXAMPLE 2 Add and subtract complex numbers

c. 10 – (6 + 7i) + 4i =

[(10 – 6) – 7i] + 4i

Definition of complex subtraction

= (4 – 7i) + 4i Simplify.

= 4 + (–7 + 4)i Definition of complex addition

= 4 – 3i Write in standard form.



3 + 6i

7. (9 – i) + (–6 + 7i)

8. (3 + 7i) – (8 – 2i)

–5 + 9i

9. –4 – (1 + i) – (5 + 9i)

–10 – 10iANSWER

ANSWER

ANSWER

EXAMPLE 4 Multiply complex numbers

Write the expression as a complex number in standardform.

a. 4i(–6 + i) b. (9 – 2i)(–4 + 7i)

SOLUTION

a. 4i(–6 + i) = –24i + 4i2 Distributive property

= –24i + 4(–1) Use i2 = –1.

= –24i – 4 Simplify.= –4 – 24i Write in standard form.

EXAMPLE 4 Multiply complex numbers

b. (9 – 2i)(–4 + 7i)

Multiply using FOIL.= –36 + 63i + 8i – 14i2

= –36 + 71i – 14(–1) Simplify and use i2 = – 1 .

= –36 + 71i + 14 Simplify.

= –22 + 71i Write in standard form.

EXAMPLE 5 Divide complex numbers

Write the quotient in standard form.7 + 5i 1 4i

7 + 5i 1 – 4i

7 + 5i 1 – 4i

= 1 + 4i 1 + 4i

Multiply numerator and denominator by 1 + 4i, the complex conjugate of 1 – 4i.

7 + 28i + 5i + 20i2

1 + 4i – 4i – 16i2= Multiply using FOIL.

7 + 33i + 20(–1)1 – 16(–1)

= Simplify and use i2 = 1.

–13 + 33i 17

= Simplify.

EXAMPLE 5 Divide complex numbers

1317

–= +3317

i Write in standard form.

WHAT IF? In Example 3, what is the impedance of the circuit if the given capacitor is replaced with one having a reactance of 7 ohms?

GUIDED PRACTICE for Examples 3, 4 and 5

10.

5 – 4i ohms.

ANSWER


11.

1 + 9i

i(9 – i)

12. (3 + i)(5 – i)

16 + 2i

13. 5 1 + i

52

–52

i

1113

+1613

i

14. 5 + 2i 3 – 2i


ANSWER

ANSWER ANSWER

ANSWER

Essential question

How do you perform operations on complex numbers?

Add/subtract – add or subtract their real parts and their imaginary parts

separately. Multiply – use the distributive property

or the FOIL method.DIVIDE – multiply the numerator and

denominator by the complex conjugate of the denominator

FACTOR THE EXPRESSION:

A. X2 + 18X + 81

B. X2 – 22X + 121

LESSON 7:

COMPLETE THE SQUARE

How is the process of completing the square used

to solve quadratic equations?

Essential question

VOCABULARY Completing the square – The process

of adding a term to a quadratic expression of the form x2 + bx to make it a perfect square trinomial

EXAMPLE 1 Solve a quadratic equation by finding square roots

Solve x2 – 8x + 16 = 25.

x2 – 8x + 16 = 25 Write original equation.

(x – 4)2 = 25 Write left side as a binomial squared.

x – 4 = +5 Take square roots of each side.

x = 4 + 5 Solve for x.

The solutions are 4 + 5 = 9 and 4 –5 = – 1.

ANSWER

EXAMPLE 2 Make a perfect square trinomial

Find the value of c that makes x2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial.

SOLUTION

STEP 1

Find half the coefficient of x.

STEP 2

162 = 8

Square the result of Step 1. 82 = 64

STEP 3

Replace c with the result of Step 2. x2 + 16x + 64

EXAMPLE 2 Make a perfect square trinomial

The trinomial x2 + 16x + c is a perfect square when c = 64. Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.

ANSWER


1. x2 + 6x + 9 = 36.

3 and –9.ANSWER

Solve the equation by finding square roots.

2. x2 – 10x + 25 = 1.

4 and 6.ANSWER

3. x2 – 24x + 144 = 100.

2 and 22.ANSWER


Find the value of c that makes the expression a perfect square trinomial.Then write the expression as the square of a binomial.

4.

x2 + 14x + c

49 ; (x + 7)2ANSWER

5.

x2 + 22x + c

121 ; (x + 11)2ANSWER

6.

x2 – 9x + c

ANSWER ; (x – )2.814

92

EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1

Solve x2 – 12x + 4 = 0 by completing the square.

x2 – 12x + 4 = 0 Write original equation.

x2 – 12x = –4 Write left side in the form x2 + bx.

x2 – 12x + 36 = –4 + 36 Add –122

2( )=(–6)2=36 to each side.

(x – 6)2 = 32 Write left side as a binomial squared.

Solve for x.

Take square roots of each side.x – 6 = + 32

x = 6 + 32

x = 6 + 4 2 Simplify: 32 = 16 2=4 2

The solutions are 6 + 4 and 6 – 42 2

ANSWER


CHECK

You can use algebra or a graph.

Algebra Substitute each solution in the original equation to verify that it is correct.

Graph Use a graphing calculator to graph

y = x2 – 12x + 4. The x-intercepts are about 0.34 6 – 4 2 and 11.66 6 + 4 2


Solve 2x2 + 8x + 14 = 0 by completing the square.

2x2 + 8x + 14 = 0 Write original equation.

x2 + 4x + 7 = 0

Write left side in the form x2 + bx.

x2 – 4x + 4 = –7 + 4 Add 42

2( )=22=4 to each side.

(x + 2)2 = –3 Write left side as a binomial squared.

Solve for x.

Take square roots of each side.x + 2 = + –3

x = –2 + –3

x = –2 + i 3

x2 + 4x = –7

Divide each side by the coefficient of x2.

Write in terms of the imaginary unit i.


The solutions are –2 + i 3 and –2 – i 3 .

ANSWER


SOLUTION

Use the formula for the area of a rectangle to write an equation.


3x(x + 2) = 72

3x2 + 6x = 72

x2 – 2x + 1 = 24 + 1 Add 22

2( )=12=1 to each side.

(x + 1)2 = 25 Write left side as a binomial squared.

Solve for x.

Take square roots of each side.x + 1 = + 5

x = –1 + 5

x2 + 2x = 24 Divide each side by the coefficient of x2.

Length Width = Area

Distributive property


So, x = –1 + 5 = 4 or x = – 1 – 5 = –6. You can reject x = –6 because the side lengths would be –18 and –4, and side lengths cannot be negative.

The value of x is 4. The correct answer is B.

ANSWER


x2 + 6x + 4 = 0

–3+ 5ANSWER

7.

Solve the equation by completing the square.

x2 – 10x + 8 = 0

5 + 17ANSWER

8.

2n2 – 4n – 14 = 0

1 + 2 2ANSWER

9.

3x2 + 12x – 18 = 010.

–2 + 10ANSWER

11. 6x(x + 8) = 12

–4 +3 2ANSWER

1 + 26ANSWER

12. 4p(p – 2) = 100

Essential question

How is the process of completing the square used to solve quadratic

equations?You complete the square so that one

side of the equation can be written as the square of a binomial. Then you take the square roots of both sides

and simplify the results.

EVALUATE B2 – 4AC WHEN A = 3, B = -6 AND C = 5.

LESSON 8:

USE THE QUADRATIC FORMULA AND THE DISCRIMINANT

How do you use the quadratic formula and the

discriminant?

Essential question

VOCABULARY Quadratic formula – The formula x =

used to find the solution of the quadratic equation ax2 + bx + c = 0

Discriminant – The expression b2 + bx + c = 0; also the expression under the radical sign of the quadratic formula

EXAMPLE 4 Use the discriminant

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

a. x2 – 8x + 17 = 0 b. x2 – 8x + 16 = 0 c. x2 – 8x + 15 = 0

SOLUTION

Equation Discriminant Solution(s)

ax2 + bx + c = 0 b2 – 4ac x =– b+ b2– 4ac2ac

a. x2 – 8x + 17 = 0 (–8)2 – 4(1)(17) = –4 Two imaginary: 4 + i

b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(16) = 0 One real: 4

b. x2 – 8x + 15 = 0 (–8)2 – 4(1)(15) = 0 Two real: 3,5


Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

4. 2x2 + 4x – 4 = 0

48 ; Two real solutions

5.

0 ; One real solution

3x2 + 12x + 12 = 0

6. 8x2 = 9x – 11

–271 ; Two imaginary solutionsANSWER

ANSWER

ANSWER

7.


7x2 – 2x = 5

144 ; Two real solutions

8. 4x2 + 3x + 12 = 3 – 3x

–108 ; Two imaginary solutions

9. 3x – 5x2 + 1 = 6 – 7x

0 ; One real solutionANSWER

ANSWER

ANSWER

Essential question

How do you use the quadratic formula and the discriminant?

Substitute the three coefficients from the standard form into the formula

and simplify the result. The discriminant gives the number and

type of solutions of a quadratic equation.

Documents

On Page 234, complete the Prerequisite skills #1-14