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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
GUIDED PRACTICE for Examples 1 and 2
2. y = – x2 – 5
ANSWER
Same axis of symmetry, vertex is shifted down 5 units, and opens down
GUIDED PRACTICE for Examples 1 and 2
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.
EXAMPLE 3 Graph a function of the form y = ax2 + bx + c
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.
EXAMPLE 3 Graph a function of the form y = ax2 + bx + c
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
GUIDED PRACTICE for Example 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.
GUIDED PRACTICE for Examples 1 and 2
Graph the function. Label the vertex and axis of symmetry.
1. y = (x + 2)2 – 3 2. y = –(x + 1)2 + 5
GUIDED PRACTICE for Examples 1 and 2
3. f(x) = (x – 3)2 – 412
GUIDED PRACTICE for Examples 1 and 2
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.
GUIDED PRACTICE for Examples 3 and 4
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)
GUIDED PRACTICE for Examples 3 and 4
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.
GUIDED PRACTICE for Examples 5 and 6
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
GUIDED PRACTICE for Examples 5 and 6
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.
GUIDED PRACTICE for Example 1
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
Factor the expression.
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
Perfect square trinomial
= x2 – 72
= d 2 + 2(d)(6) + 62
= z2 – 2(z) (13) + 132
GUIDED PRACTICE for Example 2
4. x2 – 9
(x – 3)(x + 3)
5. q2 – 100
(q – 10)(q + 10)
6. y2 + 16y + 64
(y + 8)2
Factor the expression.
ANSWER
ANSWER
ANSWER
GUIDED PRACTICE for Example 2
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
Factor the expression.
a. 9x2 – 64
= (3x + 8)(3x – 8)Difference of two squares
b. 4y2 + 20y + 25
= (2y + 5)2
Perfect square trinomial
c. 36w2 – 12w + 1
= (6w – 1)2
Perfect square trinomial
= (3x)2 – 82
= (2y)2 + 2(2y)(5) + 52
= (6w)2 – 2(6w)(1) + (1)2
GUIDED PRACTICEGUIDED PRACTICE for Example 3
Factor the expression.
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
GUIDED PRACTICEGUIDED PRACTICE for Example 3
11. 49z2 + 4z + 9
(7z + 3)2
12. 36n2 – 9 = (3y)2
(6n – 3)(6n +3)
ANSWER
ANSWER
EXAMPLE 4 Factor out monomials first
Factor the expression.
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)
GUIDED PRACTICEGUIDED PRACTICE for Example 4
Factor the expression.
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
GUIDED PRACTICEGUIDED PRACTICE for Example 4
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
GUIDED PRACTICEGUIDED PRACTICE for Example 1
271.
3 3
982.
2 7ANSWER
ANSWER
GUIDED PRACTICEGUIDED PRACTICE for Example 1
3.
6 5
10 15
4. 8 28
14 4ANSWER
ANSWER
GUIDED PRACTICEGUIDED PRACTICE for Example 1
5.
3 8
9
64
6. 15
4
215ANSWER
ANSWER
GUIDED PRACTICEGUIDED PRACTICE for Example 1
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.
EXAMPLE 3 Solve a quadratic equation
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
EXAMPLE 4 Standardized Test Practice
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
EXAMPLE 1 Solve a quadratic equation
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 .
GUIDED PRACTICE for Example 1
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.
GUIDED PRACTICE for Example 2
Write the expression as a complex number 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
GUIDED PRACTICE for Examples 3, 4 and 5
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
Write the expression as a complex number in standard form.
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
GUIDED PRACTICE for Examples 1 and 2
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
GUIDED PRACTICE for Examples 1 and 2
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
EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1
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
EXAMPLE 4 Solve ax2 + bx + c = 0 when a = 1
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.
EXAMPLE 4 Solve ax2 + bx + c = 0 when a = 1
The solutions are –2 + i 3 and –2 – i 3 .
ANSWER
EXAMPLE 5 Standardized Test Practice
SOLUTION
Use the formula for the area of a rectangle to write an equation.
EXAMPLE 5 Standardized Test Practice
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
EXAMPLE 5 Standardized Test Practice
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
GUIDED PRACTICE for Examples 3, 4 and 5
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
GUIDED PRACTICE for Example 4
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.
GUIDED PRACTICE for Example 4
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.