Math Case Study Homework

CASE

INSTRUCTIONS: Read the references found on the Background Info page. Study the examples there, and the ones given below. Work out the problems, showing all the computational steps. This is particularly important for those problems for which the answers are given. On those problems, the correct procedure is the only thing that counts toward the assignment grade.

SOLVING QUADRATIC EQUATIONS BY FACTORING.

References: Waner, 2007

Examples:

E1:

E2:

Problems 1 through 5: Find the roots (values of x when the expression is equal to zero) using factoring. Prove your answers are correct by checking, or showing that the values of x you computed satisfy the original equation.

1. Ans:

2.

3.

4.

5;

SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE.
References: Stapel, 2004a
Examples:
E3:

E4:

Problems 6-10. Solve by completing the square. Prove your answers are correct by checking.

6. Ans:

7.

8.

9.

10.
SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA.
Reference: Stapel, 2004b.
Examples:
E5:

E6:

Problems 11 to 15:
Solve by using the quadratic formula. Prove your answer are correct by checking.

11. Ans:

12.

13.

14.

15.

THE DISCRIMINANT AND ITS USE. In the quadratic formula, the expression under the radical (b2-4ac) is so important that it has its own name. It’s called the Discriminant, abbreviated D. By examining the formula, it’s easy to see that if D=0, then the equation being solved has exactly one root; namely, . If D is positive, then the radical has two real values, and the equation has two real roots. If it is negative, then the radical has two complex values, and the equation has two complex roots.
References: Simmons, 2012; MathIsFun.com, 2012.
Examples:
E7: Use the properties of the discriminant to discover whether the following equation as one real, two real, or two complex roots: x2+2=0.

Problems 16, 17: Determine the number and type of root(s) for each equation.

16.

17.

Problems 18 through 20: Consider the quadratic equation
18. For what value of c does the equation have one real root? (Hint: Using the values of a and b found in the equation, solve D = 0 for c.)
19. For what range of values does the equation have two real roots? (Hint: The answer is an inequality. Don’t simply offer examples of c that satisfy the requirement.)
20. For what range of values does the equation have two complex roots?

2
4x25x0
-=
5
x
2
=±
2
x12x360
-+=
2
x14x450
++=
16
x8
x
+=
2
3x21x0
+=
22
2
2
2
22
2
2
2
General procedure:
Given axbxc, rewrite as xbxc.
bbb
Find . Square it: .
224
b
Add to both sides of the equation.
4
bb
xbxc
44
bb
xc
24
+++=-
æö
æöæö
=
ç÷
ç÷ç÷
ç÷
èøèø
èø
æö
ç÷
ç÷
èø
æöæö
++=-+
ç÷ç÷
ç÷ç÷
èøèø
æö
=+=-+
ç÷
èø
.
Take the root of both sides and solve fo
r values of x.
æö
ç÷
ç÷
èø
22
2
2
x(x+10)=11. Put into standard form.
x10x11. Half of 10 is 5; 525. Add to bo
th sides.
x10x251125
(x5)36
x5366
x65
x651
or
x6511
ANS: x=1 or -11
Checking:
x(x+10)=11
1(1+10)=1(11)
+==
++=+
+=
+==±
=±-
=-=
=–=-
=11
(11)(-11+10)=(-11)(-1)=11
–
2
x4x30
++=
x1,3
=–
2
x5x10
+-=
2
2x7x150
+-=
2
1
x7x16
2
-=
x(x5)4
-=-
2
2
2
2
General form of a quadratic equation: a
xbxc0.
Step 1: Identify a, b, c; b or c may =
0.
x3x0; a=1, b=-3, c=0
Quadratic formula:
(3)(3)4(1)(0)
bb4ac
x=
2a2(1)
3933
22
336
x3
22
or
3
x
++=
-=
–±–
-±-
=
±±
==
+
===
–
=
2
2
2
30
0.
22
ANS: x= 3 or x=0
Checking:
x3x0
33(3)990
0(0)0
==
-=
-=-=
-=
2
2
2
2
2
x10x30; a=1, b=10, c=-3
Quadratic formula:
(10)(10)4(1)(3)
bb4ac
x=
2a2(1)
104(28)
101001210112
222
10228
528
2
ANS: x= -528
Checking:
x10x30
(-5-28)10(-5-28)30
2510282850
+-=
-±–
-±-
=
-±
-±+-±
===
-±
==-±
±
+-=
+-=
++-
102830
(2528503)(10281028)0
00
Similarly for -5+28.
–=
+–+-=
=
2
x90
-=
x3
=±
2
x90
+=
2
x3x40
-+=
2
2x3x60
++=
(
)
2
5×1130
++=
b
2a
–
æö
ç÷
èø
2
2
x20 a=1, b=0, c=2.
D=b4ac04(1)(2)8.
Since D is negative, the equation has tw
o complex roots.
+=®
-=-=-
x(x3)2
+=
2
x40.
(x2)(x2)0.
The two factors are (x+2) and (x-2).
If EITHER is equal to zero, the correspo
nding value of x is a root.
Solving the two factors for x:
(x+2)=0x=-2
(x-2)=0x2
The two roots (solutio
-=
+-=
®
®=
2
2
ns) are x=2 and x=-2.
ANS: x=2
Checking:
(2)4440
(-2)4440
±
-=-=
-=-=
2x(x2)2
+=-
2
x4xc0.
++=
2
2
22
xx6.
xx60. (Must be in this form.)
(x+2)(x-3)=0
The two factors are (x+2) and (x-3). Ch
ecking:
(x+2)(x-3)=x3x2x6xx6.
If EITHER is equal to zero, the correspo
nding value of x is a root.
Solv
-=
–=
-+-=–
2
2
2
ing the two factors for x:
(x+2)=0x=-2
(x-3)=0x3
The two roots (solutions) are x=-2 and x
=3
ANS: x=-2 or +3.
Checking:
(x)x6
(-2)(2)426
(3)(3)936
®
®=
-=
–=+=
-=-=