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Math 140
Hw 1
Worked examples of selected recommended problems.
Find all real solutions x. Write Q if no solution.
A. 3x  2 2  3x  2
Don’t divide both sides by 3x  2, you could be
dividing by 0. First get everything on the left, 0 on the
right. Then factor or use the quadratic formula.
3x  2 2  3x  2  0
3x  23x  2  1  0
3x  23x  3  0
33x  2x  1  0.
Answer: x  23 , 1

1 1411
2

3 9449
2
1  x  6x1
0
6x11x
2
 6x1
6x1
6x 2 5x12
0
6x1
2
6x 5x1
0
6x1
6x 2 5x1
0
6x1
3x12x1
0
6x1
0
x1x2x3
x 2 3x2
x1x2x3  0
S.
32x
32x


3 11
2
1
x
32xx
0
0
0
The top doesn’t factor, use the quadratic formula.
32xx
A rational is zero iff the top is zero.
x152x
x1x2x3
3x 2 6x32x 2 3x5
62 11
4
3 11
2 ,4

2x 2 x3
 x 252x
5x6  0
First factor as much as possible and simplify.
3x 2 6x3
52x
x1x2x3  x3x2  0

6 44
4
3 11
2
0
2  x  0 iff x  2
3  2x  0 iff x  32
Key numbers: x  32 , 2
Key intervals: , 32 ,  32 , 2, 2, 
Exclude 3/2 because fx is undefined there. Include 2
because fx is defined and the inequality is >.
Pick a point from each interval and evaluate if
2x
is >0 or <0.
fx  32x
f0  0, f 74   0, f10  0
Answer: , 32   2, 
3x2x 2 32x
3x 2 6x3
x1x2x3
3x 2 6x3
x1x2x3

2x
32x
32x
1
32x  x  0
32xx
32x
32xx  32xx
Anwer: x = ƒ, ½
J.
O.
3 135
x

.
24
8
Answer: no solutions: Q.
6 36421
22
Key numbers: x =
,
Key intervals:
3 11
3 11 3 11
3 11
, 2 ,  2 , 2 ,  2 , 4, 4, 
Pick a point from each interval and evaluate if
fx  x  42x 2  6x  1 is >0 or <0.
f10  0, f0  4  0, f7/2  0, f10  0
3 11
3 11
Answer: , 2    2 , 4
1 5
2
F. (a) x  8  4
Square both sides (when you do this, you may
introduce invalid answers. Delete any invalid answers
at the end.)
x  8  16
x  24
Sticking x back into x  8  4 gives 24  8  4
which is true.
Answer: 24.
(b) x  8  2x  1
x  8  2x  1 2
x  8  4x 2  4x  1
x  8  4x 2  4x  1  0
4x 2  3x  9  0
Doesn’t factor, so use quadratic formula.
4x 2  3x  9  0
I.
Solve for x. Write your answer in interval notation.
L. x  42x 2  6x  1  0
x  4 is zero at 4.
2x 2  6x  1 is zero at, use the quadratic formula,
Doesn’t factor, so use quadratic formula.
b b 2 4ac
2a
0
Answer: x = 1.
x
D. xx  1  1
xx  1  1  0
x2  x  1  0
Answer: x 
x1x2
x1x2x3 
x1
x1x3  0
0
0
1 1423
1 23
 8
2x 2  x  3  0 iff x 
24
3
Bottom = 0 iff x  2 , 0
Key numbers:  32 , 0
Key intervals: ,  32 ,  32 , 0, 0, 
2 x3
fx  2x
32xx , f10  0, f1  0, f10  0
Answer:  32 , 0
=Q