Download Test 1 - Yeah, math, whatever.

Practice problem set 1 for first test, MAT 102
(1) Solve the equation:
(a) 4(2 x  1)  3 x  5( x  2)  13
(b)
2
1 5
x 
3
4 6
(c) 2 | x  1| 3  7
(d) | x  3 || 2 x  1|
(2) Solve and graph the inequality, and give in interval notation:
(a) 2( x  3)  4  x  5
(b) 1  2x  4
(c) 2 x  1  3  4
(d) x  5  3
(3) Graph the line with equation:
(a) 2 x  3 y  12
(b) y 
3
x 1
4
(c) f ( x)  2 x  5
(4) Sketch (draw as a translate of | x |) the absolute value function:
(a) f ( x)  x  6  2
(b) g ( x)   x  3  2
(5) Factor the polynomial:
(a) x 2  8 x  7.
(b) x 2  2 x  15 .
(c) 5x2 – 7x – 6
(d) 7x2 - 700,
(e) 27 x3  64
(6) Solve the quadratic equation:
(a) x 2  3x  5 x  8 (b) 2 x 2  50  0
(7) Solve the rational equation:
1
1
4

 2
x2 x2 x 4
Solutions:
(1) The overriding strategy here is to isolate the variable:
(a) 4(2 x  1)  3 x  5( x  2)  13
8x  4  3x  5x 10 13
5 x  4  5 x  23
5 x  4  5 x  4
0x = 27,
which has no solutions.
(distribute)
(combine)
(variable to one side)
(b) Here, we're going to begin by clearing fractions:
2
1 5
x 
(multiply by the LCD = 12)
3
4 6
2
1
5
12  x    12*( ),
3
4
6
8 x  3  10
+3 +3
13
8x =13, x 
8
(c) 2 | x  1| 3  7
+3 +3
2 | x  1| 10
2
10
| x  1| ,
2
2
| x  1| 5
x - 1 = 5, x - 1 = -5
+1 +1
+1 +1
x = 6, x = -4.
(first, isolate the absolute value:)
(now use the absolute value property:)
(d) This is a 'double absolute'.
| x  3 || 2 x  1|
x  3  2 x  1,
2 x  3  2 x  3
 x  4,
x  4,
x  3  (2 x  1)
x  3  2 x  1
 2 x 3  2 x  3
3 x  2, x 
2
3
2
x  {4, }
3
(2) In (a) and (b) the object is to isolate the variable. We have to keep in mind that if we
multiply or divide by a negative number, then we have to reverse the inequality.
(a)
2x  6  4  x  5
2x  2  x  5
x  2  x  2
x  7
Interval notation (-∞,-7]
(b)
1 2x  4
1
1
-2x < 3
2x 3

 ,
2 2
3
x
2
(divide by -2 - reverse inequality:)
 3 
Interval notation:   ,  
 2 
(c) We have to keep the special rules for absolute inequalities in mind:
2 x  1  3  4 (first, isolate the absolute value:)
2x 1  3  4
(now, with operation < we 'wedge' the inequality between 3, -3:)
3 3
| 2 x  1| 7,
7  2 x  1  7
(solve the compound inequality:)
1
1 1
6  2 x  8
6 2x 8
 
 ,
2 2 2
3  x  4.
Interval notation [-3,4]
(d) Here the operation is greater than (>), and therefore we have to 'split' the inequality
into separate ones:
(> - 'split' the inequality like: )
x 5  3
x  5  3 or x  5  3,
5 5
5 5
x  2 or x  8,
Interval notation: (, 2) U (8, )
(3) (a) Find the intercepts by setting x = 0 (y-intercept) and y = 0 (x-intercept).
2(0)  3 y  12,
12
y
 4 : (0, 4)
3
2 x  3(0)  12,
12
x
 6 : (6, 0)
2
Graph the intercepts and draw the line that runs through them:
(b) Here, because the line is written in slope-intercept form, and we use the 'up and over'
a
method. First, graph the y-intercept (0,b) = (0,-1) . Then, use the slope
to move up a
b
(if positive) or down (if negative) a units and right b units, to get to the second point on
the graph. That would put the second point at (4,2):
3
y  x 1
4
(c) f ( x)  2 x  5 . This one's written in function notation, but the method of graphing is
2
the same: the y-intercept is (0,5), and the slope is 2   , so move 2 units down and 1
1
to the right to get to (1,3).
(4) (a) f ( x)  x  6  2 (inside the absolute: 6 left (adding), outside: 2 up (adding))
(b) g ( x)   x  3  2 . (inside the absolute: 3 right (subtracting), outside: 2 down
(subtracting)). The negative lead coefficient 'flips' the graph so that the point is facing up.
(5) (a) Trinomial, lead coefficient equal to 1.
x 2  8 x  7.
(need two numbers that multiply to 7 add to 8 - there ain't
many choices since 7 is prime - only 1 and 7 are factors)
1 7
(1 + 7 = 8, so:)
= (x + 1)(x + 7).
(b) x 2  2 x  15 (need two numbers which * to -15, add to -2
list the factors of -15. The larger ones will have to be
1 -15 negative.)
3 -5 (3 + -5 = -2, so: )
= (x + 3)(x - 5).
(c) Trinomial, lead coefficient other than 1.
5x2 – 7x – 6
( grouping: Multiply the lead and constant terms:)
5(-6) = -30,
1 -30
2 -15
3 -10
2
5x - 10x + 3x - 6
gcf = 5x gcf = 3
5x(x - 2) + 3(x - 2) =
(5x + 3)(x - 2).
(d) 7x2 - 700 =
7(x2 - 100) =
(x)2 - 102
7(x - 10)(x + 10).
(3 + -10 = -7 - use them to split up -7x)
(pair off, gcf's:)
(factor gcf's, combine terms: )
(gcf = 7, factor it out:)
(DOS in pharentheses).
(e) 27 x3  64
(This is a SOC)
(3x)3 + (4)3 =
(use formula, with a = 3x, b = 4).
2
(3x + 4)(9x - 12x + 16)
(answer)
(3x)2 (3x)(4) (4)2
(6) x 2  3x  5 x  8
-5x -5x -8
-8
2
x  2x  8  0
1 8
2 -4
(x + 2)(x - 4) = 0
x + 2 = 0, x - 4 =0
-2 -2 +4 +4
x = -2, x = 4
(b) 2 x 2  50  0
2  x 2  25   0
2( x  5)( x  5)  0,
x  5  0, x  5  0
x  5,
(get right-hand side = 0)
(factor left-hand side: trinomial, lead coeff = 1)
(set factors = 0)
(factor left - gcf = 2 then DOS)
(set factors = 0)
x  5
x  {5}
1
1
4

 2
(Multiply by LCD = (x + 2)(x - 4))
x2 x2 x 4

1  
4
 1
( x  2)( x  2) 

 ( x  2)( x  2),

 x  2 x  2   ( x  2)( x  2) 
(7)
x  2  x  2  4,
2 x  4,
x2
This solution doesn't work because it makes the first term and right-hand side have a
denominator of 0. Thus, there are no solutions.