MATH 20
Answers to Quiz 8
Solving 2x2 − x − 6 = 0, using factoring method.
(1a)
=⇒ (2x + 3)(x − 2) = 0
=⇒ x = − 32 or 2
=⇒ either 2x + 3 = 0 or x − 2 = 0
Solving 2x2 − x − 6 = 0, using ”Completing the Square” method.
(1b)
=⇒ x2 − 12 x − 3 = 0;
First, divide both sides by 2 to make the leading coefficient 1
=⇒ x2 − 21 x = 3;
Then, move the constant term to the right hand side
Then, add to both sides the square of half the coefficient of degree 1 term (i.e. half of
1
2x
2
=⇒ x −
=⇒ x −
1
4
q
=±
( 14 )2
=3+
49
16
= ± 74
=⇒ (x −
=⇒ x =
1
4
±
1 2
4)
7
4
=⇒ x =
=⇒ x =
Solving:
√
−b± b− 4ac
2a
1+7
4
x2
2
√
=
−(−1)±
= 2 or x =
1−7
4
1
16
=3+
(−1)2 −4(2)(−6)
2(2)
=
√
1±
48
16
=
=⇒ x =
Solving 2x2 − x − 6 = 0, using the Quadratic formula.
(1c)
(2)
+
( 14 )2
8
4
+
=2
1−(−48)
4
1
16
=⇒ (x −
or
x=
=
√
1± 49
4
−6
4
=
1 2
4)
1
2
)
=
49
16
= − 32
1±7
4
= − 64 = − 23
+ 52 x = −1
Multiply both sides by the LCD 2 to eliminate the denominators:
=⇒ x2 + 5x = −2
=⇒ x2 + 5x + 2 = 0
Rewrite it to the standare form:
Then, use the Quadratic Formula, (or Completing the Square method) to solve it, (the left hand side
is not factorable).
2
x + 5x + 2 = 0
=⇒ x =
√
−b± b− 4ac
2a
=
−5±
√
52 −4(1)(2)
2(1)
=
√
−5± 25−8
2
=
√
−5± 17
2
(17 is not a complete square, leave it in the square root sign, don’t use a calculator to give approximations.)
(3)
Solve the quadratic equation: (2x + 3)(x + 4) = 1
(expand the left hand side and rewrite it to its standard form)
=⇒ 2x2 + 11x + 12 = 1
=⇒ 2x2 + 11x + 11 = 0
(solve it by the quadratic formula:)
=⇒ x =
√
−b± b− 4ac
2a
=
−11±
√
112 −4(2)(11)
2(2)
=
√
−11± 121−88
4
=
√
−11± 33
4
Solve the equation y 4 − 7y 2 + 12 = 0
(4)
(This is NOT a quadratic equation, but it is an equation quadratic in form.)
Let t = y 2 , the original equation becomes:
t2 − 7t + 12 = 0
=⇒ (t − 3)(t − 4) = 0
t=3
2
=⇒ y = 3
=⇒ t = 3 or t = 4
√
=⇒ y = ± 3; t = 4 =⇒ y 2 = 4
√
=⇒ y = ± 4 = ±2
(5)
Using the discriminant to determine the type of the solutions of a quadratic equation.
(5a)
x(x − 3) + 10 = 0
and c = 10,
Its discriminant is:
=⇒ Its standard form is:
x2 − 3x + 10 = 0.
=⇒ a = 1, b = −3,
b2 − 4ac = (−3)2 − 4(1)(10) = 9 − 40 = −31 < 0
This quadratic equation has no real number solutions but two complex conjugate solutions.
(5b)
9x2 + 6x + 1 = 0
Its discriminant is:
=⇒ (It is in the standard form).
2
=⇒ a = 9, b = 6, and c = 1,
2
b − 4ac = (6) − 4(9)(1) = 36 − 36 = 0
This quadratic equation has one real number solution (or double roots, or coincide real solutions.)
(5c)
2x(x − 3) = 5
and c = −5,
=⇒ Its standard form is:
Its discriminant is:
2x2 − 6x − 5 = 0.
=⇒ a = 2, b = −6,
b2 − 4ac = (−6)2 − 4(2)(−5) = 36 − (−40) = 76 > 0
This quadratic equation has two unequal real number solutions.
(6)
Determine the k values such that the solutions of 3x2 + 4x + k = 0 are real numbers.
In order that a quadratic equation has real number solutions, it must be satisfied that its discriminant
b2 − 4ac ≥ 0.
The discriminant of this quadratic equation is:
b2 − 4ac ≥ 0
=⇒ 16 − 12k ≥ 0
b2 − 4ac = (4)2 − 4(3)(k) = 16 − 12k,
=⇒ −12k ≥ −16
(divide both sides by −12, don’t forget to reverse the direction of the inequality symbol:)
=⇒ k ≤
16
12
=⇒ k ≤
4
3
(7) Among all the pairs of numbers whose sum is 42, (a) find a pair whose product is as large as possible.
(b) What is the maximum value of the product?
Let x be one of the two numbers, then the other number is 42 − x.
The product of the two numbers is x(42 − x).
=⇒ y = x(42 − x) = −x2 + 42x
This is a quadratic form, let y represent it
=⇒ y = −x2 + 42x is a quadratic function, since a = −1 < 0, y has a maximum value:
ymax = c −
b2
4a
=0−
422
4(−1)
which is realized at x =
=0+
−b
2a
=
1764
4
−42
2(−1)
= 441,
= 21
Answer: (7a) when the two numbers are 21 and 21, (7b) their product has a maximum value 441.
(8)
(a)
Rewrite the quadratic function y = −2x2 + 8x into the form of y = a(x − h)2 + k.
(use ”Completing the Square”)
y = −2x2 + 8x =⇒ y = −2(x2 − 4x)
=⇒ a = −2, h = 2 and k = 8
=⇒ y = −2(x2 − 4x + 4) − (−2)4
Note:
h=
(b)
You can also use the formulas:
−b
2a
and k = c −
b2
4a
=⇒ y = −2(x − 2)2 + 8
to find h = 2 and k = 8.
Is this parabola opening upward or downward ?
The graph of this function (parabola) is opening downward since a = −2 < 0
(c)
Where is its vertex?
Its vertex is:
(d)
(h, k) = (2, 8)
(h =
−b
2a
=
−8
2(−2)
= 2, k = c −
b2
4a
=0−
82
4(−2)
= 0 − (−8) = 8)
Does y have a maximum or minimum value? Find the maximun/minimum value of y.
y has a maximum, ymax = 8 (the k value)
Note: The paraloba is opening downward, the vertex is the highest point on the graph, the
y-coordinate of the vertex is the largest possible y-value.
(e)
Find its x-intercepts.
To find x-intercepts, let y = 0 =⇒ 0 = −2x2 + 8x =⇒ −2x(x − 4) = 0
=⇒ The graph has two x-intercepts: (0, 0) and (4, 0)
(f)
=⇒ x = 0, x = 4
Roughly graph this function.
(the parabola is opening downward, vertex is at (2, 8), the axis of symmetry is x = 2.)
(9)
2
−7x = x2 +3
Solve the quadratic equation:
Multiply both sides by 2 to eliminate the denominator and rewrite it into its standard form.
=⇒ −14x = x2 + 6
=⇒ x2 + 14x + 6 = 0
√
−b± b2 −4ac
=⇒ x =
2a
Apply the Quadratic Formula:
√
=
(10)
−14±
142 −4(1)(6)
2(1)
√
−14± 196−24
2
=
2
=
√
−14± 172
2
=
√
−14±2 43
2
= −7 ±
√
43
1
x 3 −4 x 3 −5 = 0
Solve the equation:
This is not a quadratic equation, but it can be written in quadratic form.
1
Let t = x 3 , then, the original equation can be written as t2 − 4t − 5 = 0
Solve t2 − 4t − 5 = 0 by either factoring method or the quadratic formula
=⇒ (t − 5)(t + 1) = 0
Since t = x
1
3
=⇒ t = 5 or t = −1
1
3
1
=⇒ x = 5 or x 3 = −1
Take the 3rd power of both sides:
(11)
=⇒ x = 125 or x = −1
x4 = 14x2 − 45
Solve the equation:
=⇒ x4 − 14x2 + 45 = 0
This is not a quadratic equation, but it can be written in quadratic form.
Let t = x2 , then, the original equation can be written as t2 − 14t + 45 = 0
Solve it by factoring method or the quadratic formula,
t2 − 14t + 45 = 0
Since t = x2
(12)
=⇒ (t − 9)(t − 5) = 0
=⇒ x2 = 9 or x2 = 5
=⇒ t = 9 or t = 5
√
=⇒ x = ±3 or x = ± 5
x2 + 5 < 6x and graph the solution set on the number line.
Solve the inequality:
Move all the terms to the left hand side and leave the right hand side a 0.
=⇒ x2 − 6x + 5 < 0
Solve x2 − 6x + 5 = 0 to find the ZERO’s of the left hand side expression.
=⇒ x2 − 6x + 5 = 0
=⇒ (x − 5)(x − 1) = 0
=⇒ x = 5, x = 1
The above two points partition the number line into 3 intervals, then, in each interval, pick a testing
point to check whether the inequality x2 − 6x + 5 < 0 is True or False, as follows,
Interval
(−∞, 1)
(1, 5)
(5, ∞)
x2 − 6x + 5 =?
(0)2 − 6(0) + 5 = 0 − 0 + 5 = 5
(2)2 − 6(2) + 5 = 4 − 12 + 5 = −3
(6)2 − 6(6) + 5 = 36 − 36 + 5 = 5
testing point
0
2
6
=⇒ the solution set is:
q
q
(1, 5).
q
q
Is x2 − 6x + 5 < 0 True?
5 < 0 is False.
−3 < 0 is True.
5 < 0 is False.
(or, 1 < x < 5).
q
q
0
qe
1
q
x
q
q
qe
5
-
(13)
6
x
6
x
Solve the inequality:
=⇒ x6 −2 ≤ 0
≤2
≤2
=⇒ 6−2x
≤0
x
The ZERO’s of the left hand side are x = 3 (in the numerator) and x = 0 (in the denominator).
The above two points partition the number line into 3 intervals, then, in each interval, pick a testing
point to check whether the inequality 6−2x
≤ 0 is True or False, as follows,
x
Interval
testing point
(−∞, 0)
−1
(0, 3)
1
(3, ∞)
4
=⇒ the solution set is:
(Note:
(−∞, 0)
S
6−2x
=?
x
6−2(−1)
= −8
−1
6−2(1)
=4
1
6−2(4)
= − 12
4
[3, ∞).
Is 6−2x
≤ 0 True?
x
−8 ≤ 0 is True.
4 ≤ 0 is False.
− 12 ≤ 0 is True
(in inequality notation:
x < 0 or x ≥ 3).
The boundary point 3 should be included in the solution set, since it satisfies the inequality.)
q
x
¾
q
q
q
q
x
qe
q
q
qu
0
q
q
-
-
3
2x2 −x−3
4x−1 < 0
Find the ZERO’s of the left hand side expression first.
(14)
Solve the inequality:
(in the numerator) =⇒ 2x2 − x − 3 = 0 =⇒ (2x − 3)(x + 1) = 0
(in the denominator) =⇒ 4x − 1 = 0 =⇒ x = 41
=⇒ x = −1, x =
3
2
= 1 12
The above three points partition the number line into 4 intervals, then, in each interval, pick a testing
2x2 −x−3
point to check whether the inequality 4x−1 < 0 is True or False, as follows,
Interval
testing point
(−∞, −1)
−2
(−1, 14 )
0
( 14 , 1 12 )
1
(1 21 , ∞)
2
=⇒ the solution set is:
2x2 −x−3
=?
4x−1
2(−2)2 −(−2)−3
= 8+2−3
= − 97
4(−2)−1
−8−1
2
2(0) −(0)−3
= 0−0−3
=3
4(0)−1
0−1
2
2(1) −(1)−3
= 2−1−3
= −1
4(1)−1
4−1
2(2)2 −(2)−3
8−2−3
= 8−1 = 73
4(2)−1
(−∞, −1)
S
2 −x−3
< 0 True?
Is 2x4x−1
− 79 < 0 is True.
3 < 0 is False.
−1 < 0 is True.
3
7
< 0 is False.
( 14 , 1 12 ).
(in inequality notation, x < −1 or
1
4
< x < 1 12 ).
The boundary points should be all EXcluded from the solution set, since −1 and 1 21 will make left
hand side a 0, which does not satisfy the inequality, and the point 41 will make left hand side undefined.)
(Note:
(15) y is a quadratic function in x:
and (b) its vertex.
y = −2x2 + 6x. Its graph is a parabola.
(a)
To find its opening:
(b)
To find its vertex, either use the formulas or use the Completing the Square method.
(Formula:)
=⇒ h =
since a = −2 < 0
Find (a) its opening,
=⇒ the parabola is opening downward.
a = −2, b = 6, c = 0
−b
2a
the vertex is:
=
−6
2(−2)
( 23 , 92 )
=
−6
−4
=
3
2
and k = c −
b2
4a
=0−
62
4(−2)
=0−
36
−8
=0+
9
2
=
9
2
(15)
(continued)
(Completing the Square method:)
y = −2x2 + 6x
=⇒ y = −2(x2 − 3x) = −2(x2 − 3x + ( 23 )2 ) − (−2)( 32 )2
=⇒ y = −2(x − 32 )2 + ( 29 )
the vertex is:
=⇒ h =
3
2
and k =
9
2
( 32 , 92 )
(c) To find the x-intercepts, set y = 0
=⇒ 0 = −2x2 + 6x
=⇒ −2x(x − 3) = 0
=⇒ x = 0 x = 3,
The graph of y = −2x2 + 6x has two x-intercepts,
(16)
(0, 0) and (3, 0)
Find the following arithmetic and composition of functions:
Let f (x) = 3x + 5 and g(x) = x2 − 2
(a)
(f − g)(−3) = f (−3) − g(−3) = (3(−3) + 5) − ((−3)2 − 2) = (−4) − (7) = −11
(b)
(f · g)(−5) = f (−5) · g(−5) = (3(−5) + 5)((−5)2 − 2) = (−10)(23) = −230
(c)
(f ◦ g)(x) = f (g(x)) = f (x2 − 2) = 3(x2 − 2) + 5 = 3x2 − 6 + 5 = 3x2 − 1
(d)
(g ◦ f )(x) = g(f (x)) = g(3x + 5) = (3x + 5)2 − 2 = (9x2 + 30x + 25) − 2 = 9x2 + 30x + 23
(e)
(f ◦ g)(−3) = f (g(−3)) = f ((−3)2 − 2) = f (7) = 3(7) + 5 = 21 + 5 = 26
Or, make use of the result in (c),
=⇒ (f ◦ g)(−3) = 3(−3)2 − 1 = 27 − 1 = 26
(f)
(g ◦ f )(−3) = g(f (−3)) = g(3(−3) + 5) = g(−4) = (−4)2 − 2 = 16 − 2 = 14
Or, make use of the result in (d),
=⇒ (g ◦ f )(−3) = 9(−3)2 + 30(−3) + 23 = 9 × 9 + (−90) + 23 = 81 − 90 + 23 = 14
(17)
Find the inverse function of f (x) = 11 − 5x
Replace f (x) with y
=⇒ y = 11 − 5x
Interchange x and y
=⇒ x = 11 − 5y
Solve the above equation for y
Replace y with f −1 (x)
11−x
5
11−x
5
=⇒ y =
=⇒ f −1 (x) =
(You can check the above f (x) and f −1 (x) to verify that f (f −1 (x)) = x and f −1 (f (x)) = x)