Download Math 101 Study Session Spring 2016 Test 5 Chapter 13 and

Math 101 Study Session Spring 2016 Test 5 Chapter 13
and Chapter 14 Sections 1, 2, and 3
April 27, 2016
Chapter 13 Section 1: Introduction to Radical Expressions
A square root of a positive number x is a number whose square is x.
√
The symbol is called a radical sign. The number and/or variable inside the radical sign
is called the radicand.
Square Roots of Perfect Squares
√
√
√
√
1
=
1
16
=
4
49
=
7
√
√
√100 = 10
√
√4 = 2 √25 = 5 √64 = 8 √121 = 11
9=3
36 = 6
81 = 9
144 = 12
If an integer is not a perfect square, its square root can only be approximated. For example, 2 is not a perfect square so its square root can only be approximated and is called
an irrational number.
√
2 ≈ 1.4142135 . . .
A radical expression is in simplest form when the radicand contains no factor greater than
1 that is a perfect square.
The Product Property of Square Roots
√
√ √
If a and b are positive real numbers, then ab = a · b.
The square root of a2√
√
For any real number a, a2 = |a|. If a ≥ 0, then a2 = a.
A variable or a product of variables written in exponential form is a perfect square when
each exponent is an even number.
1
Chapter 13 Section 2: Addition and Subtraction of Radical Expressions
We will use the distibutive property to simplify the sum or difference of radical expressions
with the same radicand.
√
√
√
√
5 2 + 3 2 = (5 + 3) 2 = 8 2
√
√
√
√
6 2x − 4 2x = (6 − 4) 2x = 2 2x
Chapter 13 Section 3: Multiplication and Division of
Radical Expressions
Product Property of Square Roots
√
√ √
a· b= a·b
√
The Square
of
√ 2 a
For a > 0, ( a) = a.
The Quotient Property of Square Roots
r
√
a
a
If a and b are positive real numbers, the
= √
b
b
Radical Expressions in Simplest Form
A radical expression is in simplest form if:
1. The radicand contains no factor greater than 1 that is a perfect square.
2. There is no fraction under the radical sign.
3. There is no radical in the denominator of the fraction.
Chapter 13 Section 4: Solving Equations Containing
Radical Expressions
An equation that contains a variable expression in a radicand is a radical equation.
Property of Squaring Both Sides of an Equation
If a and b are real numbers and a = b, then a2 = b2 .
2
Chapter 13 Section 5: Rational Exponents and Radical
Expressions
1
Definition of a n
1
√
If n is a positive integer, then a n = n a.
m
Definition of a n
√
1
m
If m and n are positive integers, and a n is a real number, then a n = n am .
Cube
Roots Fourth
Roots Fifth
Roots
√Square√Roots
√
√
√
3
4
5
1
=
1
36
=
6
1
=
1
1
=
1
1=1
√
√
√
√
√
3
4
5
4
=
2
49
=
7
8
=
2
16
=
2
32 = 2
√
√
√
√
√
3
4
5
27 = 3
81 = 3
243 = 3
√ 9 = 3 √64 = 8
√
√
3
4
16
=
4
81
=
9
64
=
4
256 = 4
√
√
√
√
3
4
25 = 5 100 = 10
125 = 5
625 = 5
Chapter 14 Section 1: Solving Quadratic Equations by
Factoring of Taking Square Roots
An equation that can be written in the form ax2 + bx + c = 0 where a 6= 0, is a quadratic
equation.
A quadratic equation is also called a second-degree equation.
To solve quadratic equations we will use the Principle of Zero Products “If a · b = 0 then
a = 0 or b = 0”
Principle of Taking√the Square Root of Each Side of an Equation
If x2 = a, then x = ± a.
Chapter 14 Section 2: Solving Quadratic Equations by
Completing the Square
A quadratic equation of the form x2 + bc = 0, x 6= 0, that cannot be solved by factoring can
be solved by completing the square. The procedure is:
1. Write the equation in the form x2 + bx + c = 0.
2
b
2. Add
to both sides of the equation x2 + bx = −c to get the equation x2 + bx +
2
2
2
b
b
= −c +
2
2
3
2
b
3. Factor the left hand side of the equation, so that the equation x + bx +
=
2
2
2
2
b
b
b
−c +
becomes x +
= −c +
2
2
2
2
4. Take the square root of each side of the equation.
5. Solve for x.
Chapter 14 Section 3: Solving Quadratic Equations by
Using the Quadratic Formula
If ax2 + bx + c = 0, a 6= 0, then
√
b2 − 4ac
2a
or
√
−b − b2 − 4ac
x=
2a
The quadratic formula is frequently written in the form
√
−b ± b2 − 4ac
x=
2a
x=
−b +
Below are some examples for us to try with solutions at the end.
1. Simplify.
√
−4 12a4 b7
2. Simplify.
√
√
7a 28ab2 − 7b 7a3
3. Simplify.
p
p
√
7x · 42x3 y · 6y 2
4. Divide.
√
7− 6
√
5−2 6
5. Simplify.
√
√
6a5 b4
294ab4
6. Solve the equation and check your solution.
√
√
2x + 12 = 6 − 2 2x
4
7. Simplify.
p
5
243x10 y 40
8. Simplify.
5
5
14x 2 y 4
5
−35xy 4
9. Solve by factoring.
8y 2 + 10y = 12
10. Solve by factoring.
t2 + 5t − 14 = 0
11. Solve the equation by taking the square root.
3 (x − 6)2 = 12
12. Solve the equation by completing the square.
x2 + 6x + 6 = 0
13. Solve the equation by using the Quadratic Formula
3x2 − 6x − 1 = 0
Solutions
Solution to 1:
√
√
−4 12a4 b7 = −4 4 · 3 · a4 · b6 · b
√
√
= −4 4a4 b6 3b
√
= −4(2)(a2 )(b3 ) 3b
√
= −8a2 b3 3b
Solution to 2:
√
√
√
√
7a 28ab2 − 7b 7a3 = 7a 7 · 4 · a · b2 − 7b 7 · a2 · a
√ √
√ √
= 7a 4b2 7a − 7b a2 7a
√
√
= 7a(2b) 7a − 7b(a) 7a
√
√
= 14ab 7a − 7ab 7a
√
= (14ab − 7ab) 7a
√
= 7ab 7a
5
Solution to 3:
p
p
p
p
7xy · 42x3 y · 6y 2 = (7xy)(42x3 y)(6y 2 )
p
= 42 · 42 · x4 · y 4
√ p
p
= (42)2 x4 y 4
= 42x2 y 2
Solution to 4:
√
√
√
7− 6
7− 6 5+2 6
√ =
√ ·
√
5−2 6
5−2 6 5+2 6
√
√
(7 − 6)(5 + 2 6)
√
√
=
(5 − 2 6)(5 + 2 6)
√
√
√
√
7(5) + 7(2 6) + (− 6)(5) + (− 6)(2 6)
√
√
√
√
=
5(5) + 5(2 6) + (−2 6)(5) + (−2 6)(2 6)
√
√
√
35 + 14 6 − 5 6 − 2( 6)2
√
√
√
=
25 + 10 6 − 10 6 − 4( 6)2
√
35 + 9 6 − 2(6)
=
25 − 4(6)
√
35 − 12 + 9 6
=
25 √
− 24
= 23 + 9 6
Solution to 5:
√
√
6a5 b4
294ab4
r
=
r
6a5 b4
294ab4
6a5−1
r 6 · 49
a4
=
√ 49
a4
=√
49
2
a
=
7
=
6
Solution to 6:
√
√
2x + 12 = 6 − 2x
√
2 √ 2
2x + 12 = 6 − 2x
√ √ 2x + 12 = 6 − 2x 6 − 2x
√
√
√
√
2x + 12 = 6(6) + 6(− 2x) + (− 2x)(6) + (− 2x)(− 2x)
√
√
√
2x + 12 = 36 − 6 2x − 6 2x + ( 2x)2
√
2x + 12 = 36 − 12 2x + 2x
√
12 = 36 − 12 2x
√
12 − 36 = −12 2x
√
−24 = −12 2x
−24 √
= 2x
−12 √
2 = 2x
√
22 = ( 2x)2
4 = 2x
2x
4
=
2
2
2=x
Solution to 7:
Note: 243 = 3 · 81 = 3 · 9 · 9 = 3 · 3 · 3 · 3 · 3 = 35
p
p
5
243x10 y 40 = 5 35 x10 y 40
√
10
40
5
= 35 · x 5 y 5
= 3x2 y 8
Solution to 8:
5
5
5
7 · 2x 2 −1
=
5
−7 · 5
−35xy 4
5
2
2x 2 − 2
=
−5
3
2x 2
=−
5
14x 2 y 4
Solution to 9:
8y 2 + 10y − 12 = 0
7
Note: a · c = 8 · (−12) = −96. We seek to answer the question, “What two numbers multiply
to give -96 AND add to give 10?” Note that 16·(−6) = −96 and 16+(−6) = 10. We will split
the middle term 10y into 16y+−6y and use the factor by grouping method or the AC method.
8y 2 + 10y − 12 = 0
8y 2 + 16y + −6y − 12 = 0
8y 2 + 16y + (−6y − 12) = 0
8y (y + 2) + −6 (y + 2) = 0
(y + 2) (8y − 6) = 0
Using the principle of zero products, we will solve the two equations
y + 2 = 0 8y − 6 = 0
3
6
Hence y = −2 and y = = .
8
4
Solution to 10:
To solve this equation, we answer the question “What two numbers multiply to give -14 and
add to give 5?” Note that 7 · −2 = −14 and 7 + (−2) = 5.
t2 + 5t − 14 = 0
(t + 7) (t − 2) = 0
Using the principle of zero products, we will solve the two equations
t+7=0 t−2=0
Hence, t = −7 and t = 2.
Solution to 11:
3 (x − 6)2 = 12
3 (x − 6)2
12
=
3
3
2
(x − 6) = 4
q
√
(x − 6)2 = ± 4
x − 6 = ±2
To find the two solutions, we solve the two equations
x − 6 = 2 x − 6 = −2
8
Hence the two solutions are x = 8, x = 4.
Solution to 12:
Please not that we cannot solve this equation by factoring. We cannot find two numbers
that multiply to give 6 and add to give 6. To complete the square, we first move the constant
term 6 to the right hand side of the equation.
x2 + 6x = −6
6
b
Next we identify b (the number in front of x), b = 6. Then we divide b in half = = 3,
2
2
2
b
= 32 = 9. We will add 9 to both sides of the equation.
and square b over 2
2
x2 + 6x + 9 = 9 − 6
x2 + 6x + 9 is a perfect square. Two numbers that multiply to give 9 and add to give 6 are
both 3. Hence, x2 + 6x + 9 = (x + 3)(x + 3).
x2 + 6x + 9 = 3
(x + 3) (x + 3) = 3
(x + 3)2 = 3
q
√
(x + 3)2 = ± 3
√
x+3=± 3
x = −3 ±
√
3
√
√
Hence, the two solutions are x = −3 + 3 and x = −3 − 3.
Solution to 13:
To solve the equation 3x2 − 6x − 1 = 0 we will use the quadratic formula with a = 3, b = −6,
and c = −1.
9
√
b2 − 4ac
2a p
−(−6) ± (−6)2 − 4(3)(−1)
=
2(3)
√
6 ± 36 + 12
=
√6
6 ± 48
=
6√
6 ± 16 · 3
=
6√
6±4 3
=
6√
6 4 3
= ±
6
√6
2 3
=1±
3
√
√
2 3
2 3
and x = 1 −
.
Hence, the two solutions are x = 1 +
3
3
x=
−b ±
10