Download Math 1320, Section 10 Quiz IV Solutions 20 Points Please answer

Math 1320, Section 10
Quiz IV Solutions
20 Points
Please answer each question. To receive full credit you must show all work and give
answers in simplest form. Cell phones and graphing calculators are not allowed.
Name and Date:
1. (2 points) Solve the quadratic equation by factoring; credit will not be given if another
method is used.
x2 − 3x − 10 = 0
Solution: Start by factoring the left side
x2 − 3x − 10 = 0
⇒
(x − 5)(x + 2) = 0.
Since there is a zero on the right, we can use the zero factor property. Hence, x−5 = 0
or x + 2 = 0. Solving these equations for x we get
x=5
or
x = −2.
2. (2 points) Solve the equation by extracting square root; credit will not be given if another method is used. Make sure your answer is written in simplest form.
3x2 = 81.
Solution: This was homework problem 29 on page 117.
Dividing on both sides by 3 we have x2 = 27. Taking the square root on both
sides (i.e. extracting the root) we have
√
|x| = 27.
When dropping the absolute value bars we obtain the following equations
√
√
x = 27
or
x = − 27.
Now, we can simplify what’s under the root sign in both equations, because 9 is a
perfect square which divides 27. Then
√
√
√
√
√
√
x = 27 = 9 · 3 = 3 3
or
x = − 27 = − 9 · 3 = −3 3.
Hence, the final solutions are
√
x=3 3
or
Page 1 of 5
√
x = −3 3.
3. (4 points) Solve the quadratic equation by completing the square; credit will not be
given if another method is used.
x2 + 6x + 4 = 0.
Solution: The first step is to get the constant (in this case 4) to the right side.
Subtracting 4 from both sides we get
x2 + 6x = −4
(∗)
The second step is to make sure the coefficient of x2 is 1. In this case it is 1 and we
don’t need to do anything (if it were not 1, we would need to divide both sides of
the equation by whatever the coefficient happened to be). We now identify b = 6,
i.e. b is the coefficient of x. The next step is to divide b by 2. Doing this we have
6
b
= = 3.
2
2
Next, we square the number we just computed, so 32 = 9. Now take 9 and add it to
both sides of the equation in (∗). Then
x2 + 6x + 9 = −4 + 9
⇒
x2 + 6x + 9 = 5.
The nice thing about completing the square is that, if done properly, the left side
will always factor. Factoring the left side we have
(x + 3)2 = 5.
Now take the square root on both sides (i.e. extract the root) in the last equation
to get
√
|x + 3| = 5.
Solving the absolute value equation by breaking it in two pieces we have
√
√
or
x + 3 = − 5.
x+3= 5
By subtracting 3 in the two previous equations we have as our solutions
√
√
x = −3 + 5
or
x = −3 − 5.
If you prefer, you may also write the final solutions as
√
x = −3 ± 5.
Page 2 of 5
4. (3 points) Use the Quadratic Formula to solve the equation; credit will not be given if
another method is used.
2x2 + x − 1 = 0.
Solution: This was homework problem 73 on page 118. Recall that the Quadratic
Formula (QF) is
√
−b ± b2 − 4ac
.
x=
2a
The first thing to check, before using the formula, is that there is a zero on the
right side; in this case zero is on the right side, so we can go straight to identifying
a, b and c. Here we have a = 2, b = 1 and c = −1. Using parenthesis, we plug these
values into the QF to get
p
−(1) ± (1)2 − 4(2)(−1)
x=
2(2)
√
−1 ± 1 + 8
=
4
√
−1 ± 9
=
4
−1 ± 3
=
4
Splitting the above into two equations we have
x=
−1 + 3
4
or
x=
−1 − 3
.
4
We simplify the fractions above to obtain the final solutions
x=
1
2
or
x = −1.
5. (2 points) Subtract the complex numbers.
(6 − 4i) − (−11 + 7i).
Page 3 of 5
Solution: This problem is exactly like subtracting
(6 − 4x) − (−11 + 7x).
However, in this case the i’s are like terms instead of the x’s. So, begin by distributing
the negative sign to get
(6 − 4i) − (−11 + 7i) = 6 − 4i + 11 − 7i.
Combining the ’like terms’ we have
(6 − 4i) − (−11 + 7i) = 6 − 4i + 11 − 7i = 17 − 11i.
6. (3 points) Perform the multiplication.
(1 + i)(3 − 2i).
Solution: This problem is just like FOILing two binomials, the only difference being
that we will use the equation i2 = −1 when we simplify our final answer. So, from
FOILing we have
O
I
L
F
z}|{ z }| { z }| { z }| {
(1 + i)(3 − 2i) = 1 · 3 + (1)(−2i) + (3)(i) + (i)(−2i) .
Simplifying the previous equation we have
(1 + i)(3 − 2i) = 3 − 2i + 3i − 2i2
= 3 + i − 2i2 .
Plugging in −1 for i2 (because i2 = −1) into the last line in the above equation we
have
(1 + i)(3 − 2i) = 3 + i − 2(−1) = 3 + i + 2 = 5 + i.
Hence, the final answer is
(1 + i)(3 − 2i) = 5 + i.
7. (1 point) Write the complex conjugate of the complex number 17 + i.
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Solution: The complex conjugate of 17 + i is 17 − i.
8. (3 points) Write the quotient in standard form by multiplying the numerator and denominator by the complex conjugate of the denominator.
3
2+i
Solution: The conjugate of the denominator is 2 − i. Multiplying by 2 − i in the
numerator and denominator we get
3(2 − i)
(2 + i)(2 − i)
Distributing 3 in the numerator and FOILing the denominator we get
3(2 − i)
6 − 3i
=
.
(2 + i)(2 − i)
4 − 2i + 2i − i2
Canceling −2i & 2i and using i2 = −1 in the denominator we have
3(2 − i)
6 − 3i
=
(2 + i)(2 − i)
4 − 2i + 2i − i2
6 − 3i
=
4 − (−1)
6 − 3i
=
5
6 3
= − i
5 5
Hence, the standard form for the complex quotient
6 3
− i.
5 5
Page 5 of 5
3
is
2+i