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nikklg [1K]
3 years ago
7

A polynomial function upper P left parenthesis x right parenthesis with rational coefficients has the given roots. Find two addi

tional roots of upper P left parenthesis x right parenthesis equals 0. i and 7 + 8i
Mathematics
1 answer:
mylen [45]3 years ago
5 0

Given that a polynomial function P(x) has rational coefficients.

Two roots are already given which are i and 7+8i,

Now we have to find two additional roots of P(x)=0

Given roots i and 7+8i are complex roots and we know that complex roots always occur in conjugate pairs so that means conjugate of given roots will also be the roots.

conjugate of a+bi is given by a-bi

So using that logic, conjugate of i is i

also conjugate of 7+8i is 7-8i

Hence final answer for the remaining roots are (-i) and (7-8i).

You might be interested in
MATH WORD PROBLEM
Sloan [31]

cost of one regular admission ticket = $ 10

cost of one senior citizen ticket = $ 5

<h3><u>Solution:</u></h3>

Let "r" be the cost of one regular admission ticket

Let "s" be the cost of one senior citizen ticket

Given that,

<em><u>On day 1 you sold 30 Regular Admission tickets and  20 Senior Citizen tickets for a total of $400</u></em>

So we can frame a equation as:

30 Regular Admission tickets x cost of one regular admission ticket + 20 Senior Citizen tickets x cost of one senior citizen ticket = $ 400

30 \times r + 20 \times s = 400

30r + 20s = 400 ----- eqn 1

<em><u>Day two you sell 40 Regular Admission tickets and only 10 Senior  Citizen tickets for a total of $450</u></em>

So we can frame a equation as:

40 Regular Admission tickets x cost of one regular admission ticket + 10 Senior Citizen tickets x cost of one senior citizen ticket = $ 450

40 \times r + 10 \times s = 450

40r + 10s = 450 ---- eqn 2

<em><u>Let us solve eqn 1 and eqn 2 to find values of "r" and "s"</u></em>

Multiply eqn 2 by 2

80r + 20s = 900 --- eqn 3

Subtract eqn 1 from eqn 3

80r + 20s = 900

30r + 20s = 400

(-) ---------------------

50r = 500

<h3>r = 10</h3>

Substitute r = 10 in eqn 1

30r + 20s = 400

30(10) + 20s = 400

300 + 20s = 400

20s = 100

<h3>s = 5</h3>

<em><u>Thus we have:</u></em>

cost of one regular admission ticket = $ 10

cost of one senior citizen ticket = $ 5

4 0
4 years ago
Its awpicture please help if you can
ASHA 777 [7]
Someone help me... please
4 0
3 years ago
15(15b−7)=3b−9 plz help meee
Romashka-Z-Leto [24]

Answer:

B=16over37

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

15(15b−7)=3b−9

(15)(15b)+(15)(−7)=3b+−9(Distribute)

225b+−105=3b+−9

225b−105=3b−9

Step 2: Subtract 3b from both sides.

225b−105−3b=3b−9−3b

222b−105=−9

Step 3: Add 105 to both sides.

222b−105+105=−9+105

222b=96

Step 4: Divide both sides by 222.

222b

222

=

96

222

Then simplify

4 0
3 years ago
Read 2 more answers
The hypotenuse of a right triangle is 52 in. One leg of the triangle is 8 in. More than twice the lenght of the other. What is t
shusha [124]

Call the short leg a.  The Pythagorean Theorem says:

a^2 + (8 + 2 a)^2 = 52^2

a^2 + 8^2 + 32 a + 4a^2 = 52^2

5a^2 + 32 a + 8^2-52^2 = 0

5a^2 + 32 a + (8-52)(8+52) = 0

5a^2 + 32 a - (44)(60) = 0

(5a + 132)(a - 20) = 0

Only the positive root a=20 is relevant here.

The other side is 2a+8=48

The perimeter is 20 + 48 + 52 = 120 inches



3 0
3 years ago
Fiona’s Number:
Amanda [17]

Answer:

He should dial 289-743-1560

Step-by-step explanation:

The first three digits:  "My first three digits are the square of an integer less than twenty"

Step 1:   We can get the first three by looking at the squares of the integers from 1 to 19.  We need a three digit number, so starting with 10, since 10² = 100, we list the squares of the integers from 10 to 19

10² = 100

11² = 121

12² = 144

13² = 169

14² = 196

15² = 225

16² = 256

17² = 289

18² = 324

19² = 361

Step 2:  "The first three digits are in ascending order" and no digits can repeat, so we eliminate the answers above that don't satisfy both of these condintions.  This leaves just 3 choices:

13² = 169

16² = 256

17² = 289

*We will come back to this once we have more information.

The last 2 numbers: "the last two digits are multiples of sixty".  

The only 2 digit number that is a multiple of 60 is 60, so 6 and 0 are the last 2 digits.  This rules out two answers from Step 1 since no digits can repeat.  Only 289 works since 169 and 256 each have a 6 in them.  So now we know the first 3, and last 2 numbers!

                                     2 8 9 _ _ _ _ _ 6 0

The last 4 numbers: "the last 4 digits are multiples of sixty"

We want to find a 4 four digit number that has 60 as the last 2 digits.  Notice a pattern when you multiply 60 by certain digits...

60 x 1 = 60

60 x 2 = 120

...

60 x 6 = 360

60 x 11 = 660

60 x 16 = 960     (these all don't work since they are not 4 digits)

60 x 21 = 1260     (this is the first 4 digit result, but it doesn't work because

                              there is a '2' in it, and 2 has already been used as the

                               first digit of the number)

60 x 26 = 1560     (this number could work, so lets try it and see if we can

                               satisfy the rest of the statements)

Now our number is

  2 8 9 _ _ _ 1 5 6 0

The second three digits:  "The second three digits are in descending order"  and "The sum of the second three digits is 14"

If we assume that 1, 5, 6 and 0 are the last four digits, we know that 2, 8 and 9 are the first three digits, this leaves 3, 4 and 7 as the three remaining digits.  

3 + 4 + 7 = 14     so that condition is satisfied.   We just need to write then in descending order, so our final number is

289-743-1560

3 0
3 years ago
Read 2 more answers
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