Question

Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
x4 + 6x3 + 6x2 = 0 @zbay

Answer 1

Give us your thoughts, darling.

Answer 2

So for this one it looks like you need to find a value for x that will make the left side equal the right side.

Answer 3

yes i believe so

Answer 4

Because you have a 4th order polynomial you will have 4 answers. However keep in mind that you may have multiple roots at the same value.

Answer 5

So I would start by factoring out as much as you can and hopefully leaving you with a quadratic

Answer 6

I'm terrible at factoring.

Answer 7

Ok I will help you and hopefully you will see the process. When looking at the polynomial look at your lowest order term. This will hopefully allow you to identify the largest variable you can factor out.

Answer 8

would x^4 be my lowest order term ?

Answer 9

\[x^4+6x^3+6x^2=0\]

We look at the last term and notice its x^2, and that we can factor that out of the rest of the terms.

\[x^2(x^2+6x+6)=0\]

Answer 10

okay i understand that part now

Answer 11

X^4 would be your largest or highest order term. After factoring out x^2 you can then set both to zero

\[x^2=0 \]

\[x^2+6x+6=0\]

at this point you can solve for x and this will give you the solutions for the problem

Answer 12

I am guessing that you will need to use the quadratic equation for the 2nd term.

Answer 13

i dont know what the second term is

Answer 14

sorry I ment the 2nd equation \[x^2+6x+6=0\]

Answer 15

are you talking about using the quadratic formula ?

Answer 16

for the second equation

Answer 17

Yes I would use the quadratic formula to solve for x on that equation. Or you can complete the square it looks like

Answer 18

yeah ill use the quad. formula because i dont really remember how to do the completing the square

Answer 19

I always tend to use the quadratic formula also

Answer 20

remember your equation \[x^2=0\] will have 2 answers

Answer 21

ok im working on it now

Answer 22

ok great let me know if you have any problems

Answer 23

my final answer is -6 (square root) 6 is that correct

Answer 24

?

Answer 25

I'm getting x=-3-sqrt(3), and x=sqrt(3)-3

Answer 26

Let me work it out for myself really quick

Answer 27

I am going to upload a picture of my work and hopefully it will allow you to find your mistake. I'm guessing you forgot to divide by 2a

Answer 28

okay i see what i did wrong , your right i didnt divide the whole equation by two

Answer 29

awesome, did you have any other questions for me tonight?

Answer 30

1. Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
x9 − 81x5 = 0

2.Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)

( square root) 6 − x
+ 1 = x − 3

Answer 31

well this one is a little bit more tricky but we can apply the same idea

Answer 32

1. x^9 -81x^5=0 **
2.(sqaure root) 6-x + 1= x-3 **

Answer 33

\[x^9-81x^5=0\]

factor out x^5

\[x^5(x^4-81)=0\]

Answer 34

So at this point we know we have 5 of the possible solutions with the term x^5=0 leaving us with 4 remaining solutions to find.

\[x^4-81=0\]

Answer 35

is the answer is 9 ??

Answer 36

So now what do we get

\[(x^2+9)(x^2-9)=0\]

Answer 37

and we can factor out the term \[x^2-9=0\]

to
\[(x-3)(x+3)=0\]

Answer 38

So now we just have to find the values of x for \[x^2+9=0\]

I believe this will have an imaginary answer.

Answer 39

imaginary number means it'll be something like 3i ?

Answer 40

exactly

Answer 41

So the values for x are

x=0,0,0,0,0,-3,3

then the imaginary roots

Answer 42

so there are gonna be more than 3 answers ? what are all the zeros from ?

Answer 43

Yes to find out how many answers there will be for a given polynomial look at the highest power in the polynomial.

Answer 44

so for this one we have x^9 meaning there are nine given values of x that will satisfy the polynomial. Remember though some can repeat themselves for example when we factored out x^5=0 we know that the solution x=0 is repeated 5 times.

Answer 45

(sqaure root) 6-x + 1= x-3

does it equal

\[\sqrt{6-x+1}=x-3\]

Answer 46

the sqaure root is just over 6-x

Answer 47

\[\sqrt{6-x}+1=x-3\]

Answer 48

right

Answer 49

Alright, move the 1 to the right hand side (RHS) isolating the radical on the left hand side (LHS). Square both sides eliminating the radical on the LHS giving you the term

\[(x-4)^2\] that you will need to foil out. at this point in time you can move all terms to one side or the other and solve using the quadratic equation.

Answer 50

im having trouble with that one , but i do have two more questions for you

Answer 51

you there ?

Answer 52

yes im still here

Answer 53

1. Investments: if ben invest $4000 at 4% intrest per year how much additional money must he invest at 5 1/2 % annual interest to ensure that the intrest he recieves each year is 4 1/2 % ?

Answer 54

2. Dimensions of a lot : A half acre bilding lot is five times as long as it is wide. What are its dimensions? ( 1 acre= 43,560ft^2)

Answer 55

what interest formula have you been working with?

Answer 56

I'm assuming the simple interest formula?

Answer 57

yes

Answer 58

\[I=PRT\]

Answer 59

yeahh

Answer 60

1. Investments: if ben invest $4000 at 4% intrest per year how much additional money must he invest at 5 1/2 % annual interest to ensure that the intrest he recieves each year is 4 1/2 % ? ignore this just moving it down so I don't have to keep scorlling up

Answer 61

okay

Answer 62

Ok give me a min I'm working it out

Answer 63

okay thankss

Answer 64

Ok one down I will upload my work and you can look it over and ask any questions

Answer 65

alright :)

Answer 66

Dimensions of a lot : A half acre bilding lot is five times as long as it is wide. What are its dimensions? ( 1 acre= 43,560ft^2)

Answer 67

your such a life saver. when i did it i was just dividing all types of numbers in the wrong spots .

Answer 68

Yea the interest one throw me off for a min also

Answer 69

I will be back in a few min if you have any questions

Answer 70

okay thanks soo much

Answer 71

not a problem do you understand how I worked out those last two?

Answer 72

yes i see now , im more of a visual learner, if its just explained to me without working it out , i will not get it what so ever

Answer 73

Ok great would you like for me to work out the problem with the square root in it? Didn't you say you were having problems with that one?

Answer 74

\[\sqrt{6-x}+1=x-3\]

\[\sqrt{6-x}=x-4\]

\[(\sqrt{6-x})^2=(x-4)^2\]
foil RHS, radical is eliminated
\[6-x=x^2-8x+16\]

\[0=x^2-7x+10\]

\[0=(x-2)(x-5)\]

\[x=2,5\]

Answer 75

ehh im a little iffy about this problem, im confused but i dont really know what im confused out

Answer 76

about*

Answer 77

Well the only real trick to this problem is that radical, and whenever you square a radical it goes away leaving whats under it. After that part just group terms and factor with one of the methods you know.

Answer 78

So are you good to go?