Question

Need help with a few problems

Answer 1

Solve the polynomial equation by factoring and then using the zero-product principle.

36y^3-5=y-180y^2

Answer 2

@Nnesha

Answer 3

First step would be to use some algebra to get every non-zero term on one side.

Answer 4

36y^3 +18y^2-y-5=0

Answer 5

is 180 suppose to be 18
or was the 18 suppose to be 180?

Answer 6

180 my bad

Answer 7

\[36y^3+180y^2-y-5=0\]
we could try factoring by grouping

Answer 8

looking like at 36 and 180 what common factor(s) do you notice?

Answer 9

y^2

Answer 10

ok well yes y^3 and y^2 have common factor y^2
but what about 36 and 180?

Answer 11

Um 9

Answer 12

36 and 180 are even so they also have to at least have a common factor of 2
you should eventually see the common factor between 36 and 180 is 36
since 36=36(1) and 180=36(5)

Answer 13

so the greatest common factor between 36y^3 and 180y^2 is 36y^2
since 36y^3=36y^2*y
and 180y^2=36y^2*5

Answer 14

so you should be able to write 36y^3+180y^2 as:
\[36y^3+180y^2=36y^2(y+5)\]

Answer 15

-y-5 can also be factored

Answer 16

do you think you can factor out a -1?

Answer 17

Okay so...

36y^2(y+5) -1(y+5)=0

Answer 18

right

Answer 19

ok Do you know how to factor something like:
\[36y^2\cdot Y-1 \cdot Y\]
notice both terms have a common factor of ...

Answer 20

notice the common factor in both terms is Y

Answer 21

You can factor out Y

Answer 22

y= -5, -1/6, 1/6

Now I have to check them

Answer 23

\[36y^2 \cdot Y-1 \cdot Y \\ Y(36y^2-1) \\ \text{ try the same thing with } \\ 36y^2(y-5)-1(y-5)\]

Answer 24

Oops no checking

Answer 25

oh you have gotten to the answer part

Answer 26

Yeah.. Can you help with a few more???

Answer 27

I can try

Answer 28

Solve the radical equation then check.

Square root x+24 - square root x-8 = 4

Answer 29

\[\sqrt{x+24}-\sqrt{x-8}=4 \text{ is this right? }\]

Answer 30

Yes

Answer 31

Ok the way I solve these is:
Isolate a square root then square
(since you have two square roots you will have to do that twice)

Answer 32

So can get sqrt(x+24) by itself

Answer 33

and then square the equation once you have it by itself

Answer 34

Square root x+24 = 4- square root x-8

Answer 35

well one correction

Answer 36

Then i put them in parenthesis and square them by 2

Answer 37

that minus should be a plus

Answer 38

Ohh yeah I just saw that

Answer 39

\[\sqrt{x+24}-\sqrt{x-8}=4 \\ \text{ Add } \sqrt{x-8 } \text{ on both sides } \\ \sqrt{x+24}=4+\sqrt{x-8} \\ \text{ then square both sides } \\ (\sqrt{x+24})^2=(4+\sqrt{x-8})^2\]

Answer 40

now the left hand side is easy it is just x+24

Answer 41

right hand side you have to do a bit of multiplying there (distributive property twice)
and I know the street language is to use foil

Answer 42

\[(4+\sqrt{x-8})^2 =(4+\sqrt{x-8})(4+\sqrt{x-8})\]

Answer 43

Yeah after this I get a bit mixed up

Answer 44

\[(a+b)(a+b) \\ a(a+b)+b(a+b) \\ a^2+ab+ba+b^2 \\ a^2+2ab+b^2 \\ \text{ Comparing } (a+b)(a+b) \text{ \to } (4+\sqrt{x-8})(4+\sqrt{x-8}) \\ \text{ we see that } a=4 \text{ and } b=\sqrt{x-8} \\ \text{ so you could plug in into the expanded form of } a^2+2ab+b^2 \]

Answer 45

replace a with 4
replace b with sqrt(x-8)
\[(4)^2+2(4)(\sqrt{x-8})+(\sqrt{x-8})^2 \\ 16+8\sqrt{x-8}+(x-8)\]
combine like term terms that is 16-8=?

Answer 46

do you understand what I did to multiply the right hand side? please let me know if you are still confused about it

Answer 47

Not really. I have-

16+4 square root x-8 +4 square root x-8.....

So you added the 4s to get 8.. But im confused by the end part

Answer 48

oh you are confused about combining like terms

Answer 49

you know if you have 4y+4y this is 8y right?

Answer 50

4y+4y=4y(1+1)=4y(2)=4(2)y=8y

Answer 51

or you can just look at it as 4 apples plus 4 apples then you have 8 apples

Answer 52

Yes.. Im confused on why square root x-8 isnt x^2+64 .. Why did it stay the same?

Answer 53

I know how to do distributive property.

Answer 54

you mean this:
\[(\sqrt{x-8})^2=x-8\]?

Answer 55

Nevermind. Whats the next step?

Answer 56

i guess we can come back to it
\[\sqrt{x+24}-\sqrt{x-8}=4 \\ \sqrt{x+24}=4+\sqrt{x-8} \\ (\sqrt{x+24})^2=(4+\sqrt{x-
8})^2 \\ x+24=(4)^2+2(4)\sqrt{x-8}+(\sqrt{x-8})^2 \\ x+24=16+8\sqrt{x-8}+x-8 \\ x+24=8+8 \sqrt{x-8}+x\]

well remember how I said we would have to isolate the square root term and then square both sides
(we would have to do this twice)

Answer 57

that is what we have to do here isolate the 8sqrt(x-8) term

Answer 58

so we need to subtract x and subtract 8 on both sides to do this

Answer 59

Okay

Answer 60

x+24-8-x= 8 square root x-8

Answer 61

ok and we do have like terms on on the left

Answer 62

so we can do
x-x
and
24-8

Answer 63

16= 8 square root x-8

Answer 64

\[16=8 \sqrt{x-8}\]
if you want to you could divide both sides by 8 first then square
or you can square then divide both sides by 8^2

honestly you would working with smaller numbers if you go ahead and divide both sides by 8

Answer 65

\[\frac{16}{8}=\sqrt{x-8} \\ 2=\sqrt{x-8}\]
square both sides

Answer 66

Okay

Answer 67

4= x-8

Answer 68

right !

Answer 69

x=12

Answer 70

right and then you check solution

Answer 71

Its true

Answer 72

hey @Destinyyyy so do we need to talk about this part:
\[(4+\sqrt{x-8})^2 \]

Answer 73

Nope I figured it out

Answer 74

Next problem... I couldn't find any examples to help me I wasn't sure if Im suppose to minus the x to the other side or what..

x- square root 4x-8 =5

Answer 75

ok sometimes it is easier to look at it in a not so ugly way
well in my opinion it is easier to look at it like this:
\[(a+b)^2=(a+b)(a+b)=a(a+b)+b(a+b) \\ (a+b)^2=a^2+ab+ba+b^2 \\ (a+b)^2=a^2+2ab+b^2 \ \text{ and after doing the multiplication, plug \in the terms }\]
ok but anyways on to next problem

Answer 76

\[x-\sqrt{4x-8}=5\]
again you want to isolate the square root term

Answer 77

and then square both sides

Answer 78

you could subtract x on both sides to do this
but then you would have a - in front of the square root
which is fine
you can can square both sides without multiplying both sides by -1

Answer 79

\[-\sqrt{4x-8}=5-x\]
like you can go ahead and square both sides here

Answer 80

Ohh okay.. I was wondering about the negative

Answer 81

or
\[\sqrt{4x-8}=x-5 \text{ and then \square both sides } \]
pick your favorite

Answer 82

-4x-8= 25-10x+1

Answer 83

\[- \sqrt{4x-8}=5-x \\ \text{ the \square both sides } \\ (-\sqrt{4x-8})^2=(5-x)^2 \\ (-1 \sqrt{4x-8})^2=(5-x)^2 \\ (-1)^2 (\sqrt{4x-8})^2=(5-x)^2 \\ (1) (4x-8)=(5-x)^2 \\ 4x-8=(5-x)^2 \]
it looks like you might have expanded the right hand side a bit off

Answer 84

\[(5-x)^2=(5-x)(5-x)=5(5-x)-x(5-x)=25-5x-5x+x^2 \\=25-10x+x^2 \]

Answer 85

your equation should look like this instead:
\[4x-8=25-10x+x^2 \]

Answer 86

now if you put all non-zero terms on one side you will see you have a quadratic equation to solve (you might already see you have a quadratic before even doing that)

Answer 87

x^2 -14x+33=0

Answer 88

right and this problem made it easy on us
because x^2-14x+33 is factorable

Answer 89

x=11,3 now to check

Answer 90

which of those are actually a solution?

Answer 91

3

Answer 92

hmm...

Answer 93

are you sure?

Answer 94

\[x-\sqrt{4x-8}=5 \\ \text{ put in 11 }\\ 11-\sqrt{4(11)-8}=11-\sqrt{44-8}=11-\sqrt{36}=11-6=5 \text{ check mark } \\ \text{ now put \in 3} \\ 3 -\sqrt{4(3)-8}=3-\sqrt{12-8}=3-\sqrt{4}=3-2=1 \text{ no check mark }\]

Answer 95

the only one that gives you a true equation is x=11

Answer 96

since you end up with 5=5

Answer 97

x=3 is an extraneous solution (meaning not actually a solution but it is a solution to an equation we "reduced" our equation to)
since 1=5 isn't true

Answer 98

Yeah. I didnt bring the negative over