Need help with a few problems

- Destinyyyy

Need help with a few problems

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Destinyyyy

Solve the polynomial equation by factoring and then using the zero-product principle.
36y^3-5=y-180y^2

- Destinyyyy

- freckles

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Destinyyyy

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

- freckles

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

- Destinyyyy

180 my bad

- freckles

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

- freckles

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

- Destinyyyy

y^2

- freckles

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

- Destinyyyy

Um 9

- freckles

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)

- freckles

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

- freckles

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

- freckles

-y-5 can also be factored

- freckles

do you think you can factor out a -1?

- Destinyyyy

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

- freckles

right

- freckles

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

- freckles

notice the common factor in both terms is Y

- freckles

You can factor out Y

- Destinyyyy

y= -5, -1/6, 1/6
Now I have to check them

- freckles

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

- Destinyyyy

Oops no checking

- freckles

oh you have gotten to the answer part

- Destinyyyy

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

- freckles

I can try

- Destinyyyy

Solve the radical equation then check.
Square root x+24 - square root x-8 = 4

- freckles

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

- Destinyyyy

Yes

- freckles

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)

- freckles

So can get sqrt(x+24) by itself

- freckles

and then square the equation once you have it by itself

- Destinyyyy

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

- freckles

well one correction

- Destinyyyy

Then i put them in parenthesis and square them by 2

- freckles

that minus should be a plus

- Destinyyyy

Ohh yeah I just saw that

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- Destinyyyy

Yeah after this I get a bit mixed up

- freckles

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

- freckles

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=?

- freckles

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

- Destinyyyy

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

- freckles

oh you are confused about combining like terms

- freckles

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

- freckles

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

- freckles

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

- Destinyyyy

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

- Destinyyyy

I know how to do distributive property.

- freckles

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

- Destinyyyy

Nevermind. Whats the next step?

- freckles

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)

- freckles

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

- freckles

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

- Destinyyyy

Okay

- Destinyyyy

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

- freckles

ok and we do have like terms on on the left

- freckles

so we can do
x-x
and
24-8

- Destinyyyy

16= 8 square root x-8

- freckles

\[16=8 \sqrt{x-8}\]
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

- freckles

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

- Destinyyyy

Okay

- Destinyyyy

4= x-8

- freckles

right !

- Destinyyyy

x=12

- freckles

right and then you check solution

- Destinyyyy

Its true

- freckles

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

- Destinyyyy

Nope I figured it out

- Destinyyyy

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

- freckles

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

- freckles

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

- freckles

and then square both sides

- freckles

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

- freckles

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

- Destinyyyy

Ohh okay.. I was wondering about the negative

- freckles

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

- Destinyyyy

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

- freckles

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

- freckles

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

- freckles

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

- freckles

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)

- Destinyyyy

x^2 -14x+33=0

- freckles

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

- Destinyyyy

x=11,3 now to check

- freckles

which of those are actually a solution?

- Destinyyyy

3

- freckles

hmm...

- freckles

are you sure?

- freckles

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

- freckles

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

- freckles

since you end up with 5=5

- freckles

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

- Destinyyyy

Yeah. I didnt bring the negative over

- freckles

any questions on those 3 we did together?

- Destinyyyy

Nope

- freckles

ok
enjoy your fun math adventures

Looking for something else?

Not the answer you are looking for? Search for more explanations.