Need help with a few problems

- Destinyyyy

Need help with a few problems

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- Destinyyyy

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

- Destinyyyy

@Nnesha

- freckles

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

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## 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

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