Destinyyyy
  • Destinyyyy
Need help with a few problems
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Destinyyyy
  • Destinyyyy
Solve the polynomial equation by factoring and then using the zero-product principle. 36y^3-5=y-180y^2
Destinyyyy
  • Destinyyyy
@Nnesha
freckles
  • freckles
First step would be to use some algebra to get every non-zero term on one side.

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Destinyyyy
  • Destinyyyy
36y^3 +18y^2-y-5=0
freckles
  • freckles
is 180 suppose to be 18 or was the 18 suppose to be 180?
Destinyyyy
  • Destinyyyy
180 my bad
freckles
  • freckles
\[36y^3+180y^2-y-5=0\] we could try factoring by grouping
freckles
  • freckles
looking like at 36 and 180 what common factor(s) do you notice?
Destinyyyy
  • Destinyyyy
y^2
freckles
  • freckles
ok well yes y^3 and y^2 have common factor y^2 but what about 36 and 180?
Destinyyyy
  • Destinyyyy
Um 9
freckles
  • 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
  • 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
  • freckles
so you should be able to write 36y^3+180y^2 as: \[36y^3+180y^2=36y^2(y+5)\]
freckles
  • freckles
-y-5 can also be factored
freckles
  • freckles
do you think you can factor out a -1?
Destinyyyy
  • Destinyyyy
Okay so... 36y^2(y+5) -1(y+5)=0
freckles
  • freckles
right
freckles
  • 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
  • freckles
notice the common factor in both terms is Y
freckles
  • freckles
You can factor out Y
Destinyyyy
  • Destinyyyy
y= -5, -1/6, 1/6 Now I have to check them
freckles
  • 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
  • Destinyyyy
Oops no checking
freckles
  • freckles
oh you have gotten to the answer part
Destinyyyy
  • Destinyyyy
Yeah.. Can you help with a few more???
freckles
  • freckles
I can try
Destinyyyy
  • Destinyyyy
Solve the radical equation then check. Square root x+24 - square root x-8 = 4
freckles
  • freckles
\[\sqrt{x+24}-\sqrt{x-8}=4 \text{ is this right? }\]
Destinyyyy
  • Destinyyyy
Yes
freckles
  • 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
  • freckles
So can get sqrt(x+24) by itself
freckles
  • freckles
and then square the equation once you have it by itself
Destinyyyy
  • Destinyyyy
Square root x+24 = 4- square root x-8
freckles
  • freckles
well one correction
Destinyyyy
  • Destinyyyy
Then i put them in parenthesis and square them by 2
freckles
  • freckles
that minus should be a plus
Destinyyyy
  • Destinyyyy
Ohh yeah I just saw that
freckles
  • 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
  • freckles
now the left hand side is easy it is just x+24
freckles
  • 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
  • freckles
\[(4+\sqrt{x-8})^2 =(4+\sqrt{x-8})(4+\sqrt{x-8})\]
Destinyyyy
  • Destinyyyy
Yeah after this I get a bit mixed up
freckles
  • 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
  • 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
  • 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
  • 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
  • freckles
oh you are confused about combining like terms
freckles
  • freckles
you know if you have 4y+4y this is 8y right?
freckles
  • freckles
4y+4y=4y(1+1)=4y(2)=4(2)y=8y
freckles
  • freckles
or you can just look at it as 4 apples plus 4 apples then you have 8 apples
Destinyyyy
  • Destinyyyy
Yes.. Im confused on why square root x-8 isnt x^2+64 .. Why did it stay the same?
Destinyyyy
  • Destinyyyy
I know how to do distributive property.
freckles
  • freckles
you mean this: \[(\sqrt{x-8})^2=x-8\]?
Destinyyyy
  • Destinyyyy
Nevermind. Whats the next step?
freckles
  • 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
  • freckles
that is what we have to do here isolate the 8sqrt(x-8) term
freckles
  • freckles
so we need to subtract x and subtract 8 on both sides to do this
Destinyyyy
  • Destinyyyy
Okay
Destinyyyy
  • Destinyyyy
x+24-8-x= 8 square root x-8
freckles
  • freckles
ok and we do have like terms on on the left
freckles
  • freckles
so we can do x-x and 24-8
Destinyyyy
  • Destinyyyy
16= 8 square root x-8
freckles
  • 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
  • freckles
\[\frac{16}{8}=\sqrt{x-8} \\ 2=\sqrt{x-8}\] square both sides
Destinyyyy
  • Destinyyyy
Okay
Destinyyyy
  • Destinyyyy
4= x-8
freckles
  • freckles
right !
Destinyyyy
  • Destinyyyy
x=12
freckles
  • freckles
right and then you check solution
Destinyyyy
  • Destinyyyy
Its true
freckles
  • freckles
hey @Destinyyyy so do we need to talk about this part: \[(4+\sqrt{x-8})^2 \]
Destinyyyy
  • Destinyyyy
Nope I figured it out
Destinyyyy
  • 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
  • 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
  • freckles
\[x-\sqrt{4x-8}=5\] again you want to isolate the square root term
freckles
  • freckles
and then square both sides
freckles
  • 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
  • freckles
\[-\sqrt{4x-8}=5-x\] like you can go ahead and square both sides here
Destinyyyy
  • Destinyyyy
Ohh okay.. I was wondering about the negative
freckles
  • freckles
or \[\sqrt{4x-8}=x-5 \text{ and then \square both sides } \] pick your favorite
Destinyyyy
  • Destinyyyy
-4x-8= 25-10x+1
freckles
  • 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
  • freckles
\[(5-x)^2=(5-x)(5-x)=5(5-x)-x(5-x)=25-5x-5x+x^2 \\=25-10x+x^2 \]
freckles
  • freckles
your equation should look like this instead: \[4x-8=25-10x+x^2 \]
freckles
  • 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
  • Destinyyyy
x^2 -14x+33=0
freckles
  • freckles
right and this problem made it easy on us because x^2-14x+33 is factorable
Destinyyyy
  • Destinyyyy
x=11,3 now to check
freckles
  • freckles
which of those are actually a solution?
Destinyyyy
  • Destinyyyy
3
freckles
  • freckles
hmm...
freckles
  • freckles
are you sure?
freckles
  • 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
  • freckles
the only one that gives you a true equation is x=11
freckles
  • freckles
since you end up with 5=5
freckles
  • 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
  • Destinyyyy
Yeah. I didnt bring the negative over
freckles
  • freckles
any questions on those 3 we did together?
Destinyyyy
  • Destinyyyy
Nope
freckles
  • freckles
ok enjoy your fun math adventures

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