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Destinyyyy
 one year ago
Need help with a few problems
Destinyyyy
 one year ago
Need help with a few problems

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Solve the polynomial equation by factoring and then using the zeroproduct principle. 36y^35=y180y^2

freckles
 one year ago
Best ResponseYou've already chosen the best response.3First step would be to use some algebra to get every nonzero term on one side.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3is 180 suppose to be 18 or was the 18 suppose to be 180?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[36y^3+180y^2y5=0\] we could try factoring by grouping

freckles
 one year ago
Best ResponseYou've already chosen the best response.3looking like at 36 and 180 what common factor(s) do you notice?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ok well yes y^3 and y^2 have common factor y^2 but what about 36 and 180?

freckles
 one year ago
Best ResponseYou've already chosen the best response.336 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
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.3so you should be able to write 36y^3+180y^2 as: \[36y^3+180y^2=36y^2(y+5)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3y5 can also be factored

freckles
 one year ago
Best ResponseYou've already chosen the best response.3do you think you can factor out a 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so... 36y^2(y+5) 1(y+5)=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ok Do you know how to factor something like: \[36y^2\cdot Y1 \cdot Y\] notice both terms have a common factor of ...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3notice the common factor in both terms is Y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y= 5, 1/6, 1/6 Now I have to check them

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[36y^2 \cdot Y1 \cdot Y \\ Y(36y^21) \\ \text{ try the same thing with } \\ 36y^2(y5)1(y5)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3oh you have gotten to the answer part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah.. Can you help with a few more???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Solve the radical equation then check. Square root x+24  square root x8 = 4

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\sqrt{x+24}\sqrt{x8}=4 \text{ is this right? }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Ok 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
 one year ago
Best ResponseYou've already chosen the best response.3So can get sqrt(x+24) by itself

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and then square the equation once you have it by itself

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Square root x+24 = 4 square root x8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then i put them in parenthesis and square them by 2

freckles
 one year ago
Best ResponseYou've already chosen the best response.3that minus should be a plus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh yeah I just saw that

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\sqrt{x+24}\sqrt{x8}=4 \\ \text{ Add } \sqrt{x8 } \text{ on both sides } \\ \sqrt{x+24}=4+\sqrt{x8} \\ \text{ then square both sides } \\ (\sqrt{x+24})^2=(4+\sqrt{x8})^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3now the left hand side is easy it is just x+24

freckles
 one year ago
Best ResponseYou've already chosen the best response.3right 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
 one year ago
Best ResponseYou've already chosen the best response.3\[(4+\sqrt{x8})^2 =(4+\sqrt{x8})(4+\sqrt{x8})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah after this I get a bit mixed up

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[(a+b)(a+b) \\ a(a+b)+b(a+b) \\ a^2+ab+ba+b^2 \\ a^2+2ab+b^2 \\ \text{ Comparing } (a+b)(a+b) \text{ \to } (4+\sqrt{x8})(4+\sqrt{x8}) \\ \text{ we see that } a=4 \text{ and } b=\sqrt{x8} \\ \text{ so you could plug in into the expanded form of } a^2+2ab+b^2 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3replace a with 4 replace b with sqrt(x8) \[(4)^2+2(4)(\sqrt{x8})+(\sqrt{x8})^2 \\ 16+8\sqrt{x8}+(x8)\] combine like term terms that is 168=?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3do you understand what I did to multiply the right hand side? please let me know if you are still confused about it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not really. I have 16+4 square root x8 +4 square root x8..... So you added the 4s to get 8.. But im confused by the end part

freckles
 one year ago
Best ResponseYou've already chosen the best response.3oh you are confused about combining like terms

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you know if you have 4y+4y this is 8y right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.34y+4y=4y(1+1)=4y(2)=4(2)y=8y

freckles
 one year ago
Best ResponseYou've already chosen the best response.3or you can just look at it as 4 apples plus 4 apples then you have 8 apples

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes.. Im confused on why square root x8 isnt x^2+64 .. Why did it stay the same?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know how to do distributive property.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you mean this: \[(\sqrt{x8})^2=x8\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nevermind. Whats the next step?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3i guess we can come back to it \[\sqrt{x+24}\sqrt{x8}=4 \\ \sqrt{x+24}=4+\sqrt{x8} \\ (\sqrt{x+24})^2=(4+\sqrt{x 8})^2 \\ x+24=(4)^2+2(4)\sqrt{x8}+(\sqrt{x8})^2 \\ x+24=16+8\sqrt{x8}+x8 \\ x+24=8+8 \sqrt{x8}+x\] 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
 one year ago
Best ResponseYou've already chosen the best response.3that is what we have to do here isolate the 8sqrt(x8) term

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so we need to subtract x and subtract 8 on both sides to do this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x+248x= 8 square root x8

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ok and we do have like terms on on the left

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so we can do xx and 248

anonymous
 one year ago
Best ResponseYou've already chosen the best response.016= 8 square root x8

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[16=8 \sqrt{x8}\] 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{16}{8}=\sqrt{x8} \\ 2=\sqrt{x8}\] square both sides

freckles
 one year ago
Best ResponseYou've already chosen the best response.3right and then you check solution

freckles
 one year ago
Best ResponseYou've already chosen the best response.3hey @Destinyyyy so do we need to talk about this part: \[(4+\sqrt{x8})^2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope I figured it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Next 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 4x8 =5

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ok 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
 one year ago
Best ResponseYou've already chosen the best response.3\[x\sqrt{4x8}=5\] again you want to isolate the square root term

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and then square both sides

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\sqrt{4x8}=5x\] like you can go ahead and square both sides here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh okay.. I was wondering about the negative

freckles
 one year ago
Best ResponseYou've already chosen the best response.3or \[\sqrt{4x8}=x5 \text{ and then \square both sides } \] pick your favorite

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[ \sqrt{4x8}=5x \\ \text{ the \square both sides } \\ (\sqrt{4x8})^2=(5x)^2 \\ (1 \sqrt{4x8})^2=(5x)^2 \\ (1)^2 (\sqrt{4x8})^2=(5x)^2 \\ (1) (4x8)=(5x)^2 \\ 4x8=(5x)^2 \] it looks like you might have expanded the right hand side a bit off

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[(5x)^2=(5x)(5x)=5(5x)x(5x)=255x5x+x^2 \\=2510x+x^2 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3your equation should look like this instead: \[4x8=2510x+x^2 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3now if you put all nonzero 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)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3right and this problem made it easy on us because x^214x+33 is factorable

freckles
 one year ago
Best ResponseYou've already chosen the best response.3which of those are actually a solution?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[x\sqrt{4x8}=5 \\ \text{ put in 11 }\\ 11\sqrt{4(11)8}=11\sqrt{448}=11\sqrt{36}=116=5 \text{ check mark } \\ \text{ now put \in 3} \\ 3 \sqrt{4(3)8}=3\sqrt{128}=3\sqrt{4}=32=1 \text{ no check mark }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the only one that gives you a true equation is x=11

freckles
 one year ago
Best ResponseYou've already chosen the best response.3since you end up with 5=5

freckles
 one year ago
Best ResponseYou've already chosen the best response.3x=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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. I didnt bring the negative over

freckles
 one year ago
Best ResponseYou've already chosen the best response.3any questions on those 3 we did together?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ok enjoy your fun math adventures
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