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Mathematics 23 Online
OpenStudy (anonymous):

algebra 2 help?! please help mee. i'll fan & give a medal.

OpenStudy (anonymous):

Two Martians, Splott and Fizzle, invite you to pilot their giant robot. The heat generated by one leg mechanism is represented by the function f(x) = 3x2 + 20. The heat generated by one arm mechanism is represented by g(x) = 3x2 + 4x - 5. Describe to Splott and Fizzle, using complete sentences, which of the operations, addition, subtraction, multiplication, and division, will result in the largest degree function, and which operation will result in the smallest degree function.

OpenStudy (anonymous):

:)

OpenStudy (abdela25):

Alright Let h(x) be the result of any operation between f(x) and g(x) What is h(x) when you do f(x) + g(x)? What is the degree of h(x)?

OpenStudy (anonymous):

idk?:( thats why i need hekp.

OpenStudy (anonymous):

help*

OpenStudy (abdela25):

okay ill tell you for this one then ill explain later.. the degree would be 2

OpenStudy (anonymous):

so its h(2) ?

OpenStudy (abdela25):

What is the degree when you subtract the two functions? do you know how t do that one

OpenStudy (anonymous):

or 2 = h(2) ?

OpenStudy (abdela25):

*to not really we have to go through all of it to figure that out so not yet

OpenStudy (abdela25):

What is the degree when you subtract the two functions? the answer for this would be 1

OpenStudy (abdela25):

same for division

OpenStudy (anonymous):

okayyy. can you write it all out for me? i dont get it. so 1 equals what? lol im lost.

OpenStudy (abdela25):

we are finding the degrees using subtraction,division,addition etc to find them were subtracting or watever it is the 2 functions

OpenStudy (abdela25):

do you understand that or are yu still lost

OpenStudy (anonymous):

okay! i understand. keep going.

OpenStudy (abdela25):

the degree when you subtract the two functions = 1 the degree when you divide the 2 functions =1 the degree when you multiply the 2 functions = 4 the operation that gives you the highest degree is multiplication and the lowest would be subtraction and division

OpenStudy (anonymous):

YAYYY I LOVE YOUU SO MUCHH THANKSS!

OpenStudy (abdela25):

lol thank yaa and your welcome ;)

OpenStudy (abdela25):

any more help

OpenStudy (anonymous):

yes!

OpenStudy (anonymous):

Mr. Jones is asked to write two types of radical equations for the SAT Prep Course. The first solution to the radical equation must be extraneous. The second solution to the radical equation must be non-extraneous. Write one equation where the solution is extraneous. Then write a second equation where the solution is non-extraneous. Using complete sentences, explain each step when solving to justify your examples.

OpenStudy (anonymous):

@abdela25

OpenStudy (abdela25):

okay im working on it

OpenStudy (anonymous):

THANKS:)

OpenStudy (abdela25):

What is the first request of the problem statement? "The first solution to the radical equation must be extraneous." That is a quick example of a radical equation with an extraneous solution.

OpenStudy (abdela25):

Can you write a radical equation with both extraneous and non-extraneous solutions?

OpenStudy (abdela25):

@kendallnicole

OpenStudy (abdela25):

Is that this \[1=\sqrt{x/3-2}\]

OpenStudy (abdela25):

and i will check it for you right now too

OpenStudy (abdela25):

hello?

OpenStudy (anonymous):

yesss

OpenStudy (anonymous):

keep goingggg.

OpenStudy (abdela25):

okay

OpenStudy (abdela25):

\[3=\sqrt{x/3}\]

OpenStudy (abdela25):

9=x/3 x=27 checking... \[3=\sqrt{27/3}\]

OpenStudy (abdela25):

\[3=\sqrt{9}\]

OpenStudy (abdela25):

which means... 3=3 that is a non-extraneous solution ;)

OpenStudy (abdela25):

Anything like this will do... \[x=\sqrt{something}\] that leads to "Something" less than zero.

OpenStudy (abdela25):

@kendallnicole

OpenStudy (anonymous):

can you tell me an extrenous solution?

OpenStudy (abdela25):

thats impossible in a problem like this as you can see after all the wrok and double checking we only got a non-extrenous solution

OpenStudy (anonymous):

@abdela25

OpenStudy (abdela25):

yeah? wassup this is how the problem will go

OpenStudy (anonymous):

but like, just any other extreneous equation. because it says that i need one

OpenStudy (abdela25):

okay wait ill see

OpenStudy (anonymous):

thankksss

OpenStudy (abdela25):

If you changed it to +2 you would have a bad one. (extreneous) Anything like this will do... \[x=\sqrt{something}\] that leads to "Something" less than zero. or more simplified.... x< 0

OpenStudy (anonymous):

so like could you give me an example?

OpenStudy (abdela25):

okay

OpenStudy (abdela25):

hold on let me make on up

OpenStudy (abdela25):

you think if i get you a video that explains all of it would be good

OpenStudy (abdela25):

@kendallnicole

OpenStudy (anonymous):

its fine i just need an example!

OpenStudy (abdela25):

https://www.khanacademy.org/math/algebra/exponent-equations/radical_equations/v/extraneous-solutions-to-radical-equations okay cool watch this vvideo explains everything ! and gives more than 1 example :)

OpenStudy (abdela25):

The best way to understand what extraneous roots are is to go through a problem where you get one. The following is a solution to a problem that I did previously which dealt with extraneous roots: \[\sqrt{2y+7} +4=y\] First, get the radical expression alone on one side of the equation: \[\sqrt{2y+7} =y-4\] Now, square both sides of the equation to get rid of the radical: \[2y+7=(y-4)^{2}\] Simplify the right hand side: \[2y+7=y ^{2} -8y+16\] Combine like terms and equate to zero: \[y ^{2} -10y+9=0\] Factor the expression: \[(y-9)\times (y-1) =0\] Equate each factor to zero, and solve: \[y=9\] AND \[y=1\] Now, this part is important...you need to plug these answers back in to the original equation to be sure they are not "extraneous." An extraneous root may be mathematically correct, but it is not the true answer. If you plug y=1 back into the original equation, you will see that the equation DOES NOT hold true. Hence, this is an extraneous root. If you plug y=9 back into the original equation, you will see that the equation DOES hold true, so that one is your answer. @kendallnicole

OpenStudy (abdela25):

sorry its all long but math is long lol my head hurts now

OpenStudy (anonymous):

lol thank you.

OpenStudy (abdela25):

yep your welcome anytime love

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