write as a single radical, then simplify: square root, of the square root of 243(x+1)
\[\sqrt{\sqrt{243(x+1)}}\]
Weird that that just posted here. Was meant to be on the last question.
\[\sqrt{\sqrt{243(x+1)}}\]
nice job!
thanks(:
\[\sqrt{\sqrt{b}}=\sqrt[4]{b}\] does that help?
uhm..not really. i dont know what to do. or where to start
what i wrote is where to start. put \[\sqrt[4]{243(x+1)}\]
then you next job is to factor 243 and see if it has any fourth powers in it. if so you can pull them out
if you are a calculating savant you may recognize \[243=3^5\]but my guess is you do not, so you have to factor. i would cheat
i love my calculator. lol.. and okay im sorta understanding more now
Nice work, Satellite.
awesome.! haha
Meg, just try to get comfortable with exponent rules.
(Because roots are really just exponents)
im just learning the ones past 3... :/ dont really know them
i am fairly sure that your calculator will not factor, so acquaint yourself with wolfram or else do it by hand
ohh alright. isee.
Like for this one, you can rewrite it as (243(x+1)) ^ (1/2) ^(1/2) Then you can know that with exponents, if you raise an exponent to another exponent you multiply them. So this gives: (243(x+1))^(1/4) And with exponents, if it's a numerator it's a power, but in the denominator it's a root. So that makes it: \[\sqrt[4]{243(x+1)}\]
now you have \[\sqrt[4]{3^5(x+1)}\] and having an exponent in the radicand larger than the index is taboo, so you think \[3^5=3^4\times 3\] and \[\sqrt[4]{3^4}=3\] (you do this part in your head) giving \[3\sqrt[4]{3(x+1)}\] as your answer
really mean you do that in your head. if you see \[\sqrt[3]{x^7}\] you go right to \[x^2\sqrt[3]{x}\]
here is the thinking: "x goes in to 7 twice, with a remainder of 1" "x^2 comes outside the radical, x^1 =x stays inside"
ahhh. i totally get that! thanks so much.!!
really? great! just to make sure though... what is \[\sqrt{50x^5y^3}\]?
(hint, the index here is 2, even though it is not written)
what do u mean index?
btw i meant here is the thinking: "3 goes in to 7 twice, with a remainder of 1" "x^2 comes outside the radical, x^1 =x stays inside"
in the expression \[\sqrt[4]{abc}\] 4 is the "index" and abc is the "radicand"
\[\sqrt[3]{x}\] reads "cubed root of x" index is 3
alright. my teacher is terrrible at explainng this
\[\sqrt{x}\] reads "square root of x" index is 2 and by convention we do not write it
yeah most just say "simplify" which is meaningless unless you know the specific rules for "simplest radical form"
if you have time read this. the rules are here
the answer... is it \[2\sqrt{25x^5y^3}\] ??
no you have a few problems here. i will talk it out. \[50=25\times 2=5^2\times 2\] so a 5 comes out and a 2 stays in 2 (the index) goes in to 5 2 times with a remainder of 1 so x^2 comes out and x stays in 2 goes in to 3 1 time with a remainder of 1, so y^1 = y comes out and y stays in you get \[\sqrt{50x^5y^3}=5x^2y\sqrt{2xy}\]
if you want we can do another example from your homework
oh my gosh thanks. and okay.. ill give u one more.. hang on
\[\sqrt[4]{562^4}\]
oooh that is like asking who is buried in grants tomb
what is the fourth root of something raised to the power of 4?
\[\sqrt[4]{x^4}=?\]
x...
ohh so is the answer just 562 ?
yup!
next...
yay! you are really good at explaining things. im so glad you helped me. (:
100^-3/2 kinda simple....
yw. be happy to do another one if you think it would help
can u help with 100^3/2
yes. first off do you know what \[x^{\frac{3}{2}}\] means?
uhmm not exactly.
the exponent is a fraction. it has a numerator of 3 and a denominator of 2 the numerator is the power, and the denominator is the root
oh okay....
so this means a) cube 100, then take the square root b) take the square root of 100, then cube second way is easiest
what is the square root of 100?
10
yes. now what is \[10^3\]?
1000
exactly and so \[100^{\frac{3}{2}}=\sqrt{100^3}=10^3=1000\] and you are done
good job
okay well i have some others like that but theyre just negative.. is there a difference? on how to do them that is.. like 32^-3/5
oh yes they are different.
Negative exponents move to the denominator and become positive. So 2^(-1) is the same as 1/2
the numerator is the power the denominator is the root and the minus sign means take the reciprocal
so just switch them around...?
so if you were going to compute \[32^{-\frac{3}{5}}\] i would first compute \[32^{\frac{3}{5}}\] and the flip the result
so lets go slow
ohh okay so same process, just flip the answer.? and oka
first lets compute \[32^{\frac{3}{5}}\]
the numerator is 2, so second power. the denominator is 5 so 5th root. i prefer method b) from above. what is the 5th root of 32 (make a guess, it will probably be right)
2 so i worked it sand got 8 as the answer...
and *
great! now we have \[32^{\frac{3}{5}}=2^3=8\] right. so now we take care of the minus sign. take the reciprocal of 8 and you are done
by which i mean just flip it. answer...
so its... -1/8 ?
close but you were thinking too hard. just flip it. don't make it negative
oh alright. haha. (:
when you raise to a power if the base is positive you NEVER get a negative answer. your doing great
got another?
thank you again. (: and no im all good now that you explained it really good. i may need help again some other day. so i may ask u then but goodnight satellite..!
g'night
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