simply. assume all variables represent positive real numbers (sqrt = quare root): 1.) sqrt. 96x^2 y 2.) ^4 sqrt. 16a^4 3.) ^3 sqrt. 64x^4 y^5
please help, i am taking an exam & dont understand it :(
i dont understand what you are supposed to find out.
like whats it on? on my exam it just says to simplify
its 3 different problems
\[\sqrt{96x ^{2}y}=\sqrt{16*6x ^{2}y}=4x \sqrt{6y}\]
\[\sqrt[4]{16a ^{4}}=\sqrt[4]{(2a)^{4}} = 2a\]
\[\sqrt[3]{64 x ^{4}y ^{5}}=\sqrt[3]{(4xy)^{3}xy ^{2}}=4xy \sqrt[3]{xy ^{2}}\]
THANK YOU SO MUCH !
can you help me out please with a couple more problems?
no, I can't do your problems for you. I have shown you how to do these problems. You can do the rest on your own if you apply the same logic. if you have doubts, feel free to ask and I will clarify.
aw okay , thanks though. but for adding and subtracting them with others would they need the same sqrt inside ?
give me an example.
[2 sqrt. 128] + [5 sqrt. 32] + [2 sqrt.200]
sqrt3 is a number. So if you had 3sqrt 3 +5sqrt3 sqrt 4, you can factor out sqrt3 as if it was a number so it would become sqrt 3 (3+5sqrt4)
you mean [2 sqrt.200] as sqrt.3 ?
no , I was giving an example of my own. In your problem you can express sqrt 200 as sqrt 8 sqrt 25. you can express sqrt 32 as sqrt4 sqrt 8. you can express sqrt 128 as sqrt 8 sqrt 16
how do you know how to express it ?
you see what is the highest common factor between 200, 32 and 128 and factor it out.
all 3 have to have the same in common ?
which would be 8 ?
yup.
so answer would be 9 sqrt.8 ?
the numbers add up to 9
nope.
2 + 5 + 2 = 9 ?
it does
so if we have the same sqrt. already than just add the numbers which give 9
yes, but 2 sqrt 200 = 2 sqrt 5 sqrt 8 = 2*5 sqrt 8= 10sqrt8
sorry 2*sqrt 200 = 2*sqrt 25*sqrt 8 = 2*5*sqrt8 = 10sqrt8
and simplfying that would give [2 sqrt. 2] ?
half of each other
yes, if you want to simply it further, sqrt 8 = 2 sqrt2
and to simplfy that is 1 ?
simplify what is 1?
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