Question

Simplify the expression
x radical 64x^3 + 2x radical x + 8 radical 2x

Answer 1

@asnaseer could ya help

Answer 2

ok - first try to simplify each term of this expression

Answer 3

lets start with the first term: \(x\sqrt{64x^3}\)

do you know what the square root of 64 is?

Answer 4

8!

Answer 5

correct.

now when you have an even power inside the square root then you know from one of your previous questions that its square root is just half of the power - remember?

Answer 6

e.g. \(\sqrt{m^{16}}=m^8\)

Answer 7

Yes. But this is odd . 3

Answer 8

yup - so we can make use of a trick here...

Answer 9

we know we can write \(x^3=x^2\times x\)

agreed?

Answer 10

Yes!!

Answer 11

therefore: \(\sqrt{x^3}=\sqrt{x^2\times x}=\sqrt{x^2}\times\sqrt{x}=?\)

can you work this out?

Answer 12

How do you multiply radical x by radical x ^2 ....?

Answer 13

lets use some numbers to illustrate this...

Answer 14

\[\sqrt{8}=\sqrt{2^3}=\sqrt{2^2\times2}=\sqrt{2^2}\times\sqrt{2}=2\sqrt{2}\]because: \(\sqrt{2^2}=2\)

Answer 15

make sense?

Answer 16

Yes

Answer 17

good, so can you work this one out?
\[\sqrt{x^3}=\sqrt{x^2\times x}=\sqrt{x^2}\times\sqrt{x}=?\]

Answer 18

So just x radical x ?

Answer 19

perfect!

Answer 20

so, going back to the first term, we get:
\(x\sqrt{64x^3}=x\times8\sqrt{x^3}=x\times8x\sqrt{x}=8x^2\sqrt{x}\)

agreed?

Answer 21

Yessir!

Answer 22

so now your original question was to simplify:\[x\sqrt{64x^3} + 2x\sqrt{x} + 8\sqrt{2x}\]is this correct?

Answer 23

Tis correct.

Answer 24

so using our last result we get:\[x\sqrt{64x^3} + 2x\sqrt{x} + 8\sqrt{2x}=8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2x}\]

Answer 25

now notice that \(\sqrt{2x}=\sqrt{2}\times{\sqrt{x}}\), so we can write this as:\[x\sqrt{64x^3} + 2x\sqrt{x} + 8\sqrt{2x}=8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2x}\]\[\qquad=8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2}\sqrt{x}\]make sense so far?

Answer 26

Yes.

Answer 27

good.

now can you see any common factors between ALL the terms?

Answer 28

TWO,! :D

Answer 29

yes - that is ONE of the factors.

can you spot another one?

Answer 30

4 or 8...? Idk

Answer 31

can you see that every term has \(\sqrt{x}\) in it?

Answer 32

Yes oh so like 2x?

Answer 33

almost - the common factor in every term is \(2\sqrt{x}\)

Answer 34

so we can take this common factor out to get:\[x\sqrt{64x^3} + 2x\sqrt{x} + 8\sqrt{2x}=8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2x}\]\[\qquad=8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2}\sqrt{x}\]\[\qquad=2\sqrt{x}(4x^2+x+4\sqrt{2})\]

Answer 35

So that is the simplified expression?

Answer 36

yes - I cannot see any further simplification for it

Answer 37

In standard form?

Answer 38

what have you been taught in terms of "standard form"?

Answer 39

my /guess/ would be that you have been taught that this is the "standard form":\[\qquad=8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2}\sqrt{x}\]

Answer 40

in which case you can stop at this step. otherwise you can go one step further and factorise as we did above.

Answer 41

Well we want the expression simplified and then written in standarn

Answer 42

Standard form *

Answer 43

you write the answer as:\[8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2}\sqrt{x}\]and then add a sentence saying that this can be further factorised to get:\[2\sqrt{x}(4x^2+x+4\sqrt{2})\]

that way you have covered all bases. :)

Answer 44

*you could write...

Answer 45

Well the question asks me to state the leading coefficient and idk if it is

Answer 46

2 radical x or 8x^2 ... @asnaseer

Answer 47

coefficients are usually just numbers, so it would be 8 in this case

Answer 48

\[8x^2\sqrt{x} + 2x\sqrt{x} + 8\sqrt{2}\sqrt{x}\]

Answer 49

hope that made sense?

Answer 50

the leading term here is: \(8x^2\sqrt{x}\)
the leading coefficient here is: 8