Question

Nakim simplified 3 times the square root of 2x plus x times the square root of 8x minus 5 times the square root of 18x and got -10x times the square root of 2x for an answer.

I have to find what Nakim did wrong and then simplify the expression. @phi

Answer 1

\[3 \sqrt{2x}+ x \sqrt{8x}- 5\sqrt{18x}\]

Answer 2

he got \[-10x \sqrt{2x}\]

Answer 3

the first step is pull out any perfect squares from out of the square roots
can you do that ?

Answer 4

sqrt 2x

Answer 5

?

you can't simplify the first term, but you can simplify the 2nd and 3rd terms
what do you get ?

Answer 6

?
2nd term: sqrt 8x^2
I don't know what to do for the third.
Is that what you mean?

Answer 7

\[3 \sqrt2x + x \sqrt8x−5 \sqrt18x\] you can see that you have the eqaution so you can take out the second term by pulling out the perfect squares like this
\[x \sqrt 8x\] turns into \[2x \sqrt2x\]

Answer 8

can you do the third term?

Answer 9

for the 2nd term you start with
\[ x \sqrt{8x} \]
break up the 8 into its prime factors. in this case 8= 2*2*2
\[ x \sqrt{2\cdot 2\cdot 2\cdot x} \]
now "pull out" a pair of 2's, and put one of them out front
\[ 2x \sqrt{ 2 x} \]

Answer 10

Ohhh, I forgot about that. Okay.
Second term is 2x sqrt 2x third is -45x sqrt 9x?

Answer 11

how did you do the last term?

First, notice that there is no x "out front" so how did an x show up there ?
also, -5 turned into -45 which means you multiplied by 9. why ?

you should do this step by step:
factor the 18 into 2*3*3
you now have
\[ -5\sqrt{2 \cdot 3\cdot 3\cdot x} \]
now you can "pull out" *pairs* .... there is only one pair

Answer 12

3 and 3, yes.

Answer 13

so that means you pull out the pair of 3's (and only the 3's) and put one of them "out front"

Answer 14

I see what I did wrong :P
So, -15 sqrt 6x ?

Answer 15

almost, you pull out a "pair"
\[-5\sqrt{2 \cdot 3\cdot 3\cdot x} = -5\cdot 3\sqrt{2 \cdot \cancel{3\cdot 3}\cdot x}\]

Answer 16

what do you have so far with all 3 terms ?

Answer 17

Oh, so you don't keep one of the 3's behind at all?

Answer 18

This is what I have so far http://www.sketchtoy.com/47619421

Answer 19

looks good so far.

as for how square roots work, you know
\[ \sqrt{4} =2 \]
but you can write that as
\[ \sqrt{2\cdot 2} = 2 \]
as a way to remember that you pull out a pair

Answer 20

now factor out the \(\sqrt{2x} \) from each term

Answer 21

Pull out the pairs?

Answer 22

factoring is different, if you have (1A +xA +yA)
for example, you can "factor out" the A to get (1+x+y)*A
notice if you "distribute the A" you get: (A+Ax+Ay)
so factoring is the opposite of distributing

in your problem, factor out the \( \sqrt{2x}\)

Answer 23

Make it all multiplied by 2?

Answer 24

you have
\[ 3 \sqrt{2x} +2x\sqrt{2x}-15\sqrt{2x}\]
factor out \(\sqrt{2x}\) from each term. that means "erase it" from each term, put parens around everything, and put the \(\sqrt{2x} \)on the outside of the parens

Answer 25

so multiplied by sqrt 2x ?

Answer 26

try it. what do you get ?

Answer 27

3+2x-15 * sqrt2x

Answer 28

yes, except put the first 3 terms in parens (because the sqrt2x is multiplying all of them)
(3+2x-15 )* sqrt2x

now you have in the parens 3+2x+ -15 or (changing the order of adding -- which is allowed) 3+ -15 +2x

can you simplify that ?

Answer 29

-10x *sqrt 2

Answer 30

that looks dubious, as Nakim got the same answer, and we know he is wrong

Answer 31

you can combine *like terms*

so you can add and subtract the pure numbers, but the 2x is not a pure number. Leave the 2x alone

Answer 32

2x-12 *sqrt2

Answer 33

remember the parens
and what happen to sqrt(2x) (not just sqrt2)

Answer 34

(2x-12)*sqrt2x

Answer 35

yes. that could be the answer. you could also factor out a 2 from both terms
can you do that ?

Answer 36

No, I don't remember how.

Answer 37

divide both terms by 2 and simplify
then put a 2 outside the parens
can you do that ?

Answer 38

http://www.sketchtoy.com/47621827

Answer 39

almost. First, you should not use x for multiply in algebra because it looks like the variable x. people either leave out the multiply sign or if you need it for clarity, use a dot \( \cdot \)

meanwhile, you divided each term by 2, but you also have to put a 2 out front of the parens

Answer 40

Can you show me? I'm a little confused

Answer 41

example: (2x+4y)z
factor out a 2 from each term: divide each term in the parens by 2
(x +2y)2z
and put the 2 outside the parens

Answer 42

Like this? http://www.sketchtoy.com/47622643

Answer 43

@ash2326

Answer 44

@amistre64

Answer 45

Click the sketchtoy link http://www.sketchtoy.com/47622643

Answer 46

Cool
you see the second last step
\[(2x-12)\times \sqrt {2x}\]
You take 2 out of 2x-12
\[2(x-6)\times \sqrt{2x}\]
Do you notice any diff here and the sketchy last line?

Answer 47

That's all?

Answer 48

yes, you missed a 2 in the square root. otherwise it's right

Answer 49

Okay, can you help me with where Nakim messed up?

Answer 50

that's very easy, if you think that there is an x out in all the terms
then you'll get Nakim's answer.
Hint lies in the fact that we have
\[2x-12\]
now if we have x along with 12, you'd get -10x, of course the square root of (2x) is there

Answer 51

I don't understand...

Answer 52

check this out
http://www.sketchtoy.com/47622643

Answer 53

oops thats yours

Answer 54

ok I'll type here
\[3 \sqrt 2x+x\sqrt{8x} - 5 \sqrt{18x}\]
He'd have thought that x is out side every term
\[3\underline x\sqrt{2x}+x\sqrt{8x}-5\underline x\sqrt{18x}\]
\[3x\sqrt{2x}+2x\sqrt{8x}-15x\sqrt{2x}\]
\[(3x+2x-15x)\sqrt{2x}\]
\[-10x\sqrt{2x}\]
Do you get this?

Answer 55

Yes, Thanks!

Answer 56

Are you sure? and you understood how it has to be done correctly?