Question

Hello!
I need help with the following question please ( I will post as a comment so that I can use the equation thing)

Answer 1

\[(\sqrt{2x}-y)(\sqrt{2x}+y)-(4x-y)^{2}\]

Answer 2

The question asks me to simplify

Answer 3

But I dont get the right answer when I try it. I get \[-14x ^{2}+8xy\]

Answer 4

show your work ?

Answer 5

So when I expand the first one I get\[2x ^{2} + \sqrt{2xy} -\sqrt{2xy}+y ^{2} -(4x -y)^{2}\]

Answer 6

Which I then simplify to\[2x ^{2} +y ^{2} - 16x ^{2}-4xy-4xy-y ^{2}\]

Answer 7

the first term shouldn't be 2x^2

Answer 8

\(\sqrt a \times \sqrt a = a\)

\(\sqrt {2x} \times \sqrt {2x} = ??\)

Answer 9

Oh! Should it be 4x^2 ?

Answer 10

its square root

Answer 11

No no thats not right..

Answer 12

\( 2x \times 2x = 4x^2 \)

Answer 13

Ok

Answer 14

\(\sqrt {2x} \times \sqrt {2x} = ??\)

Answer 15

\(\sqrt ☻ \times \sqrt ☻ = ☻\)

Answer 16

I still get 2x^2

Answer 17

how did you get the exponent of x as 2 ?

Answer 18

oooh. I thought since x multiplied by x.. but its part of the square root right? so the exponent would still be one?

Answer 19

\(\sqrt a \times \sqrt a = \sqrt {a^2 } = a\)

Answer 20

yes

Answer 21

Hooray! Was that my only mistake?

Answer 22

\( (+y) \times (-y) = ... ?\)

Answer 23

= -y^2

Answer 24

i see you wrote it as +y^2

Answer 25

Oh, so they are both negative? the two y^2 ?

Answer 26

no,
\(-(4x-y)^2\)
lets work on this separately

Answer 27

Ok

Answer 28

your attempt was
\(-16x^2 −4xy−4xy−y^2\)

did you distribute the negative sign to ALL the expanded terms of \((4x-y)^2\) ?

Answer 29

oh and yes, both the y^2 's are negatives.

Answer 30

Ok, No I didnt distribute, I thought it could be left alone which I realize is not at all correct

Answer 31

yes.
to avoid confusion, i sugges you retain the brackets while expanding...anything

Answer 32

I see, thank you

Answer 33

\(-(a-b)^2 = -(a^2 -2ab+b^2 ) = -a^2 +2ab - b^2 \)

Answer 34

So let me try it with the distribution..

Answer 35

sure, let me know your final answer, i think this time you'll get it :)

Answer 36

\[-16x ^{2} + 8xy -y ^{2}\]

Answer 37

there were 2 y^2's which were negative, right?

-y^2 - y^2 = ...?

also add the first term, 2x in that

Answer 38

\(\sqrt {2x} \times \sqrt {2x} = 2x\)

Answer 39

Yes, I got positive y^2 and then outside negative sign made it negative

Answer 40

ok good,
so we now have
\(2x -y^2 -16x^2 +8xy -y^2\)
right?

just one simplification, and we're done

Answer 41

-16x^2 + 8xy +2x -2y^2

Answer 42

correct!

Answer 43

Thank you very much!

Answer 44

welcome ^_^