Question

Really confused about applying powers outside brackets. Sometimes it works :) can you expand it out or do you do whats in the brackets first then apply the power to that result ? does (y+3)^2 = y^2+3^2

Answer 1

Oh good question c:\[\Large (y+3)^2\quad\color{red}{\ne}\quad y^2+3^2\]

\[\Large (y+3)^2 \quad=\quad (y+3)(y+3)\]
Remember how to multiply out the brackets? :o

Answer 2

About your comment: `Sometimes it works`.

It will always work if the inside of our brackets contains no addition/subtraction.

Answer 3

So if you had something like:\[\Large (3y)^2\]You could safely apply the square to each item in the brackets without any worry (since there is a single term, no addition subtraction).\[\Large =\quad 3^2y^2\quad=\quad 9y^2\]

Answer 4

thanks so this can't be right because it has an addition \[(\sqrt{4} * \sqrt{n}- \sqrt{3})^2 = 4n -3\] thats just getting it wrong, right ? :)

Answer 5

Correct :o there should be some more stuff coming out of the square.\[\Large (\sqrt4\sqrt n-\sqrt3)^2\quad=\quad 4n\color{teal}{-2\left(\sqrt4\sqrt n \sqrt3\right)}+3\]

Answer 6

Try to think of the subtraction sign NOT AS subtraction, but as a negative attached to the sqrt3.

See how your last term gave you -3?
It should be +3.

Because the last portion should give us:\[\Large \left(-\sqrt3\right)^2\quad=\quad 3\]

Answer 7

\[\Large \left[\sqrt{4n}-\sqrt3\right]^2\quad=\quad \left[\sqrt{4n}+\left(-\sqrt3\right)\right]^2\]

Answer 8

Thankyou, Im going to work on your answer for a bit before I get it, I understand how you get =4n but then it gets mind blowing !! but ill work on it thanks.

Answer 9

haha XD

Answer 10

thx got it !!!