Question

Quick tutorial on difference of squares. Let's start by multiplying \(a + b\) and \(a - b\)

\( \color{Black}{\Rightarrow (a + b)(a - b) }\)
\( \color{Black}{\Rightarrow a(a - b) + b(a - b) }\)
\( \color{Black}{\Rightarrow a^2 - ab + ab - b^2 }\)
\( \color{Black}{\Rightarrow a^2 - b^2 }\)

Do you see that? well this is just a difference of squares. Similar for \(p^2 - q^2\) and \(x^2 - y^2\) etc. etc.

We can conclude that \((a + b)(a - b) = a^2 - b^2\)

More questions accepted below.

Answer 1

Questions?

Answer 2

If we have a quadratic, we may use the zero product rule.

Answer 3

If you guys need more examples:

\( \color{Black}{\Rightarrow 9x^2 - 4 }\)

\( \color{Black}{\Rightarrow 9x^2 - 2^2 }\)

\( \color{Black}{\Rightarrow (3x)^2 - 2^2 }\)

\( \color{Black}{\Rightarrow (3x + 2)(3x - 2) }\)

Answer 4

Love me some squares

Answer 5

What if we have \(16x^2 - 36 = 0 \)? We conclude this: \((4x - 6)(4x + 6) = 0\) and then use the product rule. :)

Answer 6

I'll do another small tutorial on the product rule soon anyway.

Answer 7

Can you write the as a difference of squares:

\[-x^4-3x^2+2x-3\]
hehe :)

Answer 8

Factor by grouping I think ^

Answer 9

That's right :)

Answer 10

Is that a 2 or a 3?

Answer 11

You should be able to write it like this:
\[x^2+2x+1-(x^4+4x^2+4)\]

Answer 12

And now factor :)

Answer 13

I cheated though
I worked backwards just to make something hard for you

Answer 14

\( \color{Black}{\Rightarrow x^2 + 2x + 1 - (x^4 + 2x^2 + 2x^2 + 4) }\)

\( \color{Black}{\Rightarrow x^2 + 2x + 1 - (x^2(x^2 + 2) + 2(x^2 + 2)) }\)

Answer 15

Ah

Answer 16

\( \color{Black}{\Rightarrow (x + 1)^2 - (x^2 + 2)^2 }\)

Answer 17

Yep yep :)

Answer 18

And then apply that...

Answer 19

how about writing \[36x^2 - 252\] as a difference of two squares? i cant get how to do it :/

Answer 20

\[36x ^{2}-36*7=(6x )^{2}-(6\sqrt{7})^{2}\]
\[(6x-6\sqrt{7})(6x+6\sqrt{7})\]
i think

Answer 21

Good tutorial @ParthKohli

Answer 22

how about factoring \(x^4+4\)?

Answer 23

\[\color{red}{x^4+4+4x^2-4x^2}=\color{blue}{(x^2+2)^2-4x^2.}\]\[=\color{purple}{(x^2+2)^2-(2x)^2.}\]
Now it is the\[\color{green}{a^2-b^2.}\]form, as u r seeing it:)
now factorize^_^

Answer 24

I know satellite already know this but it is for parthkohli; if & only if he don't this thing:)

Answer 25

lol how does sat dominate my tutorials :P

Answer 26

@TheViper woah sat gave you a medal

Answer 27

what about

\[\sqrt[3]{x^{10}} - \sqrt[5]{y^6}\]

Answer 28

You just take roots again. ;_;

Answer 29

lol I don't know this advanced stuff. Though I know how to express these in exponents. I doubt that will help.

Answer 30

They ain't perfect squares though :/

Answer 31

ummm I wonder....

Answer 32

its not that advanced...

Answer 33

It isn't. It's just not pretty. :P

Answer 34

\[(\sqrt[3]{x^5}-\sqrt[5]{y^3})(\sqrt[3]{x^5} + \sqrt[5]{y^3})\]

Answer 35

\( \color{Black}{\Rightarrow \Large x^{10 \over 3} - y^{6 \over 5} }\)

Answer 36

\[\lim_{x \rightarrow 9} \frac{ 3 - \sqrt{x}}{x - 9}\]

was a question asked yesterday

Answer 37

Ahhh NO I DON'T KNOW LIMITS!

Answer 38

Maybe I do yes... We'll try it with 8.9 then 8.99 then 8.999

Answer 39

but you know how to factorise or expand... you don't need to now limits

Answer 40

He's saying you can use this tutorial on that problem.

Answer 41

thanks limitless

Answer 42

Yeah. It'd look something like \(\Large x^3 \times x^{1 \over 3} - {y^{1 \over 5} \times y}\)

Answer 43

no... I posted the answer above...

Answer 44

You're welcome, Campbell.

Answer 45

Oh yep. Got it :D

Answer 46

lol how couldn't I do that *facepalm*

Answer 47

my point is parth... there are lots of equations reducible to quadratics.... provided you know what you are looking at

Answer 48

Hmmm

Answer 49

and the limit question is another example...

Answer 50

Shall I close the tutorial now?

Answer 51

lol... its you tutorial.... I'm just making comment

Answer 52

and the limit quest\[\lim_{x \rightarrow 9}\frac{-(\sqrt{x} -3)}{(\sqrt{x} -3)(\sqrt{x}+3)}..===> \lim_{x \rightarrow 9}\frac{-1}{\sqrt{x}+3}\]ion