Question

i need quick help! (medal)

Answer 1

\[x^2+6x+9=1\] x=? and x=?

Answer 2

First step must be : Subtract 1 both sides.

Can you tell me what will you get by subtracting 1 on both sides?

Answer 3

x^2+6x+8=0

Answer 4

That's good :)
Now you can factor it

Answer 5

Okay... I'm not good at teaching how to factor...
Take that last term, for instance, 8

Give me a pair of factors of 8, that when added, give you 6 (the middle term)

Answer 6

2

Answer 7

4

Answer 8

So... that should hint you that the factorisation would be
(x+2)(x+4)= 0

try it, :)

Answer 9

Terenz is guiding well. Good to see :) ;)

Answer 10

i got x^2+4x+2x+8=0

Answer 11

Good.

Answer 12

Thanks, Mr. Ramanujan :)

@kittenbrand I didn't factor that expression just so you could bring it back into "unfactored form" :P

Go from
(x+2)(x+4) = 0

Answer 13

okay so what do i do then

Answer 14

Now, as you know, if you multiply two numbers, and their product is zero, then one or the other must be zero... in maths terms...
if
ab = 0
then either
a = 0
or
b = 0

Answer 15

i dont get this like i get law of sines so one of my x's is going to b a 0?

Answer 16

So, in this case,
(x+2)(x+4) = 0
is no exception, it's just a fancier setup...

so EITHER
x+2 = 0
OR
x+4 = 0

Please solve both equations... it should take less than a minute :P

Answer 17

x=-2 and x=-4

Answer 18

And those, are your answers :)
Nicely done :D

Answer 19

i got it wrong it said

Answer 20

Well, that can't be right...

Answer 21

this is completing the square not regular factoring

Answer 22

Well, no matter which method you choose, you should arrive at the same answer...
Okay, try completing the square then...

Answer 23

.... yeaaah i dont have any clue how to

Answer 24

I shall demonstrate completing the square, with a similar example...
\[\huge x^2 + \color{red}8x = 9\]

Answer 25

Now, see that number in red? That's the number we pay attention to.
It's the coefficient of the x WITHOUT ANY EXPONENT...
this x...(in blue)
\[\huge x^2 + \color{red}8\color{blue}x = 9\]

Answer 26

yes okay so what about it terenz?

Answer 27

Now, what we do is we take the number that's next to the x (with no exponent)
And we divide it by 2.
That gives us 4.

Answer 28

adn that isnt a perfect square trinomial though

Answer 29

And then we SQUARE that number.

Answer 30

okay

Answer 31

So, take half of 8, and then square it, what do you get?

Answer 32

so square 2?

Answer 33

No, take half of *8* and then square it.

Answer 34

oh squarre four

Answer 35

Yes, and what do you get?

Answer 36

16

Answer 37

That's right :)
So... we add 16
\[\huge x^2 + 8x\color{red}{+16} = 9\]
But this isn't really "fair" to the equation (LOL) so we also have to add 16 to the other side, to keep it the same equation.
\[\huge x^2 + 8x+16 = 9\color{red}{+16}\]

Answer 38

Catch me so far?

Answer 39

yeah im writing it down sorry okay now what

Answer 40

Well, for one thing, simplify...
\[\huge x^2 + 8x +16 = \color{blue}{25}\]

Answer 41

The point of adding 16 is that NOW we have a perfect square trinomial on the left side :)

Answer 42

ohhhh okay

Answer 43

so... the left is a perfect square trinomial... so express it as such...

Answer 44

.... ughm x^2+8x+16

Answer 45

That's what it is, but I want you to express it as the square of some binomial...

Answer 46

(x+4)^2

Answer 47

Yes. :)
\[\huge \color{blue}{(x+4)^2}= 25\]
And you can get the square root of both sides...
\[\huge x+4 = \pm 5\]

Answer 48

\[\huge x = -4 \pm 5\]
so
x = -9
or
x = 1

Answer 49

woah there whered you get the -9 and the1

Answer 50

Now, back to your first equation...
\[\huge x^2 + 6x + 9 = 1\]
Now, if you haven't noticed yet, the left side is ALREADY a perfect square trinomial... express it as such...

Answer 51

(x+3)^2=1

Answer 52

sorry...
\[\huge -4 \pm5\]
just means
\[\huge -4 + 5 \quad\quad\quad or \quad\quad\quad -4-5\]

Answer 53

oh

Answer 54

Okay, so you now have
\[\huge (x+3)^2=1\]Now get the square root of both sides...

Answer 55

\[xsqrt{3} =\sqrt{1} \]

Answer 56

what? lol....
\[\huge \sqrt{(x+3)^\color{red}{2}}= \color{red}?\]

Answer 57

oh god i screwed that up like majorly ok so now what

Answer 58

Well, what is it?
\[\huge \sqrt{(x+3)^\color{red}{2}}= \color{red}?\]

Answer 59

i dont know how to do that because the sqrt of 1 is 1

Answer 60

No, I mean the square root of (x+3)^2
Just so we're clear.

Answer 61

x+3

Answer 62

i gtg

Answer 63

help monday ?

Answer 64

aww, oh well... practice :)
Sure, if I'm there :D