This is Similar to my first question (The one Melody helped me solve),

$$\\2x^2+8x-12=0\\\\2(x^2+4x-6)\\\\x^2-2x+6x-6\\$$

So I did this^ and then I changed the bottom problem to this. V

$$\\x(x-2)+6(x-1)\\\\(x+6)(2x-3)\\$$

I am wondering if this is correct, if not where did I make a mistake?

~Purpe

Purpe
Aug 16, 2015

#1**+10 **

Best Answer

$$\\

2x^2+8x-12=0\\\\

\text{Since each coefficient is a multiple of 2, divide each term by 2}\\\\

x^2+4x-6=0\\\\

\text{This does not factorise nicely so add and subtract 4:}\\\\

x^2+4x+4-4-6=0\\\\

(x+2)^2-10=0\\\\

(x+2)^2=10\\\\

x+2=\pm\sqrt{10}\\\\

x=-2\pm\sqrt{10}$$

Alan
Aug 16, 2015

#2**+5 **

Purpe wrote "*I am wondering if this is correct, if not where did I make a mistake?*"

Your last line doesn't follow from your next-to-last-line. All lines but your last are correct; however, they don't lead towards a solution, so are not useful here.

.

Alan
Aug 17, 2015

#3**+5 **

Hi Purpe,

Alan is right. Only expressions with rational roots can be done the 'factoring in pairs' way.

Yours is

$$\\2(x^2+4x-6)=0\\\\

x^2+4x-6=0\\\\$$

Now you have to look for 2 numbers that multiply to 1*-6=-6

Since they multiply to a negative, one is pos and the other is neg

They have to add to +4

So the 'bigger' (absolute value) one is positive

There are no 2 rational numbers that meet this requirement so you cannot do it this way!

There are a number of methods that you can use.

Alan has shown you one of them.

I would just solve it using the quadratic formula :))

Melody
Aug 17, 2015

#4**+5 **

Now to answer you real question.

Where did I go wrong!

I expect you know that

$$\\3a+4a=(3+4)a\\\\

and\\\

xa+6a = (x+6)a\\\\

$say a was not $a$ but just some expression.$\\\\

3(x+2)+4(x+2\\=3 \; lots\; of\; (x+2)+4 \; lots\; of\; (x+2)\\= (3+4)\; lots\; of\; (x+2) \\= 7(x+2)\\\\

x(x+2)+6(x+2)\\=x \; lots\; of\; (x+2)+6 \; lots\; of\; (x+2)\\= (x+6)\; lots\; of\; (x+2) \\= (x+6)(x+2)\\\\$$

Now, this is what you wrote:

$$\\x(x-2)+6(x-1)\\\\(x+6)(2x-3)\\$$

You cannot do this because you have x lots of **(x -2) **and 6 lots of

The brackets are not the same - they are unlike terms - you cannot add them like you have done :/

Melody
Aug 17, 2015