Yet Another Question (Last one in this Set)

$$\\
2x^2+8x-12=0\\\\
\text{Since each coefficient is a multiple of 2, divide each term by 2}\\\\
x^2+4x-6=0\\\\
\text{This does not factorise nicely so add and subtract 4:}\\\\
x^2+4x+4-4-6=0\\\\
(x+2)^2-10=0\\\\
(x+2)^2=10\\\\
x+2=\pm\sqrt{10}\\\\
x=-2\pm\sqrt{10}$$

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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.

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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 :))

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 (x-1)  

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