+21 This is Similar to my first question (The one Melody helped me solve),
2x2+8x−12=02(x2+4x−6)x2−2x+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
+33654 2x2+8x−12=0Since each coefficient is a multiple of 2, divide each term by 2x2+4x−6=0This does not factorise nicely so add and subtract 4:x2+4x+4−4−6=0(x+2)2−10=0(x+2)2=10x+2=±√10x=−2±√10
+33654 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.
.
+118696 Hi Purpe,
Alan is right. Only expressions with rational roots can be done the 'factoring in pairs' way.
Yours is
2(x2+4x−6)=0x2+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 :))
+118696 Now to answer you real question.
Where did I go wrong!
I expect you know that
3a+4a=(3+4)aand xa+6a=(x+6)a$sayawasnot$a$butjustsomeexpression.$3(x+2)+4(x+2=3lotsof(x+2)+4lotsof(x+2)=(3+4)lotsof(x+2)=7(x+2)x(x+2)+6(x+2)=xlotsof(x+2)+6lotsof(x+2)=(x+6)lotsof(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 :/
