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# Yet Another Question (Last one in this Set)

+5
533
4
+21

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

Aug 16, 2015

#1
+27342
+10

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

.
Aug 16, 2015

#1
+27342
+10

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

Alan Aug 16, 2015
#2
+27342
+5

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

Aug 17, 2015
#3
+95179
+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 :))

Aug 17, 2015
#4
+95179
+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 (x-1)

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

Aug 17, 2015