How do you solve #-2x ^ { 2} = 2x - 4#?

2 Answers
Oct 6, 2017

Answer:

#x=1,-2#

Explanation:

rearrange to

#ax^2+bx+c=0#

#-2x^2=2x-4#

#=>2x^2+2x-4=0#

divide by 2

#x^2+x-2=0#

factorise

#(x+2)(x-1)=0#

giving

#x+2=0=>x=-2#

#x-1=0=>x=1#

Oct 6, 2017

Answer:

#x=-2 or x=1#

Explanation:

To solve this, you first need to put everything on one side:

#-2x^2=2x-4#
#2x^2+2x-4=0#

From here, we can factor #2# out:

#2(x^2+x-2)=0#

Dividing by two, we get:

#x^2+x-2=0#

This can then be factorised by finding two numbers that multiply to get #-2# and add to get #1#. Looking at the factors of #2#, you can find these numbers are #-1# and #2#.

#(x-1)(x+2)=0#

Now, you have two things multiplied together that equal #0#. This means that one of them has to equal #0#. To find the values of #x# for this to happen, we split the two parts:

#x-1=0#
#therefore x=1#

OR

#x+2=0#
#therefore x=-2#

So, the solution to this equation is #x=-2 or x=1#