Question

What are the solutions of the equation 2x2 = 2?

Answer 1

I'm afraid the question seems somewhat invalid.

Answer 2

In a a way yes, hold on.

Answer 3

Just the solutions of 2x^2 =2 would be fine.

Answer 4

On simplification, isn't it x^2=1 that is, x^2-1=0?

Answer 5

So, there are no solutions?

Answer 6

You asked a similar question... x^2-x+x-1=0
that is x(x-1)+(x-1)=0
that is (x+1)(x-1)=0

Answer 7

2x * 2 = 2
-- --
2 2

2x =0
- -
2 2
x =0

Answer 8

not 100% but pretty sure hehe

Answer 9

Michaelc96. I solved it. ^

Answer 10

that's wrong nevermind

Answer 11

My solution, you mean to say?

Answer 12

no my solution is wrong

2x * 2 = 2
-- --
2 2

2x =1
- -
2 2
x =.5

but I think that's still wrong

Answer 13

Yes. Quite. Care to look over and comprehend my solution?

Answer 14

*I shall go to sleep in the meantime, by the way*

Answer 15

umm I don't understand it when you come back tell why my work is wrong

Answer 16

Alright. So what we have here, is 2x^2=2
I divide both sides of my equation by 2. That leaves me x^2=1
Now, I subtract 1 from both sides of my equation. That gives me, x^2-1=1-1
Which is x^2-1=0
Now in both the equations I add +x-x. The expression is 0, which is how I represent in the Right side of the eqution, but on the left side, it is seen as
x^2+x-x-1=0
that is, x(x+1)-(x+1)=0
that is, x(x+1)+(-1)(x+1)=0
Now, on the left side of the equation, I take x+1 common for both x, and -1
That gives me, (x-1)(x+1)=0

This is the equation factorised.

Post this, I want a solution. Since I have left side is equal to zero, either of the two expressions, x-1=0 or x+1=0.
First, I take x-1=0
I add 1 to both sides, leaves me x=1

Now, for x+1=0
I subtract 1 from both sides, that gives me
x=-1

So, my solution is that x is either 1 or -1.

I hope it is clear. =)