Solving quadratic equation

Question

From $X^2 - 2X(1-X) - (1-X)^2 = 7$

I get $X^2 - 2X + 2X^2 - 1 - 2X + X^2 - 7 = 0$

Which then becomes $4X^2 - 4X - 8 = 0$.

However, this should become $X^2 = 4$ how is this ?

Answer 1

Be very careful with signage legerdemain.   It is a major trip hazard when doing algebraic rearrangement by hand.   Sometimes it is worth taking the time to write things down in two steps; just to be sure you don't drop something when juggling symbols in your head.

From $X^2 - 2X(1-X) - (1-X)^2 = 7$

I get $X^2 - 2X + 2X^2 - 1 - 2X + X^2 - 7 = 0$

You should have $X^2 - (2X - 2X^2) - (1 - 2X + X^2) - 7 = 0$

Which leads to: $X^2-2X+2X^2-1+2X-X^2-7=0$

Which then becomes $4X^2 - 4X - 8 = 0$.

But with he above correction it is $2X^2-8=0$

However, this should become $X^2 = 4$ how is this ?

Like that.

Answer 2

We've:

$$\text{x}^2-2\text{x}\left(1-\text{x}\right)-\left(1-\text{x}\right)^2=7$$

And:

1. $$-2\text{x}\left(1-\text{x}\right)=-2x+2xx=2x^2-2x$$
2. $$-\left(1-\text{x}\right)^2=-\left(1-\text{x}\right)\left(1-\text{x}\right)=-1+x+x-xx=-x^2+2x-1$$

So:

$$\text{x}^2+2x^2-2x-x^2+2x-1=7$$

And:

$$\text{x}^2+2x^2-2x-x^2+2x-1=0+2x^2+0-1=2x^2-1=7$$

Answer 3

$$x^2-2x(1-x)-(1-x)^2=7$$

$$=> x^2-2x+2x^2-1+2x-x^2=7$$ $$=> 2x^2=8$$ $$=>x^2=4$$