$ (x+1)^2+x(x−1)=2(x−1)^2 $

Project ID: 
3000020501
Question: 

Four students, Michal, Bob, Pavel, and Marek, solved the equation: $$ (x+1)^2+x(x−1)=2(x−1)^2 $$ Which of them proceeded correctly in simplifying the equation?

Michal: $$\begin{aligned} x^2+1+x^2-x&=2(x^2-1) \cr 2x^2-x+1&=2x^2-2 \end{aligned}$$

Bob: $$\begin{aligned} x^2+2x+1+x^2-x&=(2x-2)^2 \cr 2x^2+x+1&=4x^2-8x+4 \end{aligned}$$

Pavel: $$\begin{aligned} x^2+2x+1+x^2-1x&=2(x^2-2x+1) \cr 2x^2+x+1&=2x^2-4x+2 \end{aligned}$$

Marek: $$\begin{aligned} x^2+x+1+x^2−x&=2(x^2-x+1) \cr 2x^2+1&=2x^2-2x+2 \end{aligned}$$

Answer 1: 

Pavel

Answer 2: 

Bob

Answer 3: 

Marek

Answer 4: 

Michal

Answer 5: 

None of them

Fixed Answer: 
Last Fixed
Correct Answer: 
Answer 1
Hint: 

Pavel correctly squared $(x+1)^2=x^2+2x+1$, $(x−1)^2=x^2-2x+1$. Unlike Bob, he realized that exponentiation has higher precedence than multiplication, so he squared $(x-1)$ first and only then used the distributive property to remove the parentheses. Michal and Marek made an error when squaring $(x+1)$ and $(x-1)$.