Square of Difference: More than one factorization

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll apply this formula and expand the parentheses in the expressions in the equation:

$(x-1)^2=x^2 \\ x^2-2\cdot x\cdot1+1^2=x^2 \\ x^2-2x+1=x^2 \\$We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2-2x+1=x^2 \\ -2x=-1\hspace{8pt}\text{/}:(-2)\\ \boxed{x=\frac{1}{2}}$Therefore, the correct answer is answer A.

Let's solve the equation. First, we'll simplify the algebraic expressions using the square of binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$

We'll apply the mentioned formula and expand the parentheses in the expressions in the equation. On the right side, since we have parentheses with an exponent multiplier, we'll expand the (existing) parentheses using the square of binomial formula into additional parentheses (marked with an underline in the following calculation):

$x^2+(x-2)^2=2\underline{(x+1)^2} \\ x^2+x^2-2\cdot x\cdot2+2^2=2\underline{(x^2+2\cdot x\cdot1+1^2)} \\ x^2+x^2-4x+4=2(x^2+2x+1) \\ x^2+x^2-4x+4=2x^2+4x+2 \\$In the final stage, we expanded the parentheses on the right side using the distributive property,

We'll continue and combine like terms, by moving terms between sides. Then - we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2+x^2-4x+4=2x^2+4x+2 \\ -8x=-2\hspace{8pt}\text{/}:(-8)\\ \boxed{x=\frac{2}{8}=\frac{1}{4}}$

In the final stage, we simplified the fraction that was obtained as the solution for $x$.

Therefore, the correct answer is answer B.

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll apply this formula and expand the parentheses in the expressions in the equation:

$(x+3)^2=(x-3)^2 \\ x^2+2\cdot x\cdot3+3^2=x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9=x^2-6x+9 \\$We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2+6x+9=x^2-6x+9 \\ 12x=0\hspace{8pt}\text{/}:12\\ \boxed{x=0}$Therefore, the correct answer is answer A.