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.

To solve the equation $(x-4)^2 = (x+2)(x-1)$, follow these detailed steps:

• Step 1: Expand the left side of the equation using the square of a binomial formula: $(x-4)^2 = x^2 - 8x + 16$.
• Step 2: Expand the right side using the distributive property: $(x+2)(x-1) = x(x-1) + 2(x-1) = x^2 - x + 2x - 2 = x^2 + x - 2$.
• Step 3: Set the expanded forms equal to each other: $x^2 - 8x + 16 = x^2 + x - 2$.
• Step 4: Subtract $x^2$ from both sides to simplify: $-8x + 16 = x - 2$.
• Step 5: Move all terms involving $x$ to one side and constant terms to the other: $-8x - x = -2 - 16$.
• Step 6: Combine like terms: $-9x = -18$.
• Step 7: Solve for $x$ by dividing both sides by $-9$: $x = 2$.

Therefore, the solution to the problem is $x = 2$.

Solve the following equation. First, we'll simplify the algebraic expressions using the square of binomial formula:

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

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 expand the parentheses on the right side by using the distributive property,

Continue to combine like terms, by moving terms between sides. We observe 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.