Question

If $\left(x-\dfrac{1}{2}\right)^2 - \left(x-\dfrac{3}{2}\right)^2 = x + 2$, then the value of $x$ is:

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Hint:

Using the identity $a^2 - b^2 = (a-b)(a+b)$ can often simplify such expressions faster.

Answer 1

The Correct Option is B

Step 1: Expand both squares.
\[ \left(x-\frac{1}{2}\right)^2 = x^2 - x + \frac{1}{4}, \] \[ \left(x-\frac{3}{2}\right)^2 = x^2 - 3x + \frac{9}{4}. \]

Step 2: Subtract the expressions.
\[ \left(x-\frac{1}{2}\right)^2 - \left(x-\frac{3}{2}\right)^2 = (x^2 - x + \tfrac{1}{4}) - (x^2 - 3x + \tfrac{9}{4}) = 2x - 2. \]

Step 3: Form the equation.
\[ 2x - 2 = x + 2. \]

Step 4: Solve for $x$.
\[ 2x - x = 2 + 2 $\Rightarrow$ x = 4. \]

Step 5: Conclusion.
Thus, the correct value of $x$ is $4$.