Find the root of the equation (x - 4) squared equals (3 - x) squared.

Step-by-step solution

1. 1

   Use the equal-squares rule rather than expanding. For real numbers,

   $A^2=B^2\iff A=B\ \text{or}\ A=-B$

   Both branches must be examined; taking only $A=B$ is the standard way to lose a root.

2. 2

   Branch 1: $x-4=3-x$.

   $x+x=3+4$

   $2x=7\ \Longrightarrow\ x=\frac{7}{2}$

3. 3

   Branch 2: $x-4=-(3-x)$. Distribute the minus sign:

   $x-4=-3+x$

   Subtracting $x$ from both sides leaves

   $-4=-3$

   which is false for every $x$. This branch contributes no solutions — geometrically, $y=x-4$ and $y=x-3$ are parallel lines that never meet.

4. 4

   Confirm with the expansion route. Expanding both sides gives

   $x^2-8x+16=9-6x+x^2$

   The $x^2$ terms cancel, leaving the linear equation $-8x+16=9-6x$, i.e. $7=2x$, so $x=\tfrac72$ — the same single root. The cancellation of $x^2$ is exactly why branch 2 was empty.

5. 5

   Check the answer. With $x=\tfrac72$:

   $\left(\tfrac72-4\right)^2=\left(-\tfrac12\right)^2=\tfrac14$

   $\left(3-\tfrac72\right)^2=\left(-\tfrac12\right)^2=\tfrac14$

   Both sides equal $\tfrac14$, so $x=\tfrac72$ is the unique root.