In the coordinate plane, point P has coordinates (m, n) where m and n satisfy (m^2 + n^2 + 1)(m^2 + n^2 + 3) = 8. Find the length of OP.

Step-by-step solution

1. 1

   Connect the goal to the given expression. The distance from $P(m,n)$ to the origin is

   $OP=\sqrt{m^2+n^2}$

   So the whole problem reduces to finding the single number $m^2+n^2$; the individual values of $m$ and $n$ are never needed.

2. 2

   Substitute to hide the two variables. Let

   $t=m^2+n^2$

   Because $m$ and $n$ are real, each square is non-negative, which gives a constraint you must carry to the end:

   $t\ge 0$

3. 3

   Expand and solve the quadratic in $t$.

   $(t+1)(t+3)=8$

   $t^2+4t+3=8$

   $t^2+4t-5=0$

   $(t+5)(t-1)=0\ \Longrightarrow\ t=-5\ \text{or}\ t=1$

4. 4

   Discard the root that violates the constraint. A sum of two real squares cannot be $-5$, so $t=-5$ is rejected and

   $m^2+n^2=1$

   This is why option D ($\sqrt5$ or $1$) is a trap: it keeps a root that the substitution's own domain forbids.

5. 5

   Compute $OP$ and verify.

   $OP=\sqrt{1}=1$

   Check with the constraint: if $t=1$ then $(1+1)(1+3)=2\times 4=8$ $\checkmark$. Geometrically, $P$ lies anywhere on the unit circle — for example $(1,0)$ or $\left(\tfrac{3}{5},\tfrac{4}{5}\right)$ — and every such point is distance $1$ from the origin.