Solve for :
Fix the domain first. We need and , i.e. , which means
Every candidate root must survive this test, and squaring later will manufacture roots that do not.
Clear both denominators. Multiplying by and then by :
Since , this rearranges to
Factor the radicand as a difference of squares. This is the step that makes the problem tractable:
and at the same time . The same factor appears on both sides, so the equation is
Read off the first solution for free. If both sides are zero, so is an exact solution. Checking in the original: on the left, and on the right the radical is , giving ✓.
Handle the remaining branch. For we have , so and one factor of cancels:
(The branch is impossible: there the left side is while .)
Solve the cubic and reject the extraneous roots. Its roots are , and . Squaring required and , so the first and third are extraneous - substituting them into the original equation gives and instead of . Only
survives, and it reproduces the original equation to ✓.
State both answers. The full solution set is and . Note times the cubic reproduces the quartic exactly, confirming no root was lost.
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