MathLabs

Problem 2

Determine all real numbers xx which satisfy 3−x−x+1>12\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac12.
Step 4 of 5: Square again and obtain the quadratic
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31−32x=8x+1⟹1024x2−2048x+897=031-32x=8\sqrt{x+1}\Longrightarrow1024x^2-2048x+897=0
Detailed analysis

Squaring 3−x=1/4+x+1\sqrt{3-x}=1/4+\sqrt{x+1} and rearranging gives 31−32x=8x+131-32x=8\sqrt{x+1}. Squaring this equation produces 1024x2−2048x+897=01024x^2-2048x+897=0.