MathLabs

Problem 2

For which real numbers xx does the inequality 4x2(1−1+2x)2<2x+9\dfrac{4x^2}{\left(1-\sqrt{1+2x}\right)^2} < 2x+9 hold?
Step 3 of 5: Rewrite the entire inequality in terms of a
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(a2−1)2(1−a)2<a2+8\dfrac{(a^2-1)^2}{(1-a)^2} < a^2+8
Detailed analysis

Direct substitution gives 4x2=(a2−1)24x^2=(a^2-1)^2, 2x+9=(a2−1)+9=a2+82x+9=(a^2-1)+9=a^2+8, and 1−1+2x=1−a1-\sqrt{1+2x}=1-a, so the inequality becomes (a2−1)2(1−a)2<a2+8\dfrac{(a^2-1)^2}{(1-a)^2}<a^2+8.