Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Find the answer

4・10 Given the equation $x^{2}-2 x+\lg \left(2 a^{2}-a\right)=0$ has one positive root and one negative root, try to find the range of real values for $a$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

[Solution] From the given, we know that $\lg \left(2 a^{2}-a\right)<0$ . Therefore, $0<2 a^{2}-a<1$ ,
Solving this, we get that $a$ should be in the range
$-\frac{1}{2}<a<0 \text { or } \frac{1}{2}<a<1 \text {. }$

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