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[Solved] Inequality

 
Logorithm
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June 15, 2021 2:40 pm
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(@armor)
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The inequality \sqrt{x}>ax+\frac{3}{2}  has the solution set \left \{ x|4<x<b \right \},  find the values of a and b.

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June 15, 2021 2:40 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

\small \sqrt{x}>ax+\frac{3}{2}\Leftrightarrow a(\sqrt{x})^2-\sqrt{x}+\frac{3}{2}<0.\small 2+\sqrt{b}=\frac{1}{a} 

The solution set  \small \left \{ x|4<x<b \right \} is equivalent to \small \left \{ x|2<\sqrt{x}<\sqrt{b} \right \}.  

Vieta's formulas imply

\small 2+\sqrt{b}=\frac{1}{a}

\small 2\sqrt{b}=\frac{3}{2a}

\small a>0

\small \Rightarrow a=\frac{1}{8}, b=36.

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