• # question_answer Let ${{A}_{1}},{{A}_{2}},{{A}_{3}}$ me three arithmetic means between two positive numbers a and b (where $a>b$). If the equation ${{A}_{1}}{{x}^{2}}+2{{A}_{2}}x+{{A}_{3}}=0$ has imaginary roots, then the range of $\frac{a}{b}$is A) $\left( 2-\sqrt{2},+2+\sqrt{2} \right)$      B) $\left( 2-\sqrt{3},+2+\sqrt{3} \right)$ C) $\left( -\sqrt{2},\sqrt{2} \right)$               D) $\left( -\sqrt{3},\sqrt{3} \right)$

Correct Answer: B

Solution :

[b] ${{A}_{1}}=\frac{3a+b}{4},{{A}_{2}}\frac{a+b}{2},{{A}_{3}}=\frac{a+3b}{4}$ For equation ${{A}_{1}}{{x}^{2}}+2{{A}_{2}}x+{{A}_{3}}=0,$ $D=4({{A}_{2}}^{2}-{{A}_{1}}{{A}_{3}})$ $=4\left( {{\left( \frac{a+b}{2} \right)}^{2}}-\frac{(b+3a)}{4}\frac{(a+3b)}{4} \right)$ $=\frac{4\left( {{a}^{2}}+{{b}^{2}}-4ab \right)}{16}<0$ $\Rightarrow \,\,\,\,\,\,\,{{\left( \frac{a}{b} \right)}^{2}}-4\left( \frac{a}{b} \right)+1<0$ $\Rightarrow \,\,\,\,2-\sqrt{3}<\frac{a}{b}<2+\sqrt{3}$

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