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Question

If the $$A.M$$. of two numbers a and $$b , a > b > 0$$ is twice their $$G.M..$$ then $$\frac { a } { b } =$$ 


A
2+3
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B
7+23
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C
4+23
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D
7+43
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Solution

The correct option is B $$7 + 4 \sqrt { 3 }$$
Let the larger number be $$x$$ and smaller be $$y$$

Given, $$A.M.=2G.M$$

$$\Rightarrow\,\dfrac{a+b}{2}=2\sqrt{ab}$$
Squaring both sides,we get
$$\Rightarrow\,{\left(\dfrac{a+b}{2}\right)}^{2}={\left(2\sqrt{ab}\right)}^{2}$$
$$\Rightarrow\,\dfrac{{a}^{2}+2ab+{b}^{2}}{4}=4ab$$
$$\Rightarrow\,{a}^{2}+2ab+{b}^{2}=16ab$$
$$\Rightarrow\,{a}^{2}+{b}^{2}=16ab-2ab=14ab$$

$$\Rightarrow\,{a}^{2}+{b}^{2}-14ab=0$$

Dividing by $${b}^{2}$$ throughout we get

$$\Rightarrow\,\dfrac{{a}^{2}}{{b}^{2}}+1-14\dfrac{a}{b}=0$$
$$\Rightarrow\,\dfrac{{a}^{2}}{{b}^{2}}-14\dfrac{a}{b}+1=0$$

$$\Rightarrow\,\dfrac{a}{b}=\dfrac{14\pm\sqrt{196-4\times 1\times 1}}{2}$$

$$=\dfrac{14\pm\sqrt{192}}{2}=\dfrac{14\pm\,8\sqrt{3}}{2}=7+4\sqrt{3}$$


Mathematics

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