Yet another question 5

Algebra Level 5

Let a quadratic equation be represented as

$A(\sqrt{3} - \sqrt{2})x^2 + \dfrac {Bx}{(\sqrt{3} + \sqrt{2})} + C=0,$

where $A = (49 + 20\sqrt 6)^{1/4}$ and $B= 8\sqrt 3 + \dfrac{8\sqrt6}{\sqrt 3} + \dfrac{16}{\sqrt3} + ....$ .

Let $\alpha$ and $\beta$ be the roots of the above equation which are related to the constraint $( \displaystyle | \alpha - \beta | = (6\sqrt6)^k,$

where $\displaystyle k = \log_{6}10 - 2\log_6 \sqrt5 + \log_6\sqrt{\log_{6} 18 + \log_{6} 72}$ .

Find the value of $C$ .

The answer is 128.

2 solutions

Rishabh Jain
Apr 28, 2016

$A=((\sqrt{24}+\sqrt{25})^2)^{1/4}\\=((\sqrt 3+\sqrt 2)^2)^{1/2}=\sqrt 3+\sqrt 2$ $B$ is a sum of Infinite GP with $r=\dfrac{\sqrt 6}{3}$ . $B=\dfrac{8\sqrt 3}{1-\dfrac{\sqrt 6}{3}}=\dfrac{24}{\sqrt 3-\sqrt 2}$

Hence equation is : $\large x^2+24x+C=0$

Using property $\log_a b+\log_a c=\log_a bc$ , $\large k=\log_6 10-\log_6 5+\log_6 \sqrt{4}\\\large =\log_6 4$

We note $\large (\alpha -\beta)^2=(\alpha +\beta)^2-4\alpha \beta\\=576-4C~~~~~\text{(Using Vieta's)}$

$\large (6^{3/2})^k=(\color{#D61F06}{6}^{\log_{\color{#D61F06}{6}}4^{3/2}})=8$

Hence, $576-4C=64$ $\implies \large C=144-16=\boxed{\Large 128}$

Gauri shankar Mishra
Apr 28, 2016

Nilesh u should make a set of your YET ANOTHER EASY QUESTION ! and link it to your questions like Aniket

Yet another question 5

The answer is 128.

2 solutions

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