Q51 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
If $\alpha$ and $\beta$ are the real roots of the equation
$\sqrt{\frac{5 x}{x-2}}+\sqrt{\frac{x-2}{5 x}}=\frac{29}{10}$ and $\alpha>\beta$, then $\sqrt{\alpha^2-11^4 \beta^2}=$
- A64
- B36
- C100
- Correct answerD6
Solution
We have, $\sqrt{\frac{5 x}{x-2}}+\sqrt{\frac{x-2}{5 x}}=\frac{29}{10}$
Let $\quad \sqrt{\frac{5 x}{x-2}}=y$
Then, $y+\frac{1}{y}=\frac{29}{10} \Rightarrow \frac{y^2+1}{y}=\frac{29}{10}$
$\Rightarrow \quad 10 y^2+10=29 y$
$\Rightarrow \quad 10 y^2-29 y+10=0$
$\Rightarrow \quad 10 y^2-25 y-4 y+10=0$
$\Rightarrow \quad 5 y(2 y-5)-2(2 y-5)=0$
$\Rightarrow \quad(5 y-2)(2 y-5)=0$
$\Rightarrow \quad y=\frac{2}{5}, \frac{5}{2}$
Now, if $y=\frac{2}{5} \Rightarrow \sqrt{\frac{5 x}{x-2}}=\frac{2}{5}$
$\Rightarrow \quad \frac{5 x}{x-2}=\frac{4}{25}$
$\Rightarrow \quad 125 x=4 x-8$
$\Rightarrow \quad 121 x=-8$
$\Rightarrow \quad x=-8 / 121$
And if $y=\frac{5}{2} \Rightarrow \sqrt{\frac{5 x}{x-2}}=\frac{5}{2}$
$\Rightarrow \quad \frac{5 x}{x-2}=\frac{25}{4}$
$\Rightarrow \quad \frac{x}{x-2}=\frac{5}{4} \Rightarrow 5 x-10=4 x$
$\Rightarrow \quad x=10$
$\therefore$ Let $\alpha=10$ and $\beta=\frac{-8}{121}$ $(\because \alpha>\beta)$
$\therefore \quad \sqrt{\alpha^2-11^4 \beta^2}$
$=\sqrt{(10)^2-11^4\left(\frac{-8}{121}\right)^2}$
$=\sqrt{100-64}$
$=\sqrt{36}=6$
Hence, the correct answer is option: 'D'.