Question

The positive roots of the equation $(\sqrt{200}+\sqrt{56}) x^2+10 x-2(\sqrt{50}-\sqrt{14})=0$

• A. $\frac{\sqrt{26}}{\sqrt{14}}$
• B. $\sqrt{200}-\sqrt{56}$
• C. $\frac{5 \sqrt{2}-\sqrt{14}}{9}$
• D. $\frac{10}{\sqrt{200}-\sqrt{56}}$

Answer 1

Answer: (C)

$2(\sqrt{50}+\sqrt{14}) x^2+10 x-2(\sqrt{50}-\sqrt{14})=0$
$\Rightarrow x =\frac{-10 \pm \sqrt{100+16 \cdot 36}}{4(\sqrt{50}+\sqrt{14})}=\frac{-5 \pm \sqrt{25+144}}{2(\sqrt{50}+\sqrt{14})}=\frac{-5 \pm 13}{2(\sqrt{50}+\sqrt{14})}$
Positive root
$x=\frac{4(\sqrt{50}-\sqrt{14})}{36}=\frac{\sqrt{50}-\sqrt{14}}{9}=\frac{5 \sqrt{2}-\sqrt{14}}{9}$