The positive roots of the equation (√200+√56) x2+10 x-2(√50-√14)=0

Question

The positive roots of the equation (√200+√56) x2+10 x-2(√50-√14)=0 (A) (√26/√14) (B) √200-√56 (C) (5 √2-√14/9) (D) (10/√200-√56)

Tardigrade Admin · Accepted Answer

2(√50+√14) x2+10 x-2(√50-√14)=0 ⇒ x =(-10 ± √100+16 ⋅ 36/4(√50+√14))=(-5 ± √25+144/2(√50+√14))=(-5 ± 13/2(√50+√14)) Positive root x=(4(√50-√14)/36)=(√50-√14/9)=(5 √2-√14/9)

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

$\frac{26}{14}$

$200 - 56$

$\frac{5 2 - 14}{9}$

$\frac{10}{200 - 56}$

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

Q. The positive roots of the equation (200​+56​)x2+10x−2(50​−14​)=0

14​26​​

200​−56​

952​−14​​

200​−56​10​