Question

zeroes of
$x {}^{2} - 2 \sqrt{5 }x + 3$

Answer 1

Given that :

${x}^{2} - 2 \sqrt{5}x + 3$

Here, a = 1 b = -2√5 and c = 3

Now, By the Shridharacharya Formula :

$x = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ = > x = \frac{2 \sqrt{5} \binom{ + }{ - } \sqrt{20 - 4 \times 1 \times 3} }{2 \times 1} \\ \\ = > x = \frac{2 \sqrt{5} \binom{ + }{ - } \sqrt{20 - 12} }{2} \\ \\ = > x = \frac{2 \sqrt{5} \binom{ + }{ - } 2\sqrt{2} }{2} \\ \\ = > x = \sqrt{5} \binom{ + }{ - } \sqrt{2}$

So, the Zeroes of equation : √5+√2 and √5-√2