Question

THE ROOT OF QUADRATIC EQUATION X2 -2√5x +1 =0

Answer 1

Given equation : x^2 - 2√5 x + 1 = 0

On comparing the given equation with ax^2 + bx + c = 0 , we get that a = 1 , b = -2√5 and c = 1 .

∴ Discriminant of the given equation : b^2 - 4ac

$\implies ( - 2\sqrt{5} )^2 - 4( 1 \times 1 ) \\\\\implies 20 - 4 \\\\\implies 16$

      By Shree Dharrya Achharaya's method :

$\implies x = \dfrac{-b \pm\sqrt{Discriminant}}{2a}$

$\implies x = \dfrac{-(-2\sqrt{5} )\pm \sqrt{16}}{2\times 1}$

$\implies x = \dfrac{2\sqrt{5}\pm 4 }{2}$

$\implies x = \sqrt{5} + 2 \:\:\:Or\:\:\:\sqrt{5} - 2$

Therefore the roots of the given equation are √5 + 2 and √5 - 2

Answer 2

X² - 2√5X + 1 = 0



Here,


a = 1 , b = -2√5 and C = 1.




Discriminant ( d ) = b² - 4ac



=> ( -2√5)² - 4 × 1 × 1



=> 20 - 4


=> 16




√D = √16 = 4



Therefore,


Alpha = -b + √D/2a and Beta = - b - √D/2a


Alpha = - ( -2√5 ) + 4 / 2 × 1 and beta = - ( -2√5 ) - 4 / 2 × 1



Alpha = 2√5 + 4 / 2 and beta = 2√5 - 4 /2




Alpha = 2 ( √5 + 2 ) /2 and beta = 2 ( √5 - 2) /2


Alpha = √5 + 2 and beta = √5 - 2.