Question

what are the roots of the equation x^2+4x-41=0

Answer 1

Let p ( x ) = x² + 4x - 41


a = 1

b = 4

c = -41


by quadratic formula " shreedhracharya method " :::::

Firstly :-

D = b² - 4ac


D = ( 4 )² - 4 × 1 × ( -41 )

D = 16 + 164

D = 180

=>

x = (- b ± √D ) / 2a

x = ( - 4 ± 6√5)/2

Now,

x = ( -4 + 6√5 ) /2

$x = \frac{2( - 2 + 3 \sqrt{5)} }{2}$


x = -2 + 3√5


And


x = ( -4 - 6√5 ) /2


$x = \frac{2( - 2 - 3 \sqrt{5)} }{2}$

x = -2 - 3√5




So, the roots of equation is

x = -2+3√5

and

x = -2 - 3√5


@Altaf