Question

Solve by using quadratic formula
xsquare + 4x + 5 =0 ​

Answer 1

Answer:

i-2,-2-i( imaginary roots)

Step-by-step explanation:

x^2+4x+5=0

formula: -b+/-√b^2-4ac divided by 2a

= (-4+/-√-4 )/2

= i-2,-i-2

Answer 2

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf \longmapsto {x}^{2} + 4x + 5 = 0$

Here ,this equation is in the form of a quadratic equation so we'll use quadratic formula for finding the roots:

$\sf \boxed{\bold{x} = {\red{ \dfrac{ - b\pm \sqrt{ {b}^{2} - 4ac } }{2a}}}}$

$\sf \: a = 1 ;\: b = 4\:\&\: c = 5$

Now,put the values in the formula:

$\sf \longmapsto \: x = \dfrac{- 4 \pm \sqrt{( {4)}^{2} - 4(1)(5) } }{2(1)}$

$\sf \longmapsto \: x = \dfrac{ - 4 \pm \sqrt{16 - 20} }{2}$

$\sf \longmapsto \: x = \dfrac{ - 4 \pm \sqrt{-4} }{2}$

Here, The discriminant b²-4ac <0,so there will be two complex roots .

$\sf \longmapsto \: x = \dfrac{ - 4 \pm 2i}{2}$

$\sf \longmapsto \: x = - \dfrac{4}{2} \pm \dfrac{2i}{2}$

$\sf \longmapsto \: x = - 2 \pm1i$

Possible values of x are :

➠x=-2+1i

➠x=-2-1i