Question

Determine whether the quadratic equation x2+5x+5=0 have real roots and if so, find
the roots.

Answer 1

Step-by-step explanation:

x²-5x+5 = 0

a=1 b=-5 c=5

D = b²-4ac

= 25-4(1)(5)

=25-20

= 5

The roots are real..

Let the roots be a and b..

then,

a = -b+√D / 2a

= -5 + √5 / 2

b = -b-√D / 2a

= -5-√5 / 2

Pls mark as brainliest..

Answer 2

Answer: Roots are real and its roots are -5+√5/2 and -5-√5/2.

Given:

$\sf x^2+5x+5=0$

To find:

$\sf Roots\:of\:the\:quadratic\:equation\:if\:the\:root\:are\:real.$

Formula used:

$\boxed{\sf Discriminant(D):b^2-4ac}$

$\sf where, a=Coefficient\:of\:x^2$

         $\sf b=Coefficient\:of\:x$

         $\sf c=Value\:of\:constant.$

$\sf If,$

$\sf b^2-4ac>0\:(Two\:real\:roots\:exist)$

$\sf b^2-4ac=0\:(Two\:identical\:roots\:exist)$

$\sf b^2-4ac<0\:(Two\:imaginary\:roots\:exist)$

$\boxed{\sf Quadratic\:formula:\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}}$

Explanation:

$\sf According\:to\:the\:question,$

$\sf a=1,b=5\:and\:c=5$

$\sf\therefore D=5^2-4\times1\times5$

      $\sf =25-20$

      $\sf =5\:(Two\:real\:roots\:exist)$

$\sf Now, substituting\:the\:values\:in\:quadratic\:formula,we\:get:$

$\Rightarrow\sf \dfrac{-5\pm\sqrt{5} }{2\times1}$

$\boxed{\sf \therefore the\:roots\:are: \sf \dfrac{-5+\sqrt{5} }{2}\:and\:\dfrac{-5-\sqrt{5} }{2}}$

Hence, the roots of x²+5x+5=0 is real and its roots are -5+√5/2 and -5-√5/2 respectively.