Given:
The equation is
[tex]x^2+(2\sqrt{5})+2x=-10[/tex]
To find:
The number of roots and discriminant of the given equation.
Solution:
We have,
[tex]x^2+(2\sqrt{5})x+2x=-10[/tex]
The highest degree of given equation is 2. So, the number of roots is also 2.
It can be written as
[tex]x^2+(2\sqrt{5}+2)x+10=0[/tex]
Here, [tex]a=1, b=(2\sqrt{5}+2), c=10[/tex].
Discriminant of the given equation is
[tex]D=b^2-4ac[/tex]
[tex]D=(2\sqrt{5}+2)^2-4(1)(10)[/tex]
[tex]D=20+8\sqrt{5}+4-40[/tex]
[tex]D=8\sqrt{5}-16[/tex]
[tex]D\approx 1.89>0[/tex]
Since discriminant is [tex]8\sqrt{5}-16\approx 1.89[/tex], which is greater than 0, therefore, the given equation has two distinct real roots.
