Question

How many roots does this has?x^2+(2√5x)+2x=-10​find Discriminant

Answer 1

Given:

The equation is

$x^2+(2\sqrt{5})+2x=-10$

To find:

The number of roots and discriminant of the given equation.

Solution:

We have,

$x^2+(2\sqrt{5})x+2x=-10$

The highest degree of given equation is 2. So, the number of roots is also 2.

It can be written as

$x^2+(2\sqrt{5}+2)x+10=0$

Here, $a=1, b=(2\sqrt{5}+2), c=10$.

Discriminant of the given equation is

$D=b^2-4ac$

$D=(2\sqrt{5}+2)^2-4(1)(10)$

$D=20+8\sqrt{5}+4-40$

$D=8\sqrt{5}-16$

$D\approx 1.89>0$

Since discriminant is $8\sqrt{5}-16\approx 1.89$, which is greater than 0, therefore, the given equation has two distinct real roots.