Question

how many real number solutions does 4x^2+2x+5=0 have?

Answer 1

Therefore $4x^2+2x+5=0$ have two real number solution.

Step-by-step explanation:

$4x^2+2x+5=0$                      The solution of the equation $ax^2+bx+c=0$ is

$\Leftrightarrow x = \frac{-2\pm \sqrt{2^2+4.4.5} }{2.5}$                     $x= \frac{-b\pm\sqrt{b^2+4ac} }{2a}$

$\Leftrightarrow x =\frac{-2\pm \sqrt{84} }{10}$

$\Leftrightarrow x =\frac{-1\pm \sqrt{21} }{5}$

Therefore $4x^2+2x+5=0$ have two real number solution

Answer 2

The quadratic equation  $\mathbf{4x^2 + 2x + 5 = 0}$ has no real solution

The equation is given as:

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

A quadratic equation is represented as:

$\mathbf{ax^2 + bx + c = 0}$

The discriminant is calculated as:

$\mathbf{d =b^2 -4ac}$

By comparison: a = 4, b = 2, and c = 5

So, we have:

$\mathbf{d =2^2 -4\times 4\times 5}$

$\mathbf{d =4 -80}$

$\mathbf{d =-76}$

When the discriminant is less than 0, then the equation has no real solutions.

Hence, $\mathbf{4x^2 + 2x + 5 = 0}$ has no real solution

Read more about real solutions at:

https://brainly.com/question/4526506