Question

5=2x-x^2
how many real solutions

Answer 1

To find how many real solutions the equation \( 5 = 2x - x^2 \) has, we can rearrange it into a standard quadratic form.

Starting with:

\[
5 = 2x - x^2
\]

We rearrange it to:

\[
-x^2 + 2x - 5 = 0
\]

Multiplying through by -1 gives:

\[
x^2 - 2x + 5 = 0
\]

Next, we can use the quadratic formula to solve for \( x \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation, \( a = 1 \), \( b = -2 \), and \( c = 5 \). We first calculate the discriminant:

\[
b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16
\]

Since the discriminant is negative (\( -16 < 0 \)), this means that the quadratic \( x^2 - 2x + 5 = 0 \) has no real solutions.

Therefore, the number of real solutions to the equation \( 5 = 2x - x^2 \) is:

\[
\boxed{0}
\]