Use the discriminant to determine the number of real solutions of

5=2x-x^2

1 answer

To determine the number of real solutions for the equation \(5 = 2x - x^2\), we first rearrange it into the standard form of a quadratic equation.

Starting with the original equation: \[ 5 = 2x - x^2 \]

We can rearrange it to: \[ -x^2 + 2x - 5 = 0 \]

Multiplying the entire equation by -1 to put it in the standard form \(ax^2 + bx + c = 0\), we get: \[ x^2 - 2x + 5 = 0 \]

Here, the coefficients are:

  • \(a = 1\)
  • \(b = -2\)
  • \(c = 5\)

Next, we calculate the discriminant \(D\) using the formula: \[ D = b^2 - 4ac \]

Plugging in the values of \(a\), \(b\), and \(c\): \[ D = (-2)^2 - 4(1)(5) = 4 - 20 = -16 \]

The discriminant \(D = -16\) is less than zero.

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

Therefore, the equation \(5 = 2x - x^2\) has no real solutions.