Solve the Inequality for x x^2-4x+5<=0

$x^{2} - 4 x + 5 \leq 0$

Step 1

Convert the inequality to an equation.

$x^{2} - 4 x + 5 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = - 4$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 5}}{2 \cdot 1}$

Multiply $- 4$ by $5$ .

$x = \frac{4 \pm \sqrt{16 - 20}}{2 \cdot 1}$

Subtract $20$ from $16$ .

$x = \frac{4 \pm \sqrt{- 4}}{2 \cdot 1}$

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{4}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{4 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{4 \pm i \cdot 2}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 i}{2}$

Simplify $\frac{4 \pm 2 i}{2}$ .

$x = 2 \pm i$

Step 5

Identify the leading coefficient.

The leading term in a polynomial is the term with the highest degree.

$x^{2}$

The leading coefficient in a polynomial is the coefficient of the leading term.

$1$

Step 6

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $x^{2} - 4 x + 5$ is always greater than $0$ .

No solution

Solve the Inequality for x x^2-4x+5<=0

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