Solve the Inequality for x x^2-4x+2>0

$x^{2} - 4 x + 2 > 0$

Step 1

Convert the inequality to an equation.

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

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 $c = 2$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot 2$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8}}{2 \cdot 1}$

Subtract $8$ from $16$ .

$x = \frac{4 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

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

$x = 2 \pm \sqrt{2}$

Step 5

Consolidate the solutions.

$x = 2 + \sqrt{2}, 2 - \sqrt{2}$

Step 6

Use each root to create test intervals.

$x < 2 - \sqrt{2}$

$2 - \sqrt{2} < x < 2 + \sqrt{2}$

$x > 2 + \sqrt{2}$

Step 7

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 2 - \sqrt{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < 2 - \sqrt{2}$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} - 4 \cdot 0 + 2 > 0$

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $2 - \sqrt{2} < x < 2 + \sqrt{2}$ to see if it makes the inequality true.

Choose a value on the interval $2 - \sqrt{2} < x < 2 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

${(2)}^{2} - 4 \cdot 2 + 2 > 0$

The left side $- 2$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2 + \sqrt{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > 2 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

${(6)}^{2} - 4 \cdot 6 + 2 > 0$

The left side $14$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < 2 - \sqrt{2}$ True

$2 - \sqrt{2} < x < 2 + \sqrt{2}$ False

$x > 2 + \sqrt{2}$ True

$x < 2 - \sqrt{2}$ True

$2 - \sqrt{2} < x < 2 + \sqrt{2}$ False

$x > 2 + \sqrt{2}$ True

Step 8

The solution consists of all of the true intervals.

$x < 2 - \sqrt{2}$ or $x > 2 + \sqrt{2}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x < 2 - \sqrt{2} or x > 2 + \sqrt{2}$

Interval Notation:

$(- \infty, 2 - \sqrt{2}) \cup (2 + \sqrt{2}, \infty)$

Step 10

Solve the Inequality for x x^2-4x+2>0

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