Solve for x x^2-2x<=2

$x^{2} - 2 x \leq 2$

Step 1

Subtract $2$ from both sides of the inequality.

$x^{2} - 2 x - 2 \leq 0$

Step 2

Convert the inequality to an equation.

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

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

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

Step 5

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 2$ .

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

Add $4$ and $8$ .

$x = \frac{2 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{3}}{2}$ .

$x = 1 \pm \sqrt{3}$

Step 6

Consolidate the solutions.

$x = 1 + \sqrt{3}, 1 - \sqrt{3}$

Step 7

Use each root to create test intervals.

$x < 1 - \sqrt{3}$

$1 - \sqrt{3} < x < 1 + \sqrt{3}$

$x > 1 + \sqrt{3}$

Step 8

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 $x < 1 - \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < 1 - \sqrt{3}$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{2} - 2 \cdot - 3 \leq 2$

The left side $15$ is greater than the right side $2$ , which means that the given statement is false.

False

Test a value on the interval $1 - \sqrt{3} < x < 1 + \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $1 - \sqrt{3} < x < 1 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 2 \cdot 0 \leq 2$

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

True

Test a value on the interval $x > 1 + \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > 1 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 5$

Replace $x$ with $5$ in the original inequality.

${(5)}^{2} - 2 \cdot 5 \leq 2$

The left side $15$ is greater than the right side $2$ , which means that the given statement is false.

False

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

$x < 1 - \sqrt{3}$ False

$1 - \sqrt{3} < x < 1 + \sqrt{3}$ True

$x > 1 + \sqrt{3}$ False

$x < 1 - \sqrt{3}$ False

$1 - \sqrt{3} < x < 1 + \sqrt{3}$ True

$x > 1 + \sqrt{3}$ False

Step 9

The solution consists of all of the true intervals.

$1 - \sqrt{3} \leq x \leq 1 + \sqrt{3}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$1 - \sqrt{3} \leq x \leq 1 + \sqrt{3}$

Interval Notation:

$[1 - \sqrt{3}, 1 + \sqrt{3}]$

Step 11

Solve for x x^2-2x<=2

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