Algebra Examples

$x^{2} - 2 x \leq 1$

Step 1

Subtract $1$ from both sides of the inequality.

$x^{2} - 2 x - 1 \leq 0$

Step 2

Convert the inequality to an equation.

$x^{2} - 2 x - 1 = 0$

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 $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 5

Simplify.

Tap for more steps...

Step 5.1

Simplify the numerator.

Tap for more steps...

Step 5.1.1

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

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

Step 5.1.2

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

Tap for more steps...

Step 5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 5.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 5.1.3

Add $4$ and $4$ .

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

Step 5.1.4

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

Tap for more steps...

Step 5.1.4.1

Factor $4$ out of $8$ .

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

Step 5.1.4.2

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

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

Step 5.1.5

Pull terms out from under the radical.

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

Step 5.2

Multiply $2$ by $1$ .

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

Step 5.3

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

$x = 1 \pm \sqrt{2}$

Step 6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Tap for more steps...

Step 6.1

Simplify the numerator.

Tap for more steps...

Step 6.1.1

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

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

Step 6.1.2

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

Tap for more steps...

Step 6.1.2.1

Multiply $- 4$ by $1$ .

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

Step 6.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 6.1.3

Add $4$ and $4$ .

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

Step 6.1.4

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

Tap for more steps...

Step 6.1.4.1

Factor $4$ out of $8$ .

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

Step 6.1.4.2

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

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

Step 6.1.5

Pull terms out from under the radical.

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

Step 6.2

Multiply $2$ by $1$ .

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

Step 6.3

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

$x = 1 \pm \sqrt{2}$

Step 6.4

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{2}$

Step 7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Tap for more steps...

Step 7.1

Simplify the numerator.

Tap for more steps...

Step 7.1.1

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

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

Step 7.1.2

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

Tap for more steps...

Step 7.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 7.1.3

Add $4$ and $4$ .

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

Step 7.1.4

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

Tap for more steps...

Step 7.1.4.1

Factor $4$ out of $8$ .

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

Step 7.1.4.2

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

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

Step 7.1.5

Pull terms out from under the radical.

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

Step 7.2

Multiply $2$ by $1$ .

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

Step 7.3

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

$x = 1 \pm \sqrt{2}$

Step 7.4

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{2}$

Step 8

Consolidate the solutions.

$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

Step 9

Use each root to create test intervals.

$x < 1 - \sqrt{2}$

$1 - \sqrt{2} < x < 1 + \sqrt{2}$

$x > 1 + \sqrt{2}$

Step 10

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

Tap for more steps...

Step 10.1

Test a value on the interval $x < 1 - \sqrt{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 10.1.1

Choose a value on the interval $x < 1 - \sqrt{2}$ and see if this value makes the original inequality true.

$x = - 3$

Step 10.1.2

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

${(- 3)}^{2} - 2 \cdot - 3 \leq 1$

Step 10.1.3

The left side $15$ is greater than the right side $1$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $1 - \sqrt{2} < x < 1 + \sqrt{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 10.2.1

Choose a value on the interval $1 - \sqrt{2} < x < 1 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

${(0)}^{2} - 2 \cdot 0 \leq 1$

Step 10.2.3

The left side $0$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval $x > 1 + \sqrt{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 10.3.1

Choose a value on the interval $x > 1 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 5$

Step 10.3.2

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

${(5)}^{2} - 2 \cdot 5 \leq 1$

Step 10.3.3

The left side $15$ is greater than the right side $1$ , which means that the given statement is false.

False

Step 10.4

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

$x < 1 - \sqrt{2}$ False

$1 - \sqrt{2} < x < 1 + \sqrt{2}$ True

$x > 1 + \sqrt{2}$ False

$x < 1 - \sqrt{2}$ False

$1 - \sqrt{2} < x < 1 + \sqrt{2}$ True

$x > 1 + \sqrt{2}$ False

Step 11

The solution consists of all of the true intervals.

$1 - \sqrt{2} \leq x \leq 1 + \sqrt{2}$

Step 12

Convert the inequality to interval notation.

$[1 - \sqrt{2}, 1 + \sqrt{2}]$

Step 13