Solve the Inequality for x |(x^2-1)/2|>=1

$| \frac{x^{2} - 1}{2} | \geq 1$

Step 1

Write $| \frac{x^{2} - 1}{2} | \geq 1$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$\frac{x^{2} - 1}{2} \geq 0$

Solve the inequality.

Multiply both sides by $2$ .

$\frac{x^{2} - 1}{2} \cdot 2 \geq 0 \cdot 2$

Simplify.

Simplify the left side.

Simplify $\frac{x^{2} - 1}{2} \cdot 2$ .

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

$\frac{x^{2} - 1^{2}}{2} \cdot 2 \geq 0 \cdot 2$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{(x + 1) (x - 1)}{2} \cdot 2 \geq 0 \cdot 2$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{(x + 1) (x - 1)}{2} \cdot 2 \geq 0 \cdot 2$

Rewrite the expression.

$(x + 1) (x - 1) \geq 0 \cdot 2$

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$x (x - 1) + 1 (x - 1) \geq 0 \cdot 2$

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 (x - 1) \geq 0 \cdot 2$

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1 \geq 0 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1 \geq 0 \cdot 2$

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1 \geq 0 \cdot 2$

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x + 1 x + 1 \cdot - 1 \geq 0 \cdot 2$

Multiply $x$ by $1$ .

$x^{2} - x + x + 1 \cdot - 1 \geq 0 \cdot 2$

Multiply $- 1$ by $1$ .

$x^{2} - x + x - 1 \geq 0 \cdot 2$

Add $- x$ and $x$ .

$x^{2} + 0 - 1 \geq 0 \cdot 2$

Add $x^{2}$ and $0$ .

$x^{2} - 1 \geq 0 \cdot 2$

Simplify the right side.

Multiply $0$ by $2$ .

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

Solve for $x$ .

Add $1$ to both sides of the inequality.

$x^{2} \geq 1$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} \geq \sqrt{1}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{1}$

Simplify the right side.

Any root of $1$ is $1$ .

$| x | \geq 1$

Write $| x | \geq 1$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq 1$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq 1$

Write as a piecewise.

${\begin{cases} x \geq 1 & x \geq 0 \\ - x \geq 1 & x < 0 \end{cases}$

Find the intersection of $x \geq 1$ and $x \geq 0$ .

$x \geq 1$

Divide each term in $- x \geq 1$ by $- 1$ and simplify.

Divide each term in $- x \geq 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{1}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{1}{- 1}$

Simplify the right side.

Divide $1$ by $- 1$ .

$x \leq - 1$

Find the union of the solutions.

$x \leq - 1$ or $x \geq 1$

In the piece where $\frac{x^{2} - 1}{2}$ is non-negative, remove the absolute value.

$\frac{x^{2} - 1}{2} \geq 1$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$\frac{x^{2} - 1}{2} < 0$

Solve the inequality.

Multiply both sides by $2$ .

$\frac{x^{2} - 1}{2} \cdot 2 < 0 \cdot 2$

Simplify.

Simplify the left side.

Simplify $\frac{x^{2} - 1}{2} \cdot 2$ .

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

$\frac{x^{2} - 1^{2}}{2} \cdot 2 < 0 \cdot 2$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{(x + 1) (x - 1)}{2} \cdot 2 < 0 \cdot 2$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{(x + 1) (x - 1)}{2} \cdot 2 < 0 \cdot 2$

Rewrite the expression.

$(x + 1) (x - 1) < 0 \cdot 2$

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$x (x - 1) + 1 (x - 1) < 0 \cdot 2$

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 (x - 1) < 0 \cdot 2$

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1 < 0 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1 < 0 \cdot 2$

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1 < 0 \cdot 2$

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x + 1 x + 1 \cdot - 1 < 0 \cdot 2$

Multiply $x$ by $1$ .

$x^{2} - x + x + 1 \cdot - 1 < 0 \cdot 2$

Multiply $- 1$ by $1$ .

$x^{2} - x + x - 1 < 0 \cdot 2$

Add $- x$ and $x$ .

$x^{2} + 0 - 1 < 0 \cdot 2$

Add $x^{2}$ and $0$ .

$x^{2} - 1 < 0 \cdot 2$

Simplify the right side.

Multiply $0$ by $2$ .

$x^{2} - 1 < 0$

Solve for $x$ .

Add $1$ to both sides of the inequality.

$x^{2} < 1$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} < \sqrt{1}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | < \sqrt{1}$

Simplify the right side.

Any root of $1$ is $1$ .

$| x | < 1$

Write $| x | < 1$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x < 1$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x < 1$

Write as a piecewise.

${\begin{cases} x < 1 & x \geq 0 \\ - x < 1 & x < 0 \end{cases}$

Find the intersection of $x < 1$ and $x \geq 0$ .

$0 \leq x < 1$

Solve $- x < 1$ when $x < 0$ .

Divide each term in $- x < 1$ by $- 1$ and simplify.

Divide each term in $- x < 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{1}{- 1}$

Divide $x$ by $1$ .

$x > \frac{1}{- 1}$

Simplify the right side.

Divide $1$ by $- 1$ .

$x > - 1$

Find the intersection of $x > - 1$ and $x < 0$ .

$- 1 < x < 0$

Find the union of the solutions.

$- 1 < x < 1$

In the piece where $\frac{x^{2} - 1}{2}$ is negative, remove the absolute value and multiply by $- 1$ .

$- \frac{x^{2} - 1}{2} \geq 1$

Write as a piecewise.

${\begin{cases} \frac{x^{2} - 1}{2} \geq 1 & x \leq - 1 or x \geq 1 \\ - \frac{x^{2} - 1}{2} \geq 1 & - 1 < x < 1 \end{cases}$

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

${\begin{cases} \frac{x^{2} - 1^{2}}{2} \geq 1 & x \leq - 1 or x \geq 1 \\ - \frac{x^{2} - 1}{2} \geq 1 & - 1 < x < 1 \end{cases}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

${\begin{cases} \frac{(x + 1) (x - 1)}{2} \geq 1 & x \leq - 1 or x \geq 1 \\ - \frac{x^{2} - 1}{2} \geq 1 & - 1 < x < 1 \end{cases}$

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

${\begin{cases} \frac{(x + 1) (x - 1)}{2} \geq 1 & x \leq - 1 or x \geq 1 \\ - \frac{x^{2} - 1^{2}}{2} \geq 1 & - 1 < x < 1 \end{cases}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

${\begin{cases} \frac{(x + 1) (x - 1)}{2} \geq 1 & x \leq - 1 or x \geq 1 \\ - \frac{(x + 1) (x - 1)}{2} \geq 1 & - 1 < x < 1 \end{cases}$

Step 2

Solve $\frac{(x + 1) (x - 1)}{2} \geq 1$ for $x$ .

Multiply both sides by $2$ .

$\frac{(x + 1) (x - 1)}{2} \cdot 2 \geq 1 \cdot 2$

Simplify.

Simplify the left side.

Simplify $\frac{(x + 1) (x - 1)}{2} \cdot 2$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{(x + 1) (x - 1)}{2} \cdot 2 \geq 1 \cdot 2$

Rewrite the expression.

$(x + 1) (x - 1) \geq 1 \cdot 2$

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$x (x - 1) + 1 (x - 1) \geq 1 \cdot 2$

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 (x - 1) \geq 1 \cdot 2$

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1 \geq 1 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1 \geq 1 \cdot 2$

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1 \geq 1 \cdot 2$

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x + 1 x + 1 \cdot - 1 \geq 1 \cdot 2$

Multiply $x$ by $1$ .

$x^{2} - x + x + 1 \cdot - 1 \geq 1 \cdot 2$

Multiply $- 1$ by $1$ .

$x^{2} - x + x - 1 \geq 1 \cdot 2$

Add $- x$ and $x$ .

$x^{2} + 0 - 1 \geq 1 \cdot 2$

Add $x^{2}$ and $0$ .

$x^{2} - 1 \geq 1 \cdot 2$

Simplify the right side.

Multiply $2$ by $1$ .

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

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Add $1$ to both sides of the inequality.

$x^{2} \geq 2 + 1$

Add $2$ and $1$ .

$x^{2} \geq 3$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} \geq \sqrt{3}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{3}$

Write $| x | \geq \sqrt{3}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq \sqrt{3}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq \sqrt{3}$

Write as a piecewise.

${\begin{cases} x \geq \sqrt{3} & x \geq 0 \\ - x \geq \sqrt{3} & x < 0 \end{cases}$

Find the intersection of $x \geq \sqrt{3}$ and $x \geq 0$ .

$x \geq \sqrt{3}$

Divide each term in $- x \geq \sqrt{3}$ by $- 1$ and simplify.

Divide each term in $- x \geq \sqrt{3}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{3}}{- 1}$ .

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

Rewrite $- 1 \cdot \sqrt{3}$ as $- \sqrt{3}$ .

$x \leq - \sqrt{3}$

Find the union of the solutions.

$x \leq - \sqrt{3}$ or $x \geq \sqrt{3}$

Step 3

Solve $- \frac{(x + 1) (x - 1)}{2} \geq 1$ for $x$ .

Multiply both sides by $2$ .

$- \frac{(x + 1) (x - 1)}{2} \cdot 2 \geq 1 \cdot 2$

Simplify.

Simplify the left side.

Simplify $- \frac{(x + 1) (x - 1)}{2} \cdot 2$ .

Simplify terms.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{(x + 1) (x - 1)}{2}$ into the numerator.

$\frac{- (x + 1) (x - 1)}{2} \cdot 2 \geq 1 \cdot 2$

Cancel the common factor.

$\frac{- (x + 1) (x - 1)}{2} \cdot 2 \geq 1 \cdot 2$

Rewrite the expression.

$- (x + 1) (x - 1) \geq 1 \cdot 2$

Apply the distributive property.

$(- x - 1 \cdot 1) (x - 1) \geq 1 \cdot 2$

Multiply $- 1$ by $1$ .

$(- x - 1) (x - 1) \geq 1 \cdot 2$

Expand $(- x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$- x (x - 1) - 1 (x - 1) \geq 1 \cdot 2$

Apply the distributive property.

$- x \cdot x - x \cdot - 1 - 1 (x - 1) \geq 1 \cdot 2$

Apply the distributive property.

$- x \cdot x - x \cdot - 1 - 1 x - 1 \cdot - 1 \geq 1 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- (x \cdot x) - x \cdot - 1 - 1 x - 1 \cdot - 1 \geq 1 \cdot 2$

Multiply $x$ by $x$ .

$- x^{2} - x \cdot - 1 - 1 x - 1 \cdot - 1 \geq 1 \cdot 2$

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$- x^{2} + 1 x - 1 x - 1 \cdot - 1 \geq 1 \cdot 2$

Multiply $x$ by $1$ .

$- x^{2} + x - 1 x - 1 \cdot - 1 \geq 1 \cdot 2$

Rewrite $- 1 x$ as $- x$ .

$- x^{2} + x - x - 1 \cdot - 1 \geq 1 \cdot 2$

Multiply $- 1$ by $- 1$ .

$- x^{2} + x - x + 1 \geq 1 \cdot 2$

Subtract $x$ from $x$ .

$- x^{2} + 0 + 1 \geq 1 \cdot 2$

Add $- x^{2}$ and $0$ .

$- x^{2} + 1 \geq 1 \cdot 2$

Simplify the right side.

Multiply $2$ by $1$ .

$- x^{2} + 1 \geq 2$

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $1$ from both sides of the inequality.

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

Subtract $1$ from $2$ .

$- x^{2} \geq 1$

Divide each term in $- x^{2} \geq 1$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

Divide $1$ by $- 1$ .

$x^{2} \leq - 1$

Since the left side has an even power, it is always positive for all real numbers.

No solution

Step 4

Find the union of the solutions.

$x \leq - \sqrt{3}$ or $x \geq \sqrt{3}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \sqrt{3} or x \geq \sqrt{3}$

Interval Notation:

$(- \infty, - \sqrt{3}] \cup [\sqrt{3}, \infty)$

Step 6

Solve the Inequality for x |(x^2-1)/2|>=1

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