Simplify 3x+x>=4x-16+x^2

$3 x + x \geq 4 x - 16 + x^{2}$

Add $3 x$ and $x$ .

$4 x \geq 4 x - 16 + x^{2}$

Rewrite so $x$ is on the left side of the inequality.

$4 x - 16 + x^{2} \leq 4 x$

Move all terms containing $x$ to the left side of the inequality.

Subtract $4 x$ from both sides of the inequality.

$4 x - 16 + x^{2} - 4 x \leq 0$

Combine the opposite terms in $4 x - 16 + x^{2} - 4 x$ .

Subtract $4 x$ from $4 x$ .

$- 16 + x^{2} + 0 \leq 0$

Add $- 16 + x^{2}$ and $0$ .

$- 16 + x^{2} \leq 0$

Add $16$ to both sides of the inequality.

$x^{2} \leq 16$

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

$\sqrt{x^{2}} \leq \sqrt{16}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{16}$

Simplify the right side.

Simplify $\sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$| x | \leq \sqrt{4^{2}}$

Pull terms out from under the radical.

$| x | \leq | 4 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \leq 4$

Write $| x | \leq 4$ 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.

$x \leq 4$

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 $- 1$ .

$- x \leq 4$

Write as a piecewise.

${\begin{cases} x \leq 4 & x \geq 0 \\ - x \leq 4 & x < 0 \end{cases}$

Find the intersection of $x \leq 4$ and $x \geq 0$ .

$0 \leq x \leq 4$

Solve $- x \leq 4$ when $x < 0$ .

Divide each term in $- x \leq 4$ by $- 1$ and simplify.

Divide each term in $- x \leq 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{4}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{4}{- 1}$

Simplify the right side.

Divide $4$ by $- 1$ .

$x \geq - 4$

Find the intersection of $x \geq - 4$ and $x < 0$ .

$- 4 \leq x < 0$

Find the union of the solutions.

$- 4 \leq x \leq 4$

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq x \leq 4$

Interval Notation:

$[- 4, 4]$

Simplify 3x+x>=4x-16+x^2

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