Solve the Inequality for x -2 square root of x+1<=-4

$- 2 \sqrt{x + 1} \leq - 4$

Step 1

To remove the radical on the left side of the inequality, square both sides of the inequality.

${(- 2 \sqrt{x + 1})}^{2} \leq {(- 4)}^{2}$

Step 2

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 1}$ as ${(x + 1)}^{\frac{1}{2}}$ .

${(- 2 {(x + 1)}^{\frac{1}{2}})}^{2} \leq {(- 4)}^{2}$

Simplify the left side.

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

Apply the product rule to $- 2 {(x + 1)}^{\frac{1}{2}}$ .

${(- 2)}^{2} {({(x + 1)}^{\frac{1}{2}})}^{2} \leq {(- 4)}^{2}$

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

$4 {({(x + 1)}^{\frac{1}{2}})}^{2} \leq {(- 4)}^{2}$

Multiply the exponents in ${({(x + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 {(x + 1)}^{\frac{1}{2} \cdot 2} \leq {(- 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(x + 1)}^{\frac{1}{2} \cdot 2} \leq {(- 4)}^{2}$

Rewrite the expression.

$4 {(x + 1)}^{1} \leq {(- 4)}^{2}$

Simplify.

$4 (x + 1) \leq {(- 4)}^{2}$

Apply the distributive property.

$4 x + 4 \cdot 1 \leq {(- 4)}^{2}$

Multiply $4$ by $1$ .

$4 x + 4 \leq {(- 4)}^{2}$

Simplify the right side.

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

$4 x + 4 \leq 16$

Step 3

Solve for $x$ .

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

Subtract $4$ from both sides of the inequality.

$4 x \leq 16 - 4$

Subtract $4$ from $16$ .

$4 x \leq 12$

Divide each term in $4 x \leq 12$ by $4$ and simplify.

Divide each term in $4 x \leq 12$ by $4$ .

$\frac{4 x}{4} \leq \frac{12}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \leq \frac{12}{4}$

Divide $x$ by $1$ .

$x \leq \frac{12}{4}$

Simplify the right side.

Divide $12$ by $4$ .

$x \leq 3$

Step 4

Find the domain of $- 2 \sqrt{x + 1} + 4$ .

Set the radicand in $\sqrt{x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x + 1 \geq 0$

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

The domain is all values of $x$ that make the expression defined.

$[- 1, \infty)$

Step 5

The solution consists of all of the true intervals.

$x \geq 3$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x \geq 3$

Interval Notation:

$[3, \infty)$

Step 7

Solve the Inequality for x -2 square root of x+1<=-4

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