Solve the Inequality for x square root of 2- square root of x+6<=- square root of x

$\sqrt{2} - \sqrt{x + 6} \leq - \sqrt{x}$

Subtract $\sqrt{2}$ from both sides of the inequality.

$- \sqrt{x + 6} \leq - \sqrt{x} - \sqrt{2}$

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

${(- \sqrt{x + 6})}^{2} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 6}$ as ${(x + 6)}^{\frac{1}{2}}$ .

${(- {(x + 6)}^{\frac{1}{2}})}^{2} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Simplify the left side.

Simplify ${(- {(x + 6)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(x + 6)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(x + 6)}^{\frac{1}{2}})}^{2} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(x + 6)}^{\frac{1}{2}})}^{2} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Multiply ${({(x + 6)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(x + 6)}^{\frac{1}{2}})}^{2} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Multiply the exponents in ${({(x + 6)}^{\frac{1}{2}})}^{2}$ .

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

${(x + 6)}^{\frac{1}{2} \cdot 2} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 6)}^{\frac{1}{2} \cdot 2} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Rewrite the expression.

${(x + 6)}^{1} \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Simplify.

$x + 6 \leq {(- \sqrt{x} - \sqrt{2})}^{2}$

Simplify the right side.

Simplify ${(- \sqrt{x} - \sqrt{2})}^{2}$ .

Rewrite ${(- \sqrt{x} - \sqrt{2})}^{2}$ as $(- \sqrt{x} - \sqrt{2}) (- \sqrt{x} - \sqrt{2})$ .

$x + 6 \leq (- \sqrt{x} - \sqrt{2}) (- \sqrt{x} - \sqrt{2})$

Expand $(- \sqrt{x} - \sqrt{2}) (- \sqrt{x} - \sqrt{2})$ using the FOIL Method.

Apply the distributive property.

$x + 6 \leq - \sqrt{x} (- \sqrt{x} - \sqrt{2}) - \sqrt{2} (- \sqrt{x} - \sqrt{2})$

Apply the distributive property.

$x + 6 \leq - \sqrt{x} (- \sqrt{x}) - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x} - \sqrt{2})$

Apply the distributive property.

$x + 6 \leq - \sqrt{x} (- \sqrt{x}) - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Simplify and combine like terms.

Simplify each term.

Multiply $- \sqrt{x} (- \sqrt{x})$ .

Multiply $- 1$ by $- 1$ .

$x + 6 \leq 1 \sqrt{x} \sqrt{x} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Multiply $\sqrt{x}$ by $1$ .

$x + 6 \leq \sqrt{x} \sqrt{x} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Raise $\sqrt{x}$ to the power of $1$ .

$x + 6 \leq {\sqrt{x}}^{1} \sqrt{x} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Raise $\sqrt{x}$ to the power of $1$ .

$x + 6 \leq {\sqrt{x}}^{1} {\sqrt{x}}^{1} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x + 6 \leq {\sqrt{x}}^{1 + 1} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Add $1$ and $1$ .

$x + 6 \leq {\sqrt{x}}^{2} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

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

$x + 6 \leq {(x^{\frac{1}{2}})}^{2} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

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

$x + 6 \leq x^{\frac{1}{2} \cdot 2} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Combine $\frac{1}{2}$ and $2$ .

$x + 6 \leq x^{\frac{2}{2}} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Cancel the common factor of $2$ .

Cancel the common factor.

$x + 6 \leq x^{\frac{2}{2}} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Rewrite the expression.

$x + 6 \leq x^{1} - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Simplify.

$x + 6 \leq x - \sqrt{x} (- \sqrt{2}) - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Multiply $- \sqrt{x} (- \sqrt{2})$ .

Multiply $- 1$ by $- 1$ .

$x + 6 \leq x + 1 \sqrt{x} \sqrt{2} - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Multiply $\sqrt{x}$ by $1$ .

$x + 6 \leq x + \sqrt{x} \sqrt{2} - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Combine using the product rule for radicals.

$x + 6 \leq x + \sqrt{x \cdot 2} - \sqrt{2} (- \sqrt{x}) - \sqrt{2} (- \sqrt{2})$

Multiply $- \sqrt{2} (- \sqrt{x})$ .

Multiply $- 1$ by $- 1$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + 1 \sqrt{2} \sqrt{x} - \sqrt{2} (- \sqrt{2})$

Multiply $\sqrt{2}$ by $1$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2} \sqrt{x} - \sqrt{2} (- \sqrt{2})$

Combine using the product rule for radicals.

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} - \sqrt{2} (- \sqrt{2})$

Multiply $- \sqrt{2} (- \sqrt{2})$ .

Multiply $- 1$ by $- 1$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + 1 \sqrt{2} \sqrt{2}$

Multiply $\sqrt{2}$ by $1$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + \sqrt{2} \sqrt{2}$

Raise $\sqrt{2}$ to the power of $1$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + {\sqrt{2}}^{1} \sqrt{2}$

Raise $\sqrt{2}$ to the power of $1$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + {\sqrt{2}}^{1} {\sqrt{2}}^{1}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + {\sqrt{2}}^{1 + 1}$

Add $1$ and $1$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + {\sqrt{2}}^{2}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + {(2^{\frac{1}{2}})}^{2}$

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

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + 2^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + 2^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + 2^{\frac{2}{2}}$

Rewrite the expression.

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + 2^{1}$

Evaluate the exponent.

$x + 6 \leq x + \sqrt{x \cdot 2} + \sqrt{2 x} + 2$

Add $\sqrt{x \cdot 2}$ and $\sqrt{2 x}$ .

Reorder $x$ and $2$ .

$x + 6 \leq x + \sqrt{2 \cdot x} + \sqrt{2 x} + 2$

Add $\sqrt{2 \cdot x}$ and $\sqrt{2 x}$ .

$x + 6 \leq x + 2 \sqrt{2 \cdot x} + 2$

Solve for $2 \sqrt{2 \cdot x}$ .

Rewrite so $2 \sqrt{2 \cdot x}$ is on the left side of the inequality.

$x + 2 \sqrt{2 \cdot x} + 2 \geq x + 6$

Move all terms not containing $2 \sqrt{2 \cdot x}$ to the right side of the inequality.

Subtract $x$ from both sides of the inequality.

$2 \sqrt{2 \cdot x} + 2 \geq x + 6 - x$

Subtract $2$ from both sides of the inequality.

$2 \sqrt{2 \cdot x} \geq x + 6 - x - 2$

Combine the opposite terms in $x + 6 - x - 2$ .

Subtract $x$ from $x$ .

$2 \sqrt{2 \cdot x} \geq 0 + 6 - 2$

Add $0$ and $6$ .

$2 \sqrt{2 \cdot x} \geq 6 - 2$

Subtract $2$ from $6$ .

$2 \sqrt{2 \cdot x} \geq 4$

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

${(2 \sqrt{2 \cdot x})}^{2} \geq 4^{2}$

Simplify each side of the inequality.

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

${(2 {(2 \cdot x)}^{\frac{1}{2}})}^{2} \geq 4^{2}$

Simplify the left side.

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

Apply the product rule to $2 x$ .

${(2 (2^{\frac{1}{2}} x^{\frac{1}{2}}))}^{2} \geq 4^{2}$

Multiply $2$ by $2^{\frac{1}{2}}$ by adding the exponents.

Move $2^{\frac{1}{2}}$ .

${(2^{\frac{1}{2}} \cdot 2 x^{\frac{1}{2}})}^{2} \geq 4^{2}$

Multiply $2^{\frac{1}{2}}$ by $2$ .

Raise $2$ to the power of $1$ .

${(2^{\frac{1}{2}} \cdot 2^{1} x^{\frac{1}{2}})}^{2} \geq 4^{2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Add $1$ and $2$ .

${(2^{\frac{3}{2}} x^{\frac{1}{2}})}^{2} \geq 4^{2}$

Apply the product rule to $2^{\frac{3}{2}} x^{\frac{1}{2}}$ .

${(2^{\frac{3}{2}})}^{2} {(x^{\frac{1}{2}})}^{2} \geq 4^{2}$

Multiply the exponents in ${(2^{\frac{3}{2}})}^{2}$ .

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

$2^{\frac{3}{2} \cdot 2} {(x^{\frac{1}{2}})}^{2} \geq 4^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$2^{\frac{3}{2} \cdot 2} {(x^{\frac{1}{2}})}^{2} \geq 4^{2}$

Rewrite the expression.

$2^{3} {(x^{\frac{1}{2}})}^{2} \geq 4^{2}$

Raise $2$ to the power of $3$ .

$8 {(x^{\frac{1}{2}})}^{2} \geq 4^{2}$

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

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

$8 x^{\frac{1}{2} \cdot 2} \geq 4^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$8 x^{\frac{1}{2} \cdot 2} \geq 4^{2}$

Rewrite the expression.

$8 x^{1} \geq 4^{2}$

Simplify.

$8 x \geq 4^{2}$

Simplify the right side.

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

$8 x \geq 16$

Divide each term in $8 x \geq 16$ by $8$ and simplify.

Divide each term in $8 x \geq 16$ by $8$ .

$\frac{8 x}{8} \geq \frac{16}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} \geq \frac{16}{8}$

Divide $x$ by $1$ .

$x \geq \frac{16}{8}$

Simplify the right side.

Divide $16$ by $8$ .

$x \geq 2$

Find the domain of $\sqrt{2} - \sqrt{x + 6} + \sqrt{x}$ .

Set the radicand in $\sqrt{x + 6}$ greater than or equal to $0$ to find where the expression is defined.

$x + 6 \geq 0$

Subtract $6$ from both sides of the inequality.

$x \geq - 6$

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

$x \geq 0$

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

$[0, \infty)$

Use each root to create test intervals.

$x < 0$

$0 < x < 2$

$x > 2$

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 $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$\sqrt{2} - \sqrt{(- 2) + 6} \leq - \sqrt{- 2}$

The left side is not equal to the right side, which means that the given statement is false.

False

Test a value on the interval $0 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

$\sqrt{2} - \sqrt{(1) + 6} \leq - \sqrt{1}$

The left side $- 1.23153774$ is less than the right side $- 1$ , which means that the given statement is always true.

True

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$\sqrt{2} - \sqrt{(4) + 6} \leq - \sqrt{4}$

The left side $- 1.74806409$ 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.

$x < 0$ False

$0 < x < 2$ True

$x > 2$ False

$x < 0$ False

$0 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$0 \leq x \leq 2$

The result can be shown in multiple forms.

Inequality Form:

$0 \leq x \leq 2$

Interval Notation:

$[0, 2]$

Solve the Inequality for x square root of 2- square root of x+6<=- square root of x

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