Solve the Inequality for x square root of 1-x< square root of 2x+6

$\sqrt{1 - x} < \sqrt{2 x + 6}$

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

${\sqrt{1 - x}}^{2} < {\sqrt{2 x + 6}}^{2}$

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$1 - x < {\sqrt{2 x + 6}}^{2}$

Simplify the right side.

Simplify ${\sqrt{2 x + 6}}^{2}$ .

Factor $2$ out of $2 x + 6$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $6$ .

$1 - x < {\sqrt{2 x + 2 \cdot 3}}^{2}$

Factor $2$ out of $2 x + 2 \cdot 3$ .

$1 - x < {\sqrt{2 (x + 3)}}^{2}$

Rewrite ${\sqrt{2 (x + 3)}}^{2}$ as $2 (x + 3)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 (x + 3)}$ as ${(2 (x + 3))}^{\frac{1}{2}}$ .

$1 - x < {({(2 (x + 3))}^{\frac{1}{2}})}^{2}$

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

$1 - x < {(2 (x + 3))}^{\frac{1}{2} \cdot 2}$

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

$1 - x < {(2 (x + 3))}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$1 - x < {(2 (x + 3))}^{\frac{2}{2}}$

Rewrite the expression.

$1 - x < {(2 (x + 3))}^{1}$

Simplify.

$1 - x < 2 (x + 3)$

Apply the distributive property.

$1 - x < 2 x + 2 \cdot 3$

Multiply $2$ by $3$ .

$1 - x < 2 x + 6$

Solve for $x$ .

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

Subtract $2 x$ from both sides of the inequality.

$1 - x - 2 x < 6$

Subtract $2 x$ from $- x$ .

$1 - 3 x < 6$

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

Subtract $1$ from both sides of the inequality.

$- 3 x < 6 - 1$

Subtract $1$ from $6$ .

$- 3 x < 5$

Divide each term in $- 3 x < 5$ by $- 3$ and simplify.

Divide each term in $- 3 x < 5$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} > \frac{5}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} > \frac{5}{- 3}$

Divide $x$ by $1$ .

$x > \frac{5}{- 3}$

Simplify the right side.

Move the negative in front of the fraction.

$x > - \frac{5}{3}$

Find the domain of $\sqrt{1 - x} - \sqrt{2 (x + 3)}$ .

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

$1 - x \geq 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$- x \geq - 1$

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 $- 1$ .

$x \leq 1$

Set the radicand in $\sqrt{2 (x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$2 (x + 3) \geq 0$

Solve for $x$ .

Divide each term in $2 (x + 3) \geq 0$ by $2$ and simplify.

Divide each term in $2 (x + 3) \geq 0$ by $2$ .

$\frac{2 (x + 3)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x + 3)}{2} \geq \frac{0}{2}$

Divide $x + 3$ by $1$ .

$x + 3 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$x \geq - 3$

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

$[- 3, 1]$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - \frac{5}{3}$

$- \frac{5}{3} < x < 1$

$x > 1$

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 $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

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

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 $- 3 < x < - \frac{5}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < - \frac{5}{3}$ and see if this value makes the original inequality true.

$x = - 2$

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

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

The left side $1.7320508$ is not less than the right side $1.41421356$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{5}{3} < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{5}{3} < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$\sqrt{1 - (0)} < \sqrt{2 (0) + 6}$

The left side $1$ is less than the right side $2.44948974$ , which means that the given statement is always true.

True

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

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

$\sqrt{1 - (4)} < \sqrt{2 (4) + 6}$

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

False

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

$x < - 3$ False

$- 3 < x < - \frac{5}{3}$ False

$- \frac{5}{3} < x < 1$ True

$x > 1$ False

$x < - 3$ False

$- 3 < x < - \frac{5}{3}$ False

$- \frac{5}{3} < x < 1$ True

$x > 1$ False

The solution consists of all of the true intervals.

$- \frac{5}{3} < x < 1$

The result can be shown in multiple forms.

Inequality Form:

$- \frac{5}{3} < x < 1$

Interval Notation:

$(- \frac{5}{3}, 1)$

Solve the Inequality for x square root of 1-x< square root of 2x+6

Related Questions

Support LCMGCF.com by disabling your adblocker.