Solve the Inequality for x square root of 3-5x> square root of x

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

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

${\sqrt{3 - 5 x}}^{2} > {\sqrt{x}}^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 - 5 x}$ as ${(3 - 5 x)}^{\frac{1}{2}}$ .

${({(3 - 5 x)}^{\frac{1}{2}})}^{2} > {\sqrt{x}}^{2}$

Simplify the left side.

Simplify ${({(3 - 5 x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 - 5 x)}^{\frac{1}{2}})}^{2}$ .

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

${(3 - 5 x)}^{\frac{1}{2} \cdot 2} > {\sqrt{x}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 - 5 x)}^{\frac{1}{2} \cdot 2} > {\sqrt{x}}^{2}$

Rewrite the expression.

${(3 - 5 x)}^{1} > {\sqrt{x}}^{2}$

Simplify.

$3 - 5 x > {\sqrt{x}}^{2}$

Simplify the right side.

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

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

$3 - 5 x > {(x^{\frac{1}{2}})}^{2}$

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

$3 - 5 x > x^{\frac{1}{2} \cdot 2}$

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

$3 - 5 x > x^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 - 5 x > x^{\frac{2}{2}}$

Rewrite the expression.

$3 - 5 x > x^{1}$

Simplify.

$3 - 5 x > x$

Solve for $x$ .

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

Subtract $x$ from both sides of the inequality.

$3 - 5 x - x > 0$

Subtract $x$ from $- 5 x$ .

$3 - 6 x > 0$

Subtract $3$ from both sides of the inequality.

$- 6 x > - 3$

Divide each term in $- 6 x > - 3$ by $- 6$ and simplify.

Divide each term in $- 6 x > - 3$ by $- 6$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 6 x}{- 6} < \frac{- 3}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 x}{- 6} < \frac{- 3}{- 6}$

Divide $x$ by $1$ .

$x < \frac{- 3}{- 6}$

Simplify the right side.

Cancel the common factor of $- 3$ and $- 6$ .

Factor $- 3$ out of $- 3$ .

$x < \frac{- 3 (1)}{- 6}$

Cancel the common factors.

Factor $- 3$ out of $- 6$ .

$x < \frac{- 3 \cdot 1}{- 3 \cdot 2}$

Cancel the common factor.

$x < \frac{- 3 \cdot 1}{- 3 \cdot 2}$

Rewrite the expression.

$x < \frac{1}{2}$

Find the domain of $\sqrt{3 - 5 x} - \sqrt{x}$ .

Set the radicand in $\sqrt{3 - 5 x}$ greater than or equal to $0$ to find where the expression is defined.

$3 - 5 x \geq 0$

Solve for $x$ .

Subtract $3$ from both sides of the inequality.

$- 5 x \geq - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

$x \leq \frac{3}{5}$

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.

$[0, \frac{3}{5}]$

Use each root to create test intervals.

$x < 0$

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

$\frac{1}{2} < x < \frac{3}{5}$

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

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.

$\sqrt{3 - 5 \cdot - 2} > \sqrt{- 2}$

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 $0 < x < \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0.25$

Replace $x$ with $0.25$ in the original inequality.

$\sqrt{3 - 5 \cdot 0.25} > \sqrt{0.25}$

The left side $1.32287565$ is greater than the right side $0.5$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{1}{2} < x < \frac{3}{5}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{2} < x < \frac{3}{5}$ and see if this value makes the original inequality true.

$x = 0.55$

Replace $x$ with $0.55$ in the original inequality.

$\sqrt{3 - 5 \cdot 0.55} > \sqrt{0.55}$

The left side $0.5$ is not greater than the right side $0.74161984$ , which means that the given statement is false.

False

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

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

$x = 3$

Replace $x$ with $3$ in the original inequality.

$\sqrt{3 - 5 \cdot 3} > \sqrt{3}$

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.

$x < 0$ False

$0 < x < \frac{1}{2}$ True

$\frac{1}{2} < x < \frac{3}{5}$ False

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

$x < 0$ False

$0 < x < \frac{1}{2}$ True

$\frac{1}{2} < x < \frac{3}{5}$ False

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

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

$(0, \frac{1}{2})$

Solve the Inequality for x square root of 3-5x> square root of x

Related Questions

Support LCMGCF.com by disabling your adblocker.