Algebra Examples

$\sqrt{x^{2} - 2 x} < \sqrt{3 x + 6}$

Step 1

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

${\sqrt{x^{2} - 2 x}}^{2} < {\sqrt{3 x + 6}}^{2}$

Step 2

Simplify each side of the inequality.

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Step 2.1

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

${({(x^{2} - 2 x)}^{\frac{1}{2}})}^{2} < {\sqrt{3 x + 6}}^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

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

${(x^{2} - 2 x)}^{\frac{1}{2} \cdot 2} < {\sqrt{3 x + 6}}^{2}$

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

${(x^{2} - 2 x)}^{\frac{1}{2} \cdot 2} < {\sqrt{3 x + 6}}^{2}$

Step 2.2.1.1.2.2

Rewrite the expression.

${(x^{2} - 2 x)}^{1} < {\sqrt{3 x + 6}}^{2}$

Step 2.2.1.2

Simplify.

$x^{2} - 2 x < {\sqrt{3 x + 6}}^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify ${\sqrt{3 x + 6}}^{2}$ .

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Step 2.3.1.1

Factor $3$ out of $3 x + 6$ .

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Step 2.3.1.1.1

Factor $3$ out of $3 x$ .

$x^{2} - 2 x < {\sqrt{3 (x) + 6}}^{2}$

Step 2.3.1.1.2

Factor $3$ out of $6$ .

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

Step 2.3.1.1.3

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

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

Step 2.3.1.2

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

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

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

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

Step 2.3.1.2.3

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

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

Step 2.3.1.2.4

Cancel the common factor of $2$ .

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Step 2.3.1.2.4.1

Cancel the common factor.

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

Step 2.3.1.2.4.2

Rewrite the expression.

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

Step 2.3.1.2.5

Simplify.

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

Step 2.3.1.3

Apply the distributive property.

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

Step 2.3.1.4

Multiply $3$ by $2$ .

$x^{2} - 2 x < 3 x + 6$

Step 3

Solve for $x$ .

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Step 3.1

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

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Step 3.1.1

Subtract $3 x$ from both sides of the inequality.

$x^{2} - 2 x - 3 x < 6$

Step 3.1.2

Subtract $3 x$ from $- 2 x$ .

$x^{2} - 5 x < 6$

Step 3.2

Convert the inequality to an equation.

$x^{2} - 5 x = 6$

Step 3.3

Subtract $6$ from both sides of the equation.

$x^{2} - 5 x - 6 = 0$

Step 3.4

Factor $x^{2} - 5 x - 6$ using the AC method.

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Step 3.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Step 3.4.2

Write the factored form using these integers.

$(x - 6) (x + 1) = 0$

Step 3.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 6 = 0$

$x + 1 = 0$

Step 3.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 3.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 3.7

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.8

The final solution is all the values that make $(x - 6) (x + 1) = 0$ true.

$x = 6, - 1$

Step 4

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

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Step 4.1

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

$x (x - 2) \geq 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 2 = 0$

Step 4.2.2

Set $x$ equal to $0$ .

$x = 0$

Step 4.2.3

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 4.2.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 4.2.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.2.4

The final solution is all the values that make $x (x - 2) \geq 0$ true.

$x = 0, 2$

Step 4.2.5

Use each root to create test intervals.

$x < 0$

$0 < x < 2$

$x > 2$

Step 4.2.6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 4.2.6.1

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

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Step 4.2.6.1.1

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

$x = - 2$

Step 4.2.6.1.2

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

$(- 2) ((- 2) - 2) \geq 0$

Step 4.2.6.1.3

The left side $8$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.2.6.2

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

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Step 4.2.6.2.1

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

$x = 1$

Step 4.2.6.2.2

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

$(1) ((1) - 2) \geq 0$

Step 4.2.6.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 4.2.6.3

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

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Step 4.2.6.3.1

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

$x = 4$

Step 4.2.6.3.2

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

$(4) ((4) - 2) \geq 0$

Step 4.2.6.3.3

The left side $8$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.2.6.4

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

$x < 0$ True

$0 < x < 2$ False

$x > 2$ True

$x < 0$ True

$0 < x < 2$ False

$x > 2$ True

Step 4.2.7

The solution consists of all of the true intervals.

$x \leq 0$ or $x \geq 2$

Step 4.3

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

$3 (x + 2) \geq 0$

Step 4.4

Solve for $x$ .

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Step 4.4.1

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

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Step 4.4.1.1

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

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

Step 4.4.1.2

Simplify the left side.

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Step 4.4.1.2.1

Cancel the common factor of $3$ .

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Step 4.4.1.2.1.1

Cancel the common factor.

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

Step 4.4.1.2.1.2

Divide $x + 2$ by $1$ .

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

Step 4.4.1.3

Simplify the right side.

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Step 4.4.1.3.1

Divide $0$ by $3$ .

$x + 2 \geq 0$

Step 4.4.2

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

Step 4.5

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

$[- 2, 0] \cup [2, \infty)$

Step 5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - 1$

$- 1 < x < 0$

$0 < x < 2$

$2 < x < 6$

$x > 6$

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 6.1

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

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Step 6.1.1

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

$x = - 4$

Step 6.1.2

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

$\sqrt{{(- 4)}^{2} - 2 \cdot - 4} < \sqrt{3 (- 4) + 6}$

Step 6.1.3

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

False

Step 6.2

Test a value on the interval $- 2 < x < - 1$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $- 2 < x < - 1$ and see if this value makes the original inequality true.

$x = - 1.5$

Step 6.2.2

Replace $x$ with $- 1.5$ in the original inequality.

$\sqrt{{(- 1.5)}^{2} - 2 \cdot - 1.5} < \sqrt{3 (- 1.5) + 6}$

Step 6.2.3

The left side $2.29128784$ is not less than the right side $1.22474487$ , which means that the given statement is false.

False

Step 6.3

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 6.3.2

Replace $x$ with $- 0.5$ in the original inequality.

$\sqrt{{(- 0.5)}^{2} - 2 \cdot - 0.5} < \sqrt{3 (- 0.5) + 6}$

Step 6.3.3

The left side $1.11803398$ is less than the right side $2.12132034$ , which means that the given statement is always true.

True

Step 6.4

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

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Step 6.4.1

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

$x = 1$

Step 6.4.2

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

$\sqrt{{(1)}^{2} - 2 \cdot 1} < \sqrt{3 (1) + 6}$

Step 6.4.3

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

False

Step 6.5

Test a value on the interval $2 < x < 6$ to see if it makes the inequality true.

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Step 6.5.1

Choose a value on the interval $2 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

Step 6.5.2

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

$\sqrt{{(4)}^{2} - 2 \cdot 4} < \sqrt{3 (4) + 6}$

Step 6.5.3

The left side $2.82842712$ is less than the right side $4.24264068$ , which means that the given statement is always true.

True

Step 6.6

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 6.6.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 6.6.2

Replace $x$ with $8$ in the original inequality.

$\sqrt{{(8)}^{2} - 2 \cdot 8} < \sqrt{3 (8) + 6}$

Step 6.6.3

The left side $6.92820323$ is not less than the right side $5.47722557$ , which means that the given statement is false.

False

Step 6.7

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

$x < - 2$ False

$- 2 < x < - 1$ False

$- 1 < x < 0$ True

$0 < x < 2$ False

$2 < x < 6$ True

$x > 6$ False

$x < - 2$ False

$- 2 < x < - 1$ False

$- 1 < x < 0$ True

$0 < x < 2$ False

$2 < x < 6$ True

$x > 6$ False

Step 7

The solution consists of all of the true intervals.

$- 1 < x < 0$ or $2 < x < 6$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < 0 or 2 < x < 6$

Interval Notation:

$(- 1, 0) \cup (2, 6)$

Step 9