Solve the Inequality for x square root of x-2-1<=0

$\sqrt{x - 2} - 1 \leq 0$

Step 1

Add $1$ to both sides of the inequality.

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

Step 2

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

${\sqrt{x - 2}}^{2} \leq 1^{2}$

Step 3

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$x - 2 \leq 1^{2}$

Simplify the right side.

One to any power is one.

$x - 2 \leq 1$

Step 4

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

Add $2$ to both sides of the inequality.

$x \leq 1 + 2$

Add $1$ and $2$ .

$x \leq 3$

Step 5

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

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

$x - 2 \geq 0$

Add $2$ to both sides of the inequality.

$x \geq 2$

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

$[2, \infty)$

Step 6

Use each root to create test intervals.

$x < 2$

$2 < x < 3$

$x > 3$

Step 7

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 $x < 2$ 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.

$x = 0$

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

$\sqrt{(0) - 2} - 1 \leq 0$

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 $2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Replace $x$ with $2.5$ in the original inequality.

$\sqrt{(2.5) - 2} - 1 \leq 0$

The left side $- 0.29289321$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

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

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

False

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

$x < 2$ False

$2 < x < 3$ True

$x > 3$ False

$x < 2$ False

$2 < x < 3$ True

$x > 3$ False

Step 8

The solution consists of all of the true intervals.

$2 \leq x \leq 3$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$2 \leq x \leq 3$

Interval Notation:

$[2, 3]$

Step 10

Solve the Inequality for x square root of x-2-1<=0

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