Solve for x square root of x^2-6x+9+2<=9

$\sqrt{x^{2} - 6 x + 9} + 2 \leq 9$

Step 1

Move all terms not containing $\sqrt{x^{2} - 6 x + 9}$ to the right side of the inequality.

Subtract $2$ from both sides of the inequality.

$\sqrt{x^{2} - 6 x + 9} \leq 9 - 2$

Subtract $2$ from $9$ .

$\sqrt{x^{2} - 6 x + 9} \leq 7$

Step 2

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

${\sqrt{x^{2} - 6 x + 9}}^{2} \leq 7^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} - 6 x + 9}$ as ${(x^{2} - 6 x + 9)}^{\frac{1}{2}}$ .

${({(x^{2} - 6 x + 9)}^{\frac{1}{2}})}^{2} \leq 7^{2}$

Simplify the left side.

Simplify ${({(x^{2} - 6 x + 9)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(x^{2} - 6 x + 9)}^{\frac{1}{2}})}^{2}$ .

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

${(x^{2} - 6 x + 9)}^{\frac{1}{2} \cdot 2} \leq 7^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x^{2} - 6 x + 9)}^{\frac{1}{2} \cdot 2} \leq 7^{2}$

Rewrite the expression.

${(x^{2} - 6 x + 9)}^{1} \leq 7^{2}$

Simplify.

$x^{2} - 6 x + 9 \leq 7^{2}$

Simplify the right side.

Raise $7$ to the power of $2$ .

$x^{2} - 6 x + 9 \leq 49$

Step 4

Solve for $x$ .

Convert the inequality to an equation.

$x^{2} - 6 x + 9 = 49$

Subtract $49$ from both sides of the equation.

$x^{2} - 6 x + 9 - 49 = 0$

Subtract $49$ from $9$ .

$x^{2} - 6 x - 40 = 0$

Factor $x^{2} - 6 x - 40$ using the AC method.

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 $- 40$ and whose sum is $- 6$ .

$- 10, 4$

Write the factored form using these integers.

$(x - 10) (x + 4) = 0$

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

$x - 10 = 0$

$x + 4 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 10) (x + 4) = 0$ true.

$x = 10, - 4$

Step 5

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 10$

$x > 10$

Step 6

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 $x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

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

$\sqrt{{(- 6)}^{2} - 6 \cdot - 6 + 9} + 2 \leq 9$

The left side $11$ is greater than the right side $9$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 10$ and see if this value makes the original inequality true.

$x = 0$

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

$\sqrt{{(0)}^{2} - 6 \cdot 0 + 9} + 2 \leq 9$

The left side $5$ is less than the right side $9$ , which means that the given statement is always true.

True

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$\sqrt{{(12)}^{2} - 6 \cdot 12 + 9} + 2 \leq 9$

The left side $11$ is greater than the right side $9$ , which means that the given statement is false.

False

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

$x < - 4$ False

$- 4 < x < 10$ True

$x > 10$ False

$x < - 4$ False

$- 4 < x < 10$ True

$x > 10$ False

Step 7

The solution consists of all of the true intervals.

$- 4 \leq x \leq 10$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq x \leq 10$

Interval Notation:

$[- 4, 10]$

Step 9

Solve for x square root of x^2-6x+9+2<=9

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