Solve the Absolute Value Inequality for x square root of 4x^2-12x+9<=9

$\sqrt{4 x^{2} - 12 x + 9} \leq 9$

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

${\sqrt{4 x^{2} - 12 x + 9}}^{2} \leq 9^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 x^{2} - 12 x + 9}$ as ${(4 x^{2} - 12 x + 9)}^{\frac{1}{2}}$ .

${({(4 x^{2} - 12 x + 9)}^{\frac{1}{2}})}^{2} \leq 9^{2}$

Simplify the left side.

Simplify ${({(4 x^{2} - 12 x + 9)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(4 x^{2} - 12 x + 9)}^{\frac{1}{2}})}^{2}$ .

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

${(4 x^{2} - 12 x + 9)}^{\frac{1}{2} \cdot 2} \leq 9^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(4 x^{2} - 12 x + 9)}^{\frac{1}{2} \cdot 2} \leq 9^{2}$

Rewrite the expression.

${(4 x^{2} - 12 x + 9)}^{1} \leq 9^{2}$

Simplify.

$4 x^{2} - 12 x + 9 \leq 9^{2}$

Simplify the right side.

Raise $9$ to the power of $2$ .

$4 x^{2} - 12 x + 9 \leq 81$

Solve for $x$ .

Convert the inequality to an equation.

$4 x^{2} - 12 x + 9 = 81$

Subtract $81$ from both sides of the equation.

$4 x^{2} - 12 x + 9 - 81 = 0$

Subtract $81$ from $9$ .

$4 x^{2} - 12 x - 72 = 0$

Factor the left side of the equation.

Factor $4$ out of $4 x^{2} - 12 x - 72$ .

Factor $4$ out of $4 x^{2}$ .

$4 (x^{2}) - 12 x - 72 = 0$

Factor $4$ out of $- 12 x$ .

$4 (x^{2}) + 4 (- 3 x) - 72 = 0$

Factor $4$ out of $- 72$ .

$4 x^{2} + 4 (- 3 x) + 4 \cdot - 18 = 0$

Factor $4$ out of $4 x^{2} + 4 (- 3 x)$ .

$4 (x^{2} - 3 x) + 4 \cdot - 18 = 0$

Factor $4$ out of $4 (x^{2} - 3 x) + 4 \cdot - 18$ .

$4 (x^{2} - 3 x - 18) = 0$

Factor.

Factor $x^{2} - 3 x - 18$ 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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Write the factored form using these integers.

$4 ((x - 6) (x + 3)) = 0$

Remove unnecessary parentheses.

$4 (x - 6) (x + 3) = 0$

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

$x - 6 = 0$

$x + 3 = 0$

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $4 (x - 6) (x + 3) = 0$ true.

$x = 6, - 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 6$

$x > 6$

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 $x < - 3$ 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.

$\sqrt{4 {(- 6)}^{2} - 12 \cdot - 6 + 9} \leq 9$

The left side $15$ is greater than the right side $9$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

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

$\sqrt{4 {(0)}^{2} - 12 \cdot 0 + 9} \leq 9$

The left side $3$ is less than the right side $9$ , which means that the given statement is always true.

True

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

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

$x = 8$

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

$\sqrt{4 {(8)}^{2} - 12 \cdot 8 + 9} \leq 9$

The left side $13$ 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.

$x < - 3$ False

$- 3 < x < 6$ True

$x > 6$ False

$x < - 3$ False

$- 3 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$- 3 \leq x \leq 6$

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq x \leq 6$

Interval Notation:

$[- 3, 6]$

Solve the Absolute Value Inequality for x square root of 4x^2-12x+9<=9

Related Questions

Support LCMGCF.com by disabling your adblocker.