Solve the Inequality for x square root of x-3<=7

$\sqrt{x - 3} \leq 7$

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

${\sqrt{x - 3}}^{2} \leq 7^{2}$

Simplify each side of the inequality.

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

${({(x - 3)}^{\frac{1}{2}})}^{2} \leq 7^{2}$

Simplify the left side.

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

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

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

${(x - 3)}^{\frac{1}{2} \cdot 2} \leq 7^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x - 3)}^{\frac{1}{2} \cdot 2} \leq 7^{2}$

Rewrite the expression.

${(x - 3)}^{1} \leq 7^{2}$

Simplify.

$x - 3 \leq 7^{2}$

Simplify the right side.

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

$x - 3 \leq 49$

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

Add $3$ to both sides of the inequality.

$x \leq 49 + 3$

Add $49$ and $3$ .

$x \leq 52$

Find the domain of $\sqrt{x - 3} - 7$ .

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

$x - 3 \geq 0$

Add $3$ to both sides of the inequality.

$x \geq 3$

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

$[3, \infty)$

Use each root to create test intervals.

$x < 3$

$3 < x < 52$

$x > 52$

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.

$x = 0$

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

$\sqrt{(0) - 3} \leq 7$

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 $3 < x < 52$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 52$ and see if this value makes the original inequality true.

$x = 6$

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

$\sqrt{(6) - 3} \leq 7$

The left side $1.7320508$ is less than the right side $7$ , which means that the given statement is always true.

True

Test a value on the interval $x > 52$ to see if it makes the inequality true.

Choose a value on the interval $x > 52$ and see if this value makes the original inequality true.

$x = 54$

Replace $x$ with $54$ in the original inequality.

$\sqrt{(54) - 3} \leq 7$

The left side $7.14142842$ is greater than the right side $7$ , which means that the given statement is false.

False

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

$x < 3$ False

$3 < x < 52$ True

$x > 52$ False

$x < 3$ False

$3 < x < 52$ True

$x > 52$ False

The solution consists of all of the true intervals.

$3 \leq x \leq 52$

The result can be shown in multiple forms.

Inequality Form:

$3 \leq x \leq 52$

Interval Notation:

$[3, 52]$

Solve the Inequality for x square root of x-3<=7

Related Questions

Support LCMGCF.com by disabling your adblocker.