Evaluate 7 square root of x+1<9

$7 \sqrt{x} + 1 < 9$

Step 1

Move all terms not containing $7 \sqrt{x}$ to the right side of the inequality.

Subtract $1$ from both sides of the inequality.

$7 \sqrt{x} < 9 - 1$

Subtract $1$ from $9$ .

$7 \sqrt{x} < 8$

Step 2

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

${(7 \sqrt{x})}^{2} < 8^{2}$

Step 3

Simplify each side of the inequality.

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

${(7 x^{\frac{1}{2}})}^{2} < 8^{2}$

Simplify the left side.

Simplify ${(7 x^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $7 x^{\frac{1}{2}}$ .

$7^{2} {(x^{\frac{1}{2}})}^{2} < 8^{2}$

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

$49 {(x^{\frac{1}{2}})}^{2} < 8^{2}$

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

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

$49 x^{\frac{1}{2} \cdot 2} < 8^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$49 x^{\frac{1}{2} \cdot 2} < 8^{2}$

Rewrite the expression.

$49 x^{1} < 8^{2}$

Simplify.

$49 x < 8^{2}$

Simplify the right side.

Raise $8$ to the power of $2$ .

$49 x < 64$

Step 4

Divide each term in $49 x < 64$ by $49$ and simplify.

Divide each term in $49 x < 64$ by $49$ .

$\frac{49 x}{49} < \frac{64}{49}$

Simplify the left side.

Cancel the common factor of $49$ .

Cancel the common factor.

$\frac{49 x}{49} < \frac{64}{49}$

Divide $x$ by $1$ .

$x < \frac{64}{49}$

Step 5

Find the domain of $7 \sqrt{x} - 8$ .

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

$x \geq 0$

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

$[0, \infty)$

Step 6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{64}{49}$

$x > \frac{64}{49}$

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 $x < 0$ to see if it makes the inequality true.

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

$x = - 2$

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

$7 \sqrt{- 2} + 1 < 9$

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 $0 < x < \frac{64}{49}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{64}{49}$ and see if this value makes the original inequality true.

$x = 1$

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

$7 \sqrt{1} + 1 < 9$

The left side $8$ is less than the right side $9$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{64}{49}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{64}{49}$ and see if this value makes the original inequality true.

$x = 4$

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

$7 \sqrt{4} + 1 < 9$

The left side $15$ is not less 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.

$x < 0$ False

$0 < x < \frac{64}{49}$ True

$x > \frac{64}{49}$ False

$x < 0$ False

$0 < x < \frac{64}{49}$ True

$x > \frac{64}{49}$ False

Step 8

The solution consists of all of the true intervals.

$0 < x < \frac{64}{49}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$0 < x < \frac{64}{49}$

Interval Notation:

$(0, \frac{64}{49})$

Step 10

Evaluate 7 square root of x+1<9

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