Graph square root of 7x+14-3>=4

$\sqrt{7 x + 14} - 3 \geq 4$

Move all terms not containing $\sqrt{7 x + 14}$ to the right side of the inequality.

Add $3$ to both sides of the inequality.

$\sqrt{7 x + 14} \geq 4 + 3$

Add $4$ and $3$ .

$\sqrt{7 x + 14} \geq 7$

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

${\sqrt{7 x + 14}}^{2} \geq 7^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7 x + 14}$ as ${(7 x + 14)}^{\frac{1}{2}}$ .

${({(7 x + 14)}^{\frac{1}{2}})}^{2} \geq 7^{2}$

Simplify the left side.

Simplify ${({(7 x + 14)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(7 x + 14)}^{\frac{1}{2}})}^{2}$ .

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

${(7 x + 14)}^{\frac{1}{2} \cdot 2} \geq 7^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(7 x + 14)}^{\frac{1}{2} \cdot 2} \geq 7^{2}$

Rewrite the expression.

${(7 x + 14)}^{1} \geq 7^{2}$

Simplify.

$7 x + 14 \geq 7^{2}$

Simplify the right side.

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

$7 x + 14 \geq 49$

Solve for $x$ .

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

Subtract $14$ from both sides of the inequality.

$7 x \geq 49 - 14$

Subtract $14$ from $49$ .

$7 x \geq 35$

Divide each term in $7 x \geq 35$ by $7$ and simplify.

Divide each term in $7 x \geq 35$ by $7$ .

$\frac{7 x}{7} \geq \frac{35}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} \geq \frac{35}{7}$

Divide $x$ by $1$ .

$x \geq \frac{35}{7}$

Simplify the right side.

Divide $35$ by $7$ .

$x \geq 5$

Find the domain of $\sqrt{7 (x + 2)} - 7$ .

Set the radicand in $\sqrt{7 (x + 2)}$ greater than or equal to $0$ to find where the expression is defined.

$7 (x + 2) \geq 0$

Solve for $x$ .

Divide each term in $7 (x + 2) \geq 0$ by $7$ and simplify.

Divide each term in $7 (x + 2) \geq 0$ by $7$ .

$\frac{7 (x + 2)}{7} \geq \frac{0}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 (x + 2)}{7} \geq \frac{0}{7}$

Divide $x + 2$ by $1$ .

$x + 2 \geq \frac{0}{7}$

Simplify the right side.

Divide $0$ by $7$ .

$x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

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

$[- 2, \infty)$

The solution consists of all of the true intervals.

$x \geq 5$

Graph square root of 7x+14-3>=4

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