Solve the Inequality for a 3+ square root of 4a-5<=10

$3 + \sqrt{4 a - 5} \leq 10$

Move all terms not containing $\sqrt{4 a - 5}$ to the right side of the inequality.

Subtract $3$ from both sides of the inequality.

$\sqrt{4 a - 5} \leq 10 - 3$

Subtract $3$ from $10$ .

$\sqrt{4 a - 5} \leq 7$

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

${\sqrt{4 a - 5}}^{2} \leq 7^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 a - 5}$ as ${(4 a - 5)}^{\frac{1}{2}}$ .

${({(4 a - 5)}^{\frac{1}{2}})}^{2} \leq 7^{2}$

Simplify the left side.

Simplify ${({(4 a - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(4 a - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(4 a - 5)}^{\frac{1}{2} \cdot 2} \leq 7^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(4 a - 5)}^{\frac{1}{2} \cdot 2} \leq 7^{2}$

Rewrite the expression.

${(4 a - 5)}^{1} \leq 7^{2}$

Simplify.

$4 a - 5 \leq 7^{2}$

Simplify the right side.

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

$4 a - 5 \leq 49$

Solve for $a$ .

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

Add $5$ to both sides of the inequality.

$4 a \leq 49 + 5$

Add $49$ and $5$ .

$4 a \leq 54$

Divide each term in $4 a \leq 54$ by $4$ and simplify.

Divide each term in $4 a \leq 54$ by $4$ .

$\frac{4 a}{4} \leq \frac{54}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 a}{4} \leq \frac{54}{4}$

Divide $a$ by $1$ .

$a \leq \frac{54}{4}$

Simplify the right side.

Cancel the common factor of $54$ and $4$ .

Factor $2$ out of $54$ .

$a \leq \frac{2 (27)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$a \leq \frac{2 \cdot 27}{2 \cdot 2}$

Cancel the common factor.

$a \leq \frac{2 \cdot 27}{2 \cdot 2}$

Rewrite the expression.

$a \leq \frac{27}{2}$

Find the domain of $- 7 + \sqrt{4 a - 5}$ .

Set the radicand in $\sqrt{4 a - 5}$ greater than or equal to $0$ to find where the expression is defined.

$4 a - 5 \geq 0$

Solve for $a$ .

Add $5$ to both sides of the inequality.

$4 a \geq 5$

Divide each term in $4 a \geq 5$ by $4$ and simplify.

Divide each term in $4 a \geq 5$ by $4$ .

$\frac{4 a}{4} \geq \frac{5}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 a}{4} \geq \frac{5}{4}$

Divide $a$ by $1$ .

$a \geq \frac{5}{4}$

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

$[\frac{5}{4}, \infty)$

Use each root to create test intervals.

$a < \frac{5}{4}$

$\frac{5}{4} < a < \frac{27}{2}$

$a > \frac{27}{2}$

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 $a < \frac{5}{4}$ to see if it makes the inequality true.

Choose a value on the interval $a < \frac{5}{4}$ and see if this value makes the original inequality true.

$a = 0$

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

$3 + \sqrt{4 (0) - 5} \leq 10$

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 $\frac{5}{4} < a < \frac{27}{2}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{5}{4} < a < \frac{27}{2}$ and see if this value makes the original inequality true.

$a = 4$

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

$3 + \sqrt{4 (4) - 5} \leq 10$

The left side $6.31662479$ is less than the right side $10$ , which means that the given statement is always true.

True

Test a value on the interval $a > \frac{27}{2}$ to see if it makes the inequality true.

Choose a value on the interval $a > \frac{27}{2}$ and see if this value makes the original inequality true.

$a = 16$

Replace $a$ with $16$ in the original inequality.

$3 + \sqrt{4 (16) - 5} \leq 10$

The left side $10.68114574$ is greater than the right side $10$ , which means that the given statement is false.

False

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

$a < \frac{5}{4}$ False

$\frac{5}{4} < a < \frac{27}{2}$ True

$a > \frac{27}{2}$ False

$a < \frac{5}{4}$ False

$\frac{5}{4} < a < \frac{27}{2}$ True

$a > \frac{27}{2}$ False

The solution consists of all of the true intervals.

$\frac{5}{4} \leq a \leq \frac{27}{2}$

The result can be shown in multiple forms.

Inequality Form:

$\frac{5}{4} \leq a \leq \frac{27}{2}$

Interval Notation:

$[\frac{5}{4}, \frac{27}{2}]$

Solve the Inequality for a 3+ square root of 4a-5<=10

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