Solve the Inequality for x x^2<=8x-13

$x^{2} \leq 8 x - 13$

Subtract $8 x$ from both sides of the inequality.

$x^{2} - 8 x \leq - 13$

Add $13$ to both sides of the inequality.

$x^{2} - 8 x + 13 \leq 0$

Convert the inequality to an equation.

$x^{2} - 8 x + 13 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 8$ , and $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 13}}{2 \cdot 1}$

Multiply $- 4$ by $13$ .

$x = \frac{8 \pm \sqrt{64 - 52}}{2 \cdot 1}$

Subtract $52$ from $64$ .

$x = \frac{8 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{8 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{8 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{8 \pm 2 \sqrt{3}}{2}$ .

$x = 4 \pm \sqrt{3}$

Consolidate the solutions.

$x = 4 + \sqrt{3}, 4 - \sqrt{3}$

Use each root to create test intervals.

$x < 4 - \sqrt{3}$

$4 - \sqrt{3} < x < 4 + \sqrt{3}$

$x > 4 + \sqrt{3}$

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 $x < 4 - \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < 4 - \sqrt{3}$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} \leq 8 (0) - 13$

The left side $0$ is greater than the right side $- 13$ , which means that the given statement is false.

False

Test a value on the interval $4 - \sqrt{3} < x < 4 + \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $4 - \sqrt{3} < x < 4 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 4$

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

${(4)}^{2} \leq 8 (4) - 13$

The left side $16$ is less than the right side $19$ , which means that the given statement is always true.

True

Test a value on the interval $x > 4 + \sqrt{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > 4 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 8$

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

${(8)}^{2} \leq 8 (8) - 13$

The left side $64$ is greater than the right side $51$ , which means that the given statement is false.

False

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

$x < 4 - \sqrt{3}$ False

$4 - \sqrt{3} < x < 4 + \sqrt{3}$ True

$x > 4 + \sqrt{3}$ False

$x < 4 - \sqrt{3}$ False

$4 - \sqrt{3} < x < 4 + \sqrt{3}$ True

$x > 4 + \sqrt{3}$ False

The solution consists of all of the true intervals.

$4 - \sqrt{3} \leq x \leq 4 + \sqrt{3}$

The result can be shown in multiple forms.

Inequality Form:

$4 - \sqrt{3} \leq x \leq 4 + \sqrt{3}$

Interval Notation:

$[4 - \sqrt{3}, 4 + \sqrt{3}]$

Solve the Inequality for x x^2<=8x-13

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