Solve the Inequality for x x^2<=6x-7

$x^{2} \leq 6 x - 7$

Step 1

Subtract $6 x$ from both sides of the inequality.

$x^{2} - 6 x \leq - 7$

Step 2

Add $7$ to both sides of the inequality.

$x^{2} - 6 x + 7 \leq 0$

Step 3

Convert the inequality to an equation.

$x^{2} - 6 x + 7 = 0$

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

Substitute the values $a = 1$ , $b = - 6$ , and $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 7)}}{2 \cdot 1}$

Step 6

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 7}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 7$ .

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 7}}{2 \cdot 1}$

Multiply $- 4$ by $7$ .

$x = \frac{6 \pm \sqrt{36 - 28}}{2 \cdot 1}$

Subtract $28$ from $36$ .

$x = \frac{6 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{6 \pm \sqrt{4 (2)}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{2}}{2}$ .

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

Step 7

Consolidate the solutions.

$x = 3 + \sqrt{2}, 3 - \sqrt{2}$

Step 8

Use each root to create test intervals.

$x < 3 - \sqrt{2}$

$3 - \sqrt{2} < x < 3 + \sqrt{2}$

$x > 3 + \sqrt{2}$

Step 9

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 $x < 3 - \sqrt{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < 3 - \sqrt{2}$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} \leq 6 (0) - 7$

The left side $0$ is greater than the right side $- 7$ , which means that the given statement is false.

False

Test a value on the interval $3 - \sqrt{2} < x < 3 + \sqrt{2}$ to see if it makes the inequality true.

Choose a value on the interval $3 - \sqrt{2} < x < 3 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

${(3)}^{2} \leq 6 (3) - 7$

The left side $9$ is less than the right side $11$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3 + \sqrt{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > 3 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 7$

Replace $x$ with $7$ in the original inequality.

${(7)}^{2} \leq 6 (7) - 7$

The left side $49$ is greater than the right side $35$ , which means that the given statement is false.

False

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

$x < 3 - \sqrt{2}$ False

$3 - \sqrt{2} < x < 3 + \sqrt{2}$ True

$x > 3 + \sqrt{2}$ False

$x < 3 - \sqrt{2}$ False

$3 - \sqrt{2} < x < 3 + \sqrt{2}$ True

$x > 3 + \sqrt{2}$ False

Step 10

The solution consists of all of the true intervals.

$3 - \sqrt{2} \leq x \leq 3 + \sqrt{2}$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$3 - \sqrt{2} \leq x \leq 3 + \sqrt{2}$

Interval Notation:

$[3 - \sqrt{2}, 3 + \sqrt{2}]$

Step 12

Solve the Inequality for x x^2<=6x-7

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