Precalculus Examples

$2 x^{2} + 16 < 8 x$

Step 1

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

$2 x^{2} + 16 - 8 x < 0$

Step 2

Convert the inequality to an equation.

$2 x^{2} + 16 - 8 x = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $2$ out of $2 x^{2} + 16 - 8 x$ .

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Step 3.1.1

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 16 - 8 x = 0$

Step 3.1.2

Factor $2$ out of $16$ .

$2 x^{2} + 2 \cdot 8 - 8 x = 0$

Step 3.1.3

Factor $2$ out of $- 8 x$ .

$2 x^{2} + 2 \cdot 8 + 2 (- 4 x) = 0$

Step 3.1.4

Factor $2$ out of $2 x^{2} + 2 \cdot 8$ .

$2 (x^{2} + 8) + 2 (- 4 x) = 0$

Step 3.1.5

Factor $2$ out of $2 (x^{2} + 8) + 2 (- 4 x)$ .

$2 (x^{2} + 8 - 4 x) = 0$

Step 3.2

Reorder terms.

$2 (x^{2} - 4 x + 8) = 0$

Step 4

Divide each term in $2 (x^{2} - 4 x + 8) = 0$ by $2$ and simplify.

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Step 4.1

Divide each term in $2 (x^{2} - 4 x + 8) = 0$ by $2$ .

$\frac{2 (x^{2} - 4 x + 8)}{2} = \frac{0}{2}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{2 (x^{2} - 4 x + 8)}{2} = \frac{0}{2}$

Step 4.2.1.2

Divide $x^{2} - 4 x + 8$ by $1$ .

$x^{2} - 4 x + 8 = \frac{0}{2}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $2$ .

$x^{2} - 4 x + 8 = 0$

Step 5

Use the quadratic formula to find the solutions.

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

Step 6

Substitute the values $a = 1$ , $b = - 4$ , and $c = 8$ into the quadratic formula and solve for $x$ .

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

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Step 7.1.2

Multiply $- 4 \cdot 1 \cdot 8$ .

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 8}}{2 \cdot 1}$

Step 7.1.2.2

Multiply $- 4$ by $8$ .

$x = \frac{4 \pm \sqrt{16 - 32}}{2 \cdot 1}$

Step 7.1.3

Subtract $32$ from $16$ .

$x = \frac{4 \pm \sqrt{- 16}}{2 \cdot 1}$

Step 7.1.4

Rewrite $- 16$ as $- 1 (16)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Step 7.1.5

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Step 7.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Step 7.1.7

Rewrite $16$ as $4^{2}$ .

$x = \frac{4 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

Step 7.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{4 \pm i \cdot 4}{2 \cdot 1}$

Step 7.1.9

Move $4$ to the left of $i$ .

$x = \frac{4 \pm 4 i}{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$x = \frac{4 \pm 4 i}{2}$

Step 7.3

Simplify $\frac{4 \pm 4 i}{2}$ .

$x = 2 \pm 2 i$

Step 8

The final answer is the combination of both solutions.

$x = 2 + 2 i, 2 - 2 i$

Step 9

Identify the leading coefficient.

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Step 9.1

Move $16$ .

$2 x^{2} - 8 x + 16$

Step 9.2

The leading term in a polynomial is the term with the highest degree.

$2 x^{2}$

Step 9.3

The leading coefficient in a polynomial is the coefficient of the leading term.

$2$

Step 10

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $2 x^{2} + 16 - 8 x$ is always greater than $0$ .

No solution