Precalculus Examples

$x^{2} - 4 x + 7 \leq 0$

Step 1

Convert the inequality to an equation.

$x^{2} - 4 x + 7 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = - 4$ , and $c = 7$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 7}}{2 \cdot 1}$

Step 4.1.2

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

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 7}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $7$ .

$x = \frac{4 \pm \sqrt{16 - 28}}{2 \cdot 1}$

Step 4.1.3

Subtract $28$ from $16$ .

$x = \frac{4 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.1.6

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

$x = \frac{4 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 4.1.7

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

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Step 4.1.7.1

Factor $4$ out of $12$ .

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

Step 4.1.7.2

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 4.1.8

Pull terms out from under the radical.

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

Step 4.1.9

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{2}$

Step 4.3

Simplify $\frac{4 \pm 2 i \sqrt{3}}{2}$ .

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

Step 5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 5.1

Simplify the numerator.

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Step 5.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 7}}{2 \cdot 1}$

Step 5.1.2

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

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 7}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $7$ .

$x = \frac{4 \pm \sqrt{16 - 28}}{2 \cdot 1}$

Step 5.1.3

Subtract $28$ from $16$ .

$x = \frac{4 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 5.1.6

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

$x = \frac{4 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 5.1.7

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

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Step 5.1.7.1

Factor $4$ out of $12$ .

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

Step 5.1.7.2

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 5.1.8

Pull terms out from under the radical.

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

Step 5.1.9

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{2}$

Step 5.3

Simplify $\frac{4 \pm 2 i \sqrt{3}}{2}$ .

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

Step 5.4

Change the $\pm$ to $+$ .

$x = 2 + i \sqrt{3}$

Step 6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 6.1

Simplify the numerator.

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Step 6.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 7}}{2 \cdot 1}$

Step 6.1.2

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

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 7}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $7$ .

$x = \frac{4 \pm \sqrt{16 - 28}}{2 \cdot 1}$

Step 6.1.3

Subtract $28$ from $16$ .

$x = \frac{4 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 6.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 6.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 6.1.6

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

$x = \frac{4 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 6.1.7

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

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Step 6.1.7.1

Factor $4$ out of $12$ .

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

Step 6.1.7.2

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 6.1.8

Pull terms out from under the radical.

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

Step 6.1.9

Move $2$ to the left of $i$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 i \sqrt{3}}{2}$

Step 6.3

Simplify $\frac{4 \pm 2 i \sqrt{3}}{2}$ .

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

Step 6.4

Change the $\pm$ to $-$ .

$x = 2 - i \sqrt{3}$

Step 7

Identify the leading coefficient.

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Step 7.1

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

$x^{2}$

Step 7.2

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

$1$

Step 8

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $x^{2} - 4 x + 7$ is always greater than $0$ .

No solution