Linear Algebra Examples

$3 x^{2} + n x + 5 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 3$ , $b = n$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{- n \pm \sqrt{n^{2} - 4 \cdot (3 \cdot 5)}}{2 \cdot 3}$

Step 3

Simplify.

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Step 3.1

Multiply $- 4 \cdot 3 \cdot 5$ .

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Step 3.1.1

Multiply $- 4$ by $3$ .

$x = \frac{- n \pm \sqrt{n^{2} - 12 \cdot 5}}{2 \cdot 3}$

Step 3.1.2

Multiply $- 12$ by $5$ .

$x = \frac{- n \pm \sqrt{n^{2} - 60}}{2 \cdot 3}$

Step 3.2

Multiply $2$ by $3$ .

$x = \frac{- n \pm \sqrt{n^{2} - 60}}{6}$

Step 4

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

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Step 4.1

Multiply $- 4 \cdot 3 \cdot 5$ .

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Step 4.1.1

Multiply $- 4$ by $3$ .

$x = \frac{- n \pm \sqrt{n^{2} - 12 \cdot 5}}{2 \cdot 3}$

Step 4.1.2

Multiply $- 12$ by $5$ .

$x = \frac{- n \pm \sqrt{n^{2} - 60}}{2 \cdot 3}$

Step 4.2

Multiply $2$ by $3$ .

$x = \frac{- n \pm \sqrt{n^{2} - 60}}{6}$

Step 4.3

Change the $\pm$ to $+$ .

$x = \frac{- n + \sqrt{n^{2} - 60}}{6}$

Step 4.4

Factor $- 1$ out of $- n$ .

$x = \frac{- (n) + \sqrt{n^{2} - 60}}{6}$

Step 4.5

Factor $- 1$ out of $\sqrt{n^{2} - 60}$ .

$x = \frac{- (n) - 1 (- \sqrt{n^{2} - 60})}{6}$

Step 4.6

Factor $- 1$ out of $- (n) - 1 (- \sqrt{n^{2} - 60})$ .

$x = \frac{- (n - \sqrt{n^{2} - 60})}{6}$

Step 4.7

Rewrite $- (n - \sqrt{n^{2} - 60})$ as $- 1 (n - \sqrt{n^{2} - 60})$ .

$x = \frac{- 1 (n - \sqrt{n^{2} - 60})}{6}$

Step 4.8

Move the negative in front of the fraction.

$x = - \frac{n - \sqrt{n^{2} - 60}}{6}$

Step 5

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

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Step 5.1

Multiply $- 4 \cdot 3 \cdot 5$ .

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Step 5.1.1

Multiply $- 4$ by $3$ .

$x = \frac{- n \pm \sqrt{n^{2} - 12 \cdot 5}}{2 \cdot 3}$

Step 5.1.2

Multiply $- 12$ by $5$ .

$x = \frac{- n \pm \sqrt{n^{2} - 60}}{2 \cdot 3}$

Step 5.2

Multiply $2$ by $3$ .

$x = \frac{- n \pm \sqrt{n^{2} - 60}}{6}$

Step 5.3

Change the $\pm$ to $-$ .

$x = \frac{- n - \sqrt{n^{2} - 60}}{6}$

Step 5.4

Factor $- 1$ out of $- n$ .

$x = \frac{- (n) - \sqrt{n^{2} - 60}}{6}$

Step 5.5

Factor $- 1$ out of $- \sqrt{n^{2} - 60}$ .

$x = \frac{- (n) - (\sqrt{n^{2} - 60})}{6}$

Step 5.6

Factor $- 1$ out of $- (n) - (\sqrt{n^{2} - 60})$ .

$x = \frac{- (n + \sqrt{n^{2} - 60})}{6}$

Step 5.7

Rewrite $- (n + \sqrt{n^{2} - 60})$ as $- 1 (n + \sqrt{n^{2} - 60})$ .

$x = \frac{- 1 (n + \sqrt{n^{2} - 60})}{6}$

Step 5.8

Move the negative in front of the fraction.

$x = - \frac{n + \sqrt{n^{2} - 60}}{6}$

Step 6

The final answer is the combination of both solutions.

$x = - \frac{n - \sqrt{n^{2} - 60}}{6}, - \frac{n + \sqrt{n^{2} - 60}}{6}$

Step 7

Set the radicand in $\sqrt{n^{2} - 60}$ greater than or equal to $0$ to find where the expression is defined.

$n^{2} - 60 \geq 0$

Step 8

Solve for $n$ .

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Step 8.1

Add $60$ to both sides of the inequality.

$n^{2} \geq 60$

Step 8.2

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{n^{2}} \geq \sqrt{60}$

Step 8.3

Simplify the equation.

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Step 8.3.1

Simplify the left side.

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Step 8.3.1.1

Pull terms out from under the radical.

$| n | \geq \sqrt{60}$

Step 8.3.2

Simplify the right side.

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Step 8.3.2.1

Simplify $\sqrt{60}$ .

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Step 8.3.2.1.1

Rewrite $60$ as $2^{2} \cdot 15$ .

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Step 8.3.2.1.1.1

Factor $4$ out of $60$ .

$| n | \geq \sqrt{4 (15)}$

Step 8.3.2.1.1.2

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

$| n | \geq \sqrt{2^{2} \cdot 15}$

Step 8.3.2.1.2

Pull terms out from under the radical.

$| n | \geq | 2 | \sqrt{15}$

Step 8.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| n | \geq 2 \sqrt{15}$

Step 8.4

Write $| n | \geq 2 \sqrt{15}$ as a piecewise.

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Step 8.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$n \geq 0$

Step 8.4.2

In the piece where $n$ is non-negative, remove the absolute value.

$n \geq 2 \sqrt{15}$

Step 8.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$n < 0$

Step 8.4.4

In the piece where $n$ is negative, remove the absolute value and multiply by $- 1$ .

$- n \geq 2 \sqrt{15}$

Step 8.4.5

Write as a piecewise.

${\begin{cases} n \geq 2 \sqrt{15} & n \geq 0 \\ - n \geq 2 \sqrt{15} & n < 0 \end{cases}$

Step 8.5

Find the intersection of $n \geq 2 \sqrt{15}$ and $n \geq 0$ .

$n \geq 2 \sqrt{15}$

Step 8.6

Divide each term in $- n \geq 2 \sqrt{15}$ by $- 1$ and simplify.

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Step 8.6.1

Divide each term in $- n \geq 2 \sqrt{15}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- n}{- 1} \leq \frac{2 \sqrt{15}}{- 1}$

Step 8.6.2

Simplify the left side.

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Step 8.6.2.1

Dividing two negative values results in a positive value.

$\frac{n}{1} \leq \frac{2 \sqrt{15}}{- 1}$

Step 8.6.2.2

Divide $n$ by $1$ .

$n \leq \frac{2 \sqrt{15}}{- 1}$

Step 8.6.3

Simplify the right side.

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Step 8.6.3.1

Move the negative one from the denominator of $\frac{2 \sqrt{15}}{- 1}$ .

$n \leq - 1 \cdot (2 \sqrt{15})$

Step 8.6.3.2

Rewrite $- 1 \cdot (2 \sqrt{15})$ as $- (2 \sqrt{15})$ .

$n \leq - (2 \sqrt{15})$

Step 8.6.3.3

Multiply $2$ by $- 1$ .

$n \leq - 2 \sqrt{15}$

Step 8.7

Find the union of the solutions.

$n \leq - 2 \sqrt{15}$ or $n \geq 2 \sqrt{15}$

Step 9

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

Interval Notation:

$(- \infty, - 2 \sqrt{15}] \cup [2 \sqrt{15}, \infty)$

Set-Builder Notation:

${n | n \leq - 2 \sqrt{15}, n \geq 2 \sqrt{15}}$

Step 10