Precalculus Examples

$3 k^{2} + 1 k - 3 = 0$

Step 1

Multiply $k$ by $1$ .

$3 k^{2} + k - 3 = 0$

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 $a = 3$ , $b = 1$ , and $c = - 3$ into the quadratic formula and solve for $k$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (3 \cdot - 3)}}{2 \cdot 3}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

One to any power is one.

$k = \frac{- 1 \pm \sqrt{1 - 4 \cdot 3 \cdot - 3}}{2 \cdot 3}$

Step 4.1.2

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

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Step 4.1.2.1

Multiply $- 4$ by $3$ .

$k = \frac{- 1 \pm \sqrt{1 - 12 \cdot - 3}}{2 \cdot 3}$

Step 4.1.2.2

Multiply $- 12$ by $- 3$ .

$k = \frac{- 1 \pm \sqrt{1 + 36}}{2 \cdot 3}$

Step 4.1.3

Add $1$ and $36$ .

$k = \frac{- 1 \pm \sqrt{37}}{2 \cdot 3}$

Step 4.2

Multiply $2$ by $3$ .

$k = \frac{- 1 \pm \sqrt{37}}{6}$

Step 5

The final answer is the combination of both solutions.

$k = - \frac{1 - \sqrt{37}}{6}, - \frac{1 + \sqrt{37}}{6}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$k = - \frac{1 - \sqrt{37}}{6}, - \frac{1 + \sqrt{37}}{6}$

Decimal Form:

$k = 0.84712708 \dots, - 1.18046042 \dots$