Precalculus Examples

$11 x^{2} - 5 x + 2 = 0$

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 $a = 11$ , $b = - 5$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (11 \cdot 2)}}{2 \cdot 11}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 11 \cdot 2}}{2 \cdot 11}$

Step 3.1.2

Multiply $- 4 \cdot 11 \cdot 2$ .

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Step 3.1.2.1

Multiply $- 4$ by $11$ .

$x = \frac{5 \pm \sqrt{25 - 44 \cdot 2}}{2 \cdot 11}$

Step 3.1.2.2

Multiply $- 44$ by $2$ .

$x = \frac{5 \pm \sqrt{25 - 88}}{2 \cdot 11}$

Step 3.1.3

Subtract $88$ from $25$ .

$x = \frac{5 \pm \sqrt{- 63}}{2 \cdot 11}$

Step 3.1.4

Rewrite $- 63$ as $- 1 (63)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 63}}{2 \cdot 11}$

Step 3.1.5

Rewrite $\sqrt{- 1 (63)}$ as $\sqrt{- 1} \cdot \sqrt{63}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{63}}{2 \cdot 11}$

Step 3.1.6

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

$x = \frac{5 \pm i \cdot \sqrt{63}}{2 \cdot 11}$

Step 3.1.7

Rewrite $63$ as $3^{2} \cdot 7$ .

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Step 3.1.7.1

Factor $9$ out of $63$ .

$x = \frac{5 \pm i \cdot \sqrt{9 (7)}}{2 \cdot 11}$

Step 3.1.7.2

Rewrite $9$ as $3^{2}$ .

$x = \frac{5 \pm i \cdot \sqrt{3^{2} \cdot 7}}{2 \cdot 11}$

Step 3.1.8

Pull terms out from under the radical.

$x = \frac{5 \pm i \cdot (3 \sqrt{7})}{2 \cdot 11}$

Step 3.1.9

Move $3$ to the left of $i$ .

$x = \frac{5 \pm 3 i \sqrt{7}}{2 \cdot 11}$

Step 3.2

Multiply $2$ by $11$ .

$x = \frac{5 \pm 3 i \sqrt{7}}{22}$