Finite Math Examples

$p^{2} + b p + 1 = 0$

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 $a = 1$ , $b = b$ , and $c = 1$ into the quadratic formula and solve for $p$ .

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Rewrite $4 \cdot 1 \cdot 1$ as ${(2 \cdot 1 \cdot 1)}^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - {(2 \cdot 1 \cdot 1)}^{2}}}{2 \cdot 1}$

Step 3.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 2 \cdot 1 \cdot 1$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 1 \cdot 1) (b - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 3.1.3

Simplify.

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Step 3.1.3.1

Multiply $2 \cdot 1 \cdot 1$ .

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Step 3.1.3.1.1

Multiply $2$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 1) (b - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 3.1.3.1.2

Multiply $2$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 2) (b - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 3.1.3.2

Multiply $- (2 \cdot 1 \cdot 1)$ .

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Step 3.1.3.2.1

Multiply $2$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 2) (b - (2 \cdot 1))}}{2 \cdot 1}$

Step 3.1.3.2.2

Multiply $2$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 2) (b - 1 \cdot 2)}}{2 \cdot 1}$

Step 3.1.3.2.3

Multiply $- 1$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 2) (b - 2)}}{2 \cdot 1}$

Step 3.2

Multiply $2$ by $1$ .

$p = \frac{- b \pm \sqrt{(b + 2) (b - 2)}}{2}$

Step 4

The final answer is the combination of both solutions.

$p = - \frac{b - \sqrt{(b + 2) (b - 2)}}{2}$

$p = - \frac{b + \sqrt{(b + 2) (b - 2)}}{2}$