Finite Math Examples

$x^{2} + (p + 1) x + 2 p - 1 = 0$

Step 1

Simplify each term.

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Step 1.1

Apply the distributive property.

$x^{2} + p x + 1 x + 2 p - 1 = 0$

Step 1.2

Multiply $x$ by $1$ .

$x^{2} + p x + x + 2 p - 1 = 0$

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 $a = 1$ , $b = p + 1$ , and $c = 2 p - 1$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the distributive property.

$x = \frac{- p - 1 \cdot 1 \pm \sqrt{{(p + 1)}^{2} - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 1$ by $1$ .

$x = \frac{- p - 1 \pm \sqrt{{(p + 1)}^{2} - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.3

Rewrite ${(p + 1)}^{2}$ as $(p + 1) (p + 1)$ .

$x = \frac{- p - 1 \pm \sqrt{(p + 1) (p + 1) - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.4

Expand $(p + 1) (p + 1)$ using the FOIL Method.

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Step 4.1.4.1

Apply the distributive property.

$x = \frac{- p - 1 \pm \sqrt{p (p + 1) + 1 (p + 1) - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.4.2

Apply the distributive property.

$x = \frac{- p - 1 \pm \sqrt{p \cdot p + p \cdot 1 + 1 (p + 1) - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.4.3

Apply the distributive property.

$x = \frac{- p - 1 \pm \sqrt{p \cdot p + p \cdot 1 + 1 p + 1 \cdot 1 - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.5

Simplify and combine like terms.

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Step 4.1.5.1

Simplify each term.

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Step 4.1.5.1.1

Multiply $p$ by $p$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + p \cdot 1 + 1 p + 1 \cdot 1 - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.5.1.2

Multiply $p$ by $1$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + p + 1 p + 1 \cdot 1 - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.5.1.3

Multiply $p$ by $1$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + p + p + 1 \cdot 1 - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.5.1.4

Multiply $1$ by $1$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + p + p + 1 - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.5.2

Add $p$ and $p$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + 2 p + 1 - 4 \cdot 1 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.6

Multiply $- 4$ by $1$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + 2 p + 1 - 4 \cdot (2 p - 1)}}{2 \cdot 1}$

Step 4.1.7

Apply the distributive property.

$x = \frac{- p - 1 \pm \sqrt{p^{2} + 2 p + 1 - 4 (2 p) - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.1.8

Multiply $2$ by $- 4$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + 2 p + 1 - 8 p - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.1.9

Multiply $- 4$ by $- 1$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} + 2 p + 1 - 8 p + 4}}{2 \cdot 1}$

Step 4.1.10

Subtract $8 p$ from $2 p$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} - 6 p + 1 + 4}}{2 \cdot 1}$

Step 4.1.11

Add $1$ and $4$ .

$x = \frac{- p - 1 \pm \sqrt{p^{2} - 6 p + 5}}{2 \cdot 1}$

Step 4.1.12

Factor $p^{2} - 6 p + 5$ using the AC method.

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Step 4.1.12.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $- 6$ .

$- 5, - 1$

Step 4.1.12.2

Write the factored form using these integers.

$x = \frac{- p - 1 \pm \sqrt{(p - 5) (p - 1)}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{- p - 1 \pm \sqrt{(p - 5) (p - 1)}}{2}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{p + 1 - \sqrt{(p - 5) (p - 1)}}{2}$

$x = - \frac{p + 1 + \sqrt{(p - 5) (p - 1)}}{2}$