Calculus Examples

$p^{2} + 2 p + q = 9$

Step 1

Subtract $9$ from both sides of the equation.

$p^{2} + 2 p + q - 9 = 0$

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 $a = 1$ , $b = 2$ , and $c = q - 9$ into the quadratic formula and solve for $p$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot (q - 9))}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$p = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot (q - 9)}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4$ by $1$ .

$p = \frac{- 2 \pm \sqrt{4 - 4 \cdot (q - 9)}}{2 \cdot 1}$

Step 4.1.3

Apply the distributive property.

$p = \frac{- 2 \pm \sqrt{4 - 4 q - 4 \cdot - 9}}{2 \cdot 1}$

Step 4.1.4

Multiply $- 4$ by $- 9$ .

$p = \frac{- 2 \pm \sqrt{4 - 4 q + 36}}{2 \cdot 1}$

Step 4.1.5

Add $4$ and $36$ .

$p = \frac{- 2 \pm \sqrt{- 4 q + 40}}{2 \cdot 1}$

Step 4.1.6

Factor $4$ out of $- 4 q + 40$ .

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Step 4.1.6.1

Factor $4$ out of $- 4 q$ .

$p = \frac{- 2 \pm \sqrt{4 (- q) + 40}}{2 \cdot 1}$

Step 4.1.6.2

Factor $4$ out of $40$ .

$p = \frac{- 2 \pm \sqrt{4 (- q) + 4 (10)}}{2 \cdot 1}$

Step 4.1.6.3

Factor $4$ out of $4 (- q) + 4 (10)$ .

$p = \frac{- 2 \pm \sqrt{4 (- q + 10)}}{2 \cdot 1}$

Step 4.1.7

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

$p = \frac{- 2 \pm \sqrt{2^{2} (- q + 10)}}{2 \cdot 1}$

Step 4.1.8

Pull terms out from under the radical.

$p = \frac{- 2 \pm 2 \sqrt{- q + 10}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$p = \frac{- 2 \pm 2 \sqrt{- q + 10}}{2}$

Step 4.3

Simplify $\frac{- 2 \pm 2 \sqrt{- q + 10}}{2}$ .

$p = - 1 \pm \sqrt{- q + 10}$

Step 5

The final answer is the combination of both solutions.

$p = - 1 + \sqrt{- q + 10}$

$p = - 1 - \sqrt{- q + 10}$