Calculus Examples

$\frac{1}{2} q^{2} - 2 q + 6 = - \frac{1}{2} q^{2} + 2 q + 18$

Step 1

Combine $\frac{1}{2}$ and $q^{2}$ .

$\frac{q^{2}}{2} - 2 q + 6 = - \frac{1}{2} q^{2} + 2 q + 18$

Step 2

Combine $q^{2}$ and $\frac{1}{2}$ .

$\frac{q^{2}}{2} - 2 q + 6 = - \frac{q^{2}}{2} + 2 q + 18$

Step 3

Move all terms containing $q$ to the left side of the equation.

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Step 3.1

Add $\frac{q^{2}}{2}$ to both sides of the equation.

$\frac{q^{2}}{2} - 2 q + 6 + \frac{q^{2}}{2} = 2 q + 18$

Step 3.2

Subtract $2 q$ from both sides of the equation.

$\frac{q^{2}}{2} - 2 q + 6 + \frac{q^{2}}{2} - 2 q = 18$

Step 3.3

Combine the numerators over the common denominator.

$- 2 q + 6 - 2 q + \frac{q^{2} + q^{2}}{2} = 18$

Step 3.4

Add $q^{2}$ and $q^{2}$ .

$- 2 q + 6 - 2 q + \frac{2 q^{2}}{2} = 18$

Step 3.5

Cancel the common factor of $2$ .

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Step 3.5.1

Cancel the common factor.

$- 2 q + 6 - 2 q + \frac{2 q^{2}}{2} = 18$

Step 3.5.2

Divide $q^{2}$ by $1$ .

$- 2 q + 6 - 2 q + q^{2} = 18$

Step 3.6

Subtract $2 q$ from $- 2 q$ .

$- 4 q + 6 + q^{2} = 18$

Step 4

Subtract $18$ from both sides of the equation.

$- 4 q + 6 + q^{2} - 18 = 0$

Step 5

Subtract $18$ from $6$ .

$- 4 q + q^{2} - 12 = 0$

Step 6

Factor the left side of the equation.

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Step 6.1

Let $u = q$ . Substitute $u$ for all occurrences of $q$ .

$- 4 u + u^{2} - 12 = 0$

Step 6.2

Factor $- 4 u + u^{2} - 12$ using the AC method.

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Step 6.2.1

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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Step 6.2.2

Write the factored form using these integers.

$(u - 6) (u + 2) = 0$

Step 6.3

Replace all occurrences of $u$ with $q$ .

$(q - 6) (q + 2) = 0$

Step 7

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$q - 6 = 0$

$q + 2 = 0$

Step 8

Set $q - 6$ equal to $0$ and solve for $q$ .

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Step 8.1

Set $q - 6$ equal to $0$ .

$q - 6 = 0$

Step 8.2

Add $6$ to both sides of the equation.

$q = 6$

Step 9

Set $q + 2$ equal to $0$ and solve for $q$ .

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Step 9.1

Set $q + 2$ equal to $0$ .

$q + 2 = 0$

Step 9.2

Subtract $2$ from both sides of the equation.

$q = - 2$

Step 10

The final solution is all the values that make $(q - 6) (q + 2) = 0$ true.

$q = 6, - 2$