Precalculus Examples

$y = x^{2} - 8 x + 8$ , $y = - x^{2} - 16 x + 2$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - 8 x + 8 = - x^{2} - 16 x + 2$

Step 2

Solve $x^{2} - 8 x + 8 = - x^{2} - 16 x + 2$ for $x$ .

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Step 2.1

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

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Step 2.1.1

Add $x^{2}$ to both sides of the equation.

$x^{2} - 8 x + 8 + x^{2} = - 16 x + 2$

Step 2.1.2

Add $16 x$ to both sides of the equation.

$x^{2} - 8 x + 8 + x^{2} + 16 x = 2$

Step 2.1.3

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

$2 x^{2} - 8 x + 8 + 16 x = 2$

Step 2.1.4

Add $- 8 x$ and $16 x$ .

$2 x^{2} + 8 x + 8 = 2$

Step 2.2

Subtract $2$ from both sides of the equation.

$2 x^{2} + 8 x + 8 - 2 = 0$

Step 2.3

Subtract $2$ from $8$ .

$2 x^{2} + 8 x + 6 = 0$

Step 2.4

Factor the left side of the equation.

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Step 2.4.1

Factor $2$ out of $2 x^{2} + 8 x + 6$ .

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Step 2.4.1.1

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 8 x + 6 = 0$

Step 2.4.1.2

Factor $2$ out of $8 x$ .

$2 (x^{2}) + 2 (4 x) + 6 = 0$

Step 2.4.1.3

Factor $2$ out of $6$ .

$2 x^{2} + 2 (4 x) + 2 \cdot 3 = 0$

Step 2.4.1.4

Factor $2$ out of $2 x^{2} + 2 (4 x)$ .

$2 (x^{2} + 4 x) + 2 \cdot 3 = 0$

Step 2.4.1.5

Factor $2$ out of $2 (x^{2} + 4 x) + 2 \cdot 3$ .

$2 (x^{2} + 4 x + 3) = 0$

Step 2.4.2

Factor.

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Step 2.4.2.1

Factor $x^{2} + 4 x + 3$ using the AC method.

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Step 2.4.2.1.1

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 $3$ and whose sum is $4$ .

$1, 3 + y = - x^{2} - 16 x + 2$

Step 2.4.2.1.2

Write the factored form using these integers.

$2 ((x + 1) (x + 3)) = 0$

Step 2.4.2.2

Remove unnecessary parentheses.

$2 (x + 1) (x + 3) = 0$

Step 2.5

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

$x + 1 = 0$

$x + 3 = 0 + y = - x^{2} - 16 x + 2$

Step 2.6

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.7

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.8

The final solution is all the values that make $2 (x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

Step 3

Evaluate $y$ when $x = - 1$ .

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Step 3.1

Substitute $- 1$ for $x$ .

$y = - {(- 1)}^{2} - 16 \cdot - 1 + 2$

Step 3.2

Substitute $- 1$ for $x$ in $y = - {(- 1)}^{2} - 16 \cdot - 1 + 2$ and solve for $y$ .

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Step 3.2.1

Remove parentheses.

$y = - {(- 1)}^{2} - 16 \cdot - 1 + 2$

Step 3.2.2

Simplify $- {(- 1)}^{2} - 16 \cdot - 1 + 2$ .

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 3.2.2.1.1.1

Multiply $- 1$ by ${(- 1)}^{2}$ .

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Step 3.2.2.1.1.1.1

Raise $- 1$ to the power of $1$ .

$y = {(- 1)}^{1} {(- 1)}^{2} - 16 \cdot - 1 + 2$

Step 3.2.2.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = {(- 1)}^{1 + 2} - 16 \cdot - 1 + 2$

Step 3.2.2.1.1.2

Add $1$ and $2$ .

$y = {(- 1)}^{3} - 16 \cdot - 1 + 2$

Step 3.2.2.1.2

Raise $- 1$ to the power of $3$ .

$y = - 1 - 16 \cdot - 1 + 2$

Step 3.2.2.1.3

Multiply $- 16$ by $- 1$ .

$y = - 1 + 16 + 2$

Step 3.2.2.2

Simplify by adding numbers.

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Step 3.2.2.2.1

Add $- 1$ and $16$ .

$y = 15 + 2$

Step 3.2.2.2.2

Add $15$ and $2$ .

$y = 17$

Step 4

Evaluate $y$ when $x = - 3$ .

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Step 4.1

Substitute $- 3$ for $x$ .

$y = - {(- 3)}^{2} - 16 \cdot - 3 + 2$

Step 4.2

Substitute $- 3$ for $x$ in $y = - {(- 3)}^{2} - 16 \cdot - 3 + 2$ and solve for $y$ .

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Step 4.2.1

Remove parentheses.

$y = - {(- 3)}^{2} - 16 \cdot - 3 + 2$

Step 4.2.2

Simplify $- {(- 3)}^{2} - 16 \cdot - 3 + 2$ .

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Raise $- 3$ to the power of $2$ .

$y = - 1 \cdot 9 - 16 \cdot - 3 + 2$

Step 4.2.2.1.2

Multiply $- 1$ by $9$ .

$y = - 9 - 16 \cdot - 3 + 2$

Step 4.2.2.1.3

Multiply $- 16$ by $- 3$ .

$y = - 9 + 48 + 2$

Step 4.2.2.2

Simplify by adding numbers.

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Step 4.2.2.2.1

Add $- 9$ and $48$ .

$y = 39 + 2$

Step 4.2.2.2.2

Add $39$ and $2$ .

$y = 41$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 1, 17)$

$(- 3, 41)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 1, 17), (- 3, 41)$

Equation Form:

$x = - 1, y = 17$

$x = - 3, y = 41$

Step 7