Precalculus Examples

$y = x^{2} + 4 x$ , $(a, - 3)$

Step 1

Insert the values of $x = a$ and $y = - 3$ into the equation to find if the ordered pair is a solution.

$- 3 = {(a)}^{2} + 4 (a)$

Step 2

Since $a$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$a^{2} + 4 a = - 3$

Step 3

Add $3$ to both sides of the equation.

$a^{2} + 4 a + 3 = 0$

Step 4

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

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Step 4.1

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 $4$ .

$1, 3$

Step 4.2

Write the factored form using these integers.

$(a + 1) (a + 3) = 0$

Step 5

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

$a + 1 = 0$

$a + 3 = 0$

Step 6

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

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Step 6.1

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

$a + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$a = - 1$

Step 7

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

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Step 7.1

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

$a + 3 = 0$

Step 7.2

Subtract $3$ from both sides of the equation.

$a = - 3$

Step 8

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

$a = - 1, - 3$

Step 9

Since the equation is not true when the values are used, the ordered pair is not a solution.

The ordered pair is not a solution to the equation.