Finite Math Examples

$| x^{2} + 2 x - 15 | = | x - 3 |$

Step 1

Rewrite the equation as $| x - 3 | = | x^{2} + 2 x - 15 |$ .

$| x - 3 | = | x^{2} + 2 x - 15 |$

Step 2

Rewrite the absolute value equation as four equations without absolute value bars.

$x - 3 = x^{2} + 2 x - 15$

$x - 3 = - (x^{2} + 2 x - 15)$

$- (x - 3) = x^{2} + 2 x - 15$

$- (x - 3) = - (x^{2} + 2 x - 15)$

Step 3

After simplifying, there are only two unique equations to be solved.

$x - 3 = x^{2} + 2 x - 15$

$x - 3 = - (x^{2} + 2 x - 15)$

Step 4

Solve $x - 3 = x^{2} + 2 x - 15$ for $x$ .

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Step 4.1

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

$x^{2} + 2 x - 15 = x - 3$

Step 4.2

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

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Step 4.2.1

Subtract $x$ from both sides of the equation.

$x^{2} + 2 x - 15 - x = - 3$

Step 4.2.2

Subtract $x$ from $2 x$ .

$x^{2} + x - 15 = - 3$

Step 4.3

Add $3$ to both sides of the equation.

$x^{2} + x - 15 + 3 = 0$

Step 4.4

Add $- 15$ and $3$ .

$x^{2} + x - 12 = 0$

Step 4.5

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

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Step 4.5.1

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 $1$ .

$- 3, 4$

Step 4.5.2

Write the factored form using these integers.

$(x - 3) (x + 4) = 0$

Step 4.6

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

$x - 3 = 0$

$x + 4 = 0$

Step 4.7

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

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Step 4.7.1

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

$x - 3 = 0$

Step 4.7.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.8

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 4.8.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 4.8.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 4.9

The final solution is all the values that make $(x - 3) (x + 4) = 0$ true.

$x = 3, - 4$

Step 5

Solve $x - 3 = - (x^{2} + 2 x - 15)$ for $x$ .

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Step 5.1

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

$- (x^{2} + 2 x - 15) = x - 3$

Step 5.2

Simplify $- (x^{2} + 2 x - 15)$ .

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Step 5.2.1

Rewrite.

$0 + 0 - (x^{2} + 2 x - 15) = x - 3$

Step 5.2.2

Simplify by adding zeros.

$- (x^{2} + 2 x - 15) = x - 3$

Step 5.2.3

Apply the distributive property.

$- x^{2} - (2 x) - - 15 = x - 3$

Step 5.2.4

Simplify.

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Step 5.2.4.1

Multiply $2$ by $- 1$ .

$- x^{2} - 2 x - - 15 = x - 3$

Step 5.2.4.2

Multiply $- 1$ by $- 15$ .

$- x^{2} - 2 x + 15 = x - 3$

Step 5.3

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

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Step 5.3.1

Subtract $x$ from both sides of the equation.

$- x^{2} - 2 x + 15 - x = - 3$

Step 5.3.2

Subtract $x$ from $- 2 x$ .

$- x^{2} - 3 x + 15 = - 3$

Step 5.4

Add $3$ to both sides of the equation.

$- x^{2} - 3 x + 15 + 3 = 0$

Step 5.5

Add $15$ and $3$ .

$- x^{2} - 3 x + 18 = 0$

Step 5.6

Factor the left side of the equation.

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Step 5.6.1

Factor $- 1$ out of $- x^{2} - 3 x + 18$ .

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Step 5.6.1.1

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 3 x + 18 = 0$

Step 5.6.1.2

Factor $- 1$ out of $- 3 x$ .

$- (x^{2}) - (3 x) + 18 = 0$

Step 5.6.1.3

Rewrite $18$ as $- 1 (- 18)$ .

$- (x^{2}) - (3 x) - 1 \cdot - 18 = 0$

Step 5.6.1.4

Factor $- 1$ out of $- (x^{2}) - (3 x)$ .

$- (x^{2} + 3 x) - 1 \cdot - 18 = 0$

Step 5.6.1.5

Factor $- 1$ out of $- (x^{2} + 3 x) - 1 (- 18)$ .

$- (x^{2} + 3 x - 18) = 0$

Step 5.6.2

Factor.

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Step 5.6.2.1

Factor $x^{2} + 3 x - 18$ using the AC method.

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Step 5.6.2.1.1

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 $- 18$ and whose sum is $3$ .

$- 3, 6$

Step 5.6.2.1.2

Write the factored form using these integers.

$- ((x - 3) (x + 6)) = 0$

Step 5.6.2.2

Remove unnecessary parentheses.

$- (x - 3) (x + 6) = 0$

Step 5.7

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

$x - 3 = 0$

$x + 6 = 0$

Step 5.8

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

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Step 5.8.1

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

$x - 3 = 0$

Step 5.8.2

Add $3$ to both sides of the equation.

$x = 3$

Step 5.9

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 5.9.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 5.9.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 5.10

The final solution is all the values that make $- (x - 3) (x + 6) = 0$ true.

$x = 3, - 6$

Step 6

List all of the solutions.

$x = 3, - 4, 3, - 6$