Pre-Algebra Examples

x (x - 3) = 54

$x (x - 3) = 54$

Step 1

Simplify

x (x - 3)

$x (x - 3)$ .

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Step 1.1

Apply the distributive property.

x \cdot x + x \cdot - 3 = 54

$x \cdot x + x \cdot - 3 = 54$

Step 1.2

Simplify the expression.

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Step 1.2.1

Multiply

x

$x$ by

x

$x$ .

x^{2} + x \cdot - 3 = 54

$x^{2} + x \cdot - 3 = 54$

Step 1.2.2

Move

- 3

$- 3$ to the left of

x

$x$ .

x^{2} - 3 x = 54

$x^{2} - 3 x = 54$

x^{2} - 3 x = 54

$x^{2} - 3 x = 54$

x^{2} - 3 x = 54

$x^{2} - 3 x = 54$

Step 2

Subtract

54

$54$ from both sides of the equation.

x^{2} - 3 x - 54 = 0

$x^{2} - 3 x - 54 = 0$

Step 3

Factor

x^{2} - 3 x - 54

$x^{2} - 3 x - 54$ using the AC method.

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Step 3.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 54

$- 54$ and whose sum is

- 3

$- 3$ .

- 9, 6

$- 9, 6$

Step 3.2

Write the factored form using these integers.

(x - 9) (x + 6) = 0

$(x - 9) (x + 6) = 0$

(x - 9) (x + 6) = 0

$(x - 9) (x + 6) = 0$

Step 4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 9 = 0

$x - 9 = 0$

x + 6 = 0

$x + 6 = 0$

Step 5

Set

x - 9

$x - 9$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

x - 9

$x - 9$ equal to

0

$0$ .

x - 9 = 0

$x - 9 = 0$

Step 5.2

Add

9

$9$ to both sides of the equation.

x = 9

$x = 9$

x = 9

$x = 9$

Step 6

Set

x + 6

$x + 6$ equal to

0

$0$ and solve for

x

$x$ .

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Step 6.1

Set

x + 6

$x + 6$ equal to

0

$0$ .

x + 6 = 0

$x + 6 = 0$

Step 6.2

Subtract

6

$6$ from both sides of the equation.

x = - 6

$x = - 6$

x = - 6

$x = - 6$

Step 7

The final solution is all the values that make

(x - 9) (x + 6) = 0

$(x - 9) (x + 6) = 0$ true.

x = 9, - 6

$x = 9, - 6$