Pre-Algebra Examples

$x^{2} + {(2 x - 5)}^{2} = 610$

Step 1

Subtract $610$ from both sides of the equation.

$x^{2} + {(2 x - 5)}^{2} - 610 = 0$

Step 2

Simplify $x^{2} + {(2 x - 5)}^{2} - 610$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Rewrite ${(2 x - 5)}^{2}$ as $(2 x - 5) (2 x - 5)$ .

$x^{2} + (2 x - 5) (2 x - 5) - 610 = 0$

Step 2.1.2

Expand $(2 x - 5) (2 x - 5)$ using the FOIL Method.

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Step 2.1.2.1

Apply the distributive property.

$x^{2} + 2 x (2 x - 5) - 5 (2 x - 5) - 610 = 0$

Step 2.1.2.2

Apply the distributive property.

$x^{2} + 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x - 5) - 610 = 0$

Step 2.1.2.3

Apply the distributive property.

$x^{2} + 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 - 610 = 0$

Step 2.1.3

Simplify and combine like terms.

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Step 2.1.3.1

Simplify each term.

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Step 2.1.3.1.1

Rewrite using the commutative property of multiplication.

$x^{2} + 2 \cdot 2 x \cdot x + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 - 610 = 0$

Step 2.1.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.3.1.2.1

Move $x$ .

$x^{2} + 2 \cdot 2 (x \cdot x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 - 610 = 0$

Step 2.1.3.1.2.2

Multiply $x$ by $x$ .

$x^{2} + 2 \cdot 2 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 - 610 = 0$

Step 2.1.3.1.3

Multiply $2$ by $2$ .

$x^{2} + 4 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 - 610 = 0$

Step 2.1.3.1.4

Multiply $- 5$ by $2$ .

$x^{2} + 4 x^{2} - 10 x - 5 (2 x) - 5 \cdot - 5 - 610 = 0$

Step 2.1.3.1.5

Multiply $2$ by $- 5$ .

$x^{2} + 4 x^{2} - 10 x - 10 x - 5 \cdot - 5 - 610 = 0$

Step 2.1.3.1.6

Multiply $- 5$ by $- 5$ .

$x^{2} + 4 x^{2} - 10 x - 10 x + 25 - 610 = 0$

Step 2.1.3.2

Subtract $10 x$ from $- 10 x$ .

$x^{2} + 4 x^{2} - 20 x + 25 - 610 = 0$

Step 2.2

Add $x^{2}$ and $4 x^{2}$ .

$5 x^{2} - 20 x + 25 - 610 = 0$

Step 2.3

Subtract $610$ from $25$ .

$5 x^{2} - 20 x - 585 = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $5$ out of $5 x^{2} - 20 x - 585$ .

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Step 3.1.1

Factor $5$ out of $5 x^{2}$ .

$5 (x^{2}) - 20 x - 585 = 0$

Step 3.1.2

Factor $5$ out of $- 20 x$ .

$5 (x^{2}) + 5 (- 4 x) - 585 = 0$

Step 3.1.3

Factor $5$ out of $- 585$ .

$5 x^{2} + 5 (- 4 x) + 5 \cdot - 117 = 0$

Step 3.1.4

Factor $5$ out of $5 x^{2} + 5 (- 4 x)$ .

$5 (x^{2} - 4 x) + 5 \cdot - 117 = 0$

Step 3.1.5

Factor $5$ out of $5 (x^{2} - 4 x) + 5 \cdot - 117$ .

$5 (x^{2} - 4 x - 117) = 0$

Step 3.2

Factor.

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Step 3.2.1

Factor $x^{2} - 4 x - 117$ using the AC method.

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Step 3.2.1.1

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 $- 117$ and whose sum is $- 4$ .

$- 13, 9$

Step 3.2.1.2

Write the factored form using these integers.

$5 ((x - 13) (x + 9)) = 0$

Step 3.2.2

Remove unnecessary parentheses.

$5 (x - 13) (x + 9) = 0$

Step 4

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

$x - 13 = 0$

$x + 9 = 0$

Step 5

Set $x - 13$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 13$ equal to $0$ .

$x - 13 = 0$

Step 5.2

Add $13$ to both sides of the equation.

$x = 13$

Step 6

Set $x + 9$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Step 6.2

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 7

The final solution is all the values that make $5 (x - 13) (x + 9) = 0$ true.

$x = 13, - 9$