Algebra Examples

$5 (\frac{5 + k}{2}) - 4 = {(\frac{5 + k}{2})}^{2} - k (\frac{5 + k}{2})$

Step 1

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

${(\frac{5 + k}{2})}^{2} - k \frac{5 + k}{2} = 5 (\frac{5 + k}{2}) - 4$

Step 2

Simplify ${(\frac{5 + k}{2})}^{2} - k \frac{5 + k}{2}$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the product rule to $\frac{5 + k}{2}$ .

$\frac{{(5 + k)}^{2}}{2^{2}} - k \frac{5 + k}{2} = 5 (\frac{5 + k}{2}) - 4$

Step 2.1.2

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

$\frac{{(5 + k)}^{2}}{4} - k \frac{5 + k}{2} = 5 (\frac{5 + k}{2}) - 4$

Step 2.1.3

Combine $\frac{5 + k}{2}$ and $k$ .

$\frac{{(5 + k)}^{2}}{4} - \frac{(5 + k) k}{2} = 5 (\frac{5 + k}{2}) - 4$

Step 2.2

To write $- \frac{(5 + k) k}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{{(5 + k)}^{2}}{4} - \frac{(5 + k) k}{2} \cdot \frac{2}{2} = 5 (\frac{5 + k}{2}) - 4$

Step 2.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{(5 + k) k}{2}$ by $\frac{2}{2}$ .

$\frac{{(5 + k)}^{2}}{4} - \frac{(5 + k) k \cdot 2}{2 \cdot 2} = 5 (\frac{5 + k}{2}) - 4$

Step 2.3.2

Multiply $2$ by $2$ .

$\frac{{(5 + k)}^{2}}{4} - \frac{(5 + k) k \cdot 2}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 2.4

Combine the numerators over the common denominator.

$\frac{{(5 + k)}^{2} - (5 + k) k \cdot 2}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 2.5

Simplify the numerator.

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Step 2.5.1

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

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Step 2.5.1.1

Factor $5 + k$ out of ${(5 + k)}^{2}$ .

$\frac{(5 + k) (5 + k) - (5 + k) k \cdot 2}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 2.5.1.2

Factor $5 + k$ out of $- (5 + k) k \cdot 2$ .

$\frac{(5 + k) (5 + k) + (5 + k) (- 1 k \cdot 2)}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 2.5.1.3

Factor $5 + k$ out of $(5 + k) (5 + k) + (5 + k) (- 1 k \cdot 2)$ .

$\frac{(5 + k) (5 + k - 1 k \cdot 2)}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 2.5.2

Rewrite $- 1 k$ as $- k$ .

$\frac{(5 + k) (5 + k - k \cdot 2)}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 2.5.3

Multiply $2$ by $- 1$ .

$\frac{(5 + k) (5 + k - 2 k)}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 2.5.4

Subtract $2 k$ from $k$ .

$\frac{(5 + k) (5 - k)}{4} = 5 (\frac{5 + k}{2}) - 4$

Step 3

Simplify $5 (\frac{5 + k}{2}) - 4$ .

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Step 3.1

Combine $5$ and $\frac{5 + k}{2}$ .

$\frac{(5 + k) (5 - k)}{4} = \frac{5 (5 + k)}{2} - 4$

Step 3.2

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{(5 + k) (5 - k)}{4} = \frac{5 (5 + k)}{2} - 4 \cdot \frac{2}{2}$

Step 3.3

Simplify terms.

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Step 3.3.1

Combine $- 4$ and $\frac{2}{2}$ .

$\frac{(5 + k) (5 - k)}{4} = \frac{5 (5 + k)}{2} + \frac{- 4 \cdot 2}{2}$

Step 3.3.2

Combine the numerators over the common denominator.

$\frac{(5 + k) (5 - k)}{4} = \frac{5 (5 + k) - 4 \cdot 2}{2}$

Step 3.4

Simplify the numerator.

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Step 3.4.1

Apply the distributive property.

$\frac{(5 + k) (5 - k)}{4} = \frac{5 \cdot 5 + 5 k - 4 \cdot 2}{2}$

Step 3.4.2

Multiply $5$ by $5$ .

$\frac{(5 + k) (5 - k)}{4} = \frac{25 + 5 k - 4 \cdot 2}{2}$

Step 3.4.3

Multiply $- 4$ by $2$ .

$\frac{(5 + k) (5 - k)}{4} = \frac{25 + 5 k - 8}{2}$

Step 3.4.4

Subtract $8$ from $25$ .

$\frac{(5 + k) (5 - k)}{4} = \frac{5 k + 17}{2}$

Step 4

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

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Step 4.1

Subtract $\frac{5 k + 17}{2}$ from both sides of the equation.

$\frac{(5 + k) (5 - k)}{4} - \frac{5 k + 17}{2} = 0$

Step 4.2

To write $- \frac{5 k + 17}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{(5 + k) (5 - k)}{4} - \frac{5 k + 17}{2} \cdot \frac{2}{2} = 0$

Step 4.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.3.1

Multiply $\frac{5 k + 17}{2}$ by $\frac{2}{2}$ .

$\frac{(5 + k) (5 - k)}{4} - \frac{(5 k + 17) \cdot 2}{2 \cdot 2} = 0$

Step 4.3.2

Multiply $2$ by $2$ .

$\frac{(5 + k) (5 - k)}{4} - \frac{(5 k + 17) \cdot 2}{4} = 0$

Step 4.4

Combine the numerators over the common denominator.

$\frac{(5 + k) (5 - k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5

Simplify the numerator.

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Step 4.5.1

Expand $(5 + k) (5 - k)$ using the FOIL Method.

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Step 4.5.1.1

Apply the distributive property.

$\frac{5 (5 - k) + k (5 - k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.1.2

Apply the distributive property.

$\frac{5 \cdot 5 + 5 (- k) + k (5 - k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.1.3

Apply the distributive property.

$\frac{5 \cdot 5 + 5 (- k) + k \cdot 5 + k (- k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2

Simplify and combine like terms.

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Step 4.5.2.1

Simplify each term.

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Step 4.5.2.1.1

Multiply $5$ by $5$ .

$\frac{25 + 5 (- k) + k \cdot 5 + k (- k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2.1.2

Multiply $- 1$ by $5$ .

$\frac{25 - 5 k + k \cdot 5 + k (- k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2.1.3

Move $5$ to the left of $k$ .

$\frac{25 - 5 k + 5 \cdot k + k (- k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2.1.4

Rewrite using the commutative property of multiplication.

$\frac{25 - 5 k + 5 k - k \cdot k - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2.1.5

Multiply $k$ by $k$ by adding the exponents.

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Step 4.5.2.1.5.1

Move $k$ .

$\frac{25 - 5 k + 5 k - (k \cdot k) - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2.1.5.2

Multiply $k$ by $k$ .

$\frac{25 - 5 k + 5 k - k^{2} - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2.2

Add $- 5 k$ and $5 k$ .

$\frac{25 + 0 - k^{2} - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.2.3

Add $25$ and $0$ .

$\frac{25 - k^{2} - (5 k + 17) \cdot 2}{4} = 0$

Step 4.5.3

Apply the distributive property.

$\frac{25 - k^{2} + (- (5 k) - 1 \cdot 17) \cdot 2}{4} = 0$

Step 4.5.4

Multiply $5$ by $- 1$ .

$\frac{25 - k^{2} + (- 5 k - 1 \cdot 17) \cdot 2}{4} = 0$

Step 4.5.5

Multiply $- 1$ by $17$ .

$\frac{25 - k^{2} + (- 5 k - 17) \cdot 2}{4} = 0$

Step 4.5.6

Apply the distributive property.

$\frac{25 - k^{2} - 5 k \cdot 2 - 17 \cdot 2}{4} = 0$

Step 4.5.7

Multiply $2$ by $- 5$ .

$\frac{25 - k^{2} - 10 k - 17 \cdot 2}{4} = 0$

Step 4.5.8

Multiply $- 17$ by $2$ .

$\frac{25 - k^{2} - 10 k - 34}{4} = 0$

Step 4.5.9

Subtract $34$ from $25$ .

$\frac{- k^{2} - 10 k - 9}{4} = 0$

Step 4.5.10

Factor by grouping.

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Step 4.5.10.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 9 = 9$ and whose sum is $b = - 10$ .

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Step 4.5.10.1.1

Factor $- 10$ out of $- 10 k$ .

$\frac{- k^{2} - 10 (k) - 9}{4} = 0$

Step 4.5.10.1.2

Rewrite $- 10$ as $- 1$ plus $- 9$

$\frac{- k^{2} + (- 1 - 9) k - 9}{4} = 0$

Step 4.5.10.1.3

Apply the distributive property.

$\frac{- k^{2} - 1 k - 9 k - 9}{4} = 0$

Step 4.5.10.2

Factor out the greatest common factor from each group.

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Step 4.5.10.2.1

Group the first two terms and the last two terms.

$\frac{(- k^{2} - 1 k) - 9 k - 9}{4} = 0$

Step 4.5.10.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{k (- k - 1) + 9 (- k - 1)}{4} = 0$

Step 4.5.10.3

Factor the polynomial by factoring out the greatest common factor, $- k - 1$ .

$\frac{(- k - 1) (k + 9)}{4} = 0$

Step 4.6

Factor $- 1$ out of $- k$ .

$\frac{(- (k) - 1) (k + 9)}{4} = 0$

Step 4.7

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

$\frac{(- (k) - 1 (1)) (k + 9)}{4} = 0$

Step 4.8

Factor $- 1$ out of $- (k) - 1 (1)$ .

$\frac{- (k + 1) (k + 9)}{4} = 0$

Step 4.9

Rewrite $- (k + 1)$ as $- 1 (k + 1)$ .

$\frac{- 1 (k + 1) (k + 9)}{4} = 0$

Step 4.10

Move the negative in front of the fraction.

$- \frac{(k + 1) (k + 9)}{4} = 0$

Step 5

Set the numerator equal to zero.

$(k + 1) (k + 9) = 0$

Step 6

Solve the equation for $k$ .

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Step 6.1

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

$k + 1 = 0$

$k + 9 = 0$

Step 6.2

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

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Step 6.2.1

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

$k + 1 = 0$

Step 6.2.2

Subtract $1$ from both sides of the equation.

$k = - 1$

Step 6.3

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

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Step 6.3.1

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

$k + 9 = 0$

Step 6.3.2

Subtract $9$ from both sides of the equation.

$k = - 9$

Step 6.4

The final solution is all the values that make $(k + 1) (k + 9) = 0$ true.

$k = - 1, - 9$