Linear Algebra Examples

$5 k = 1 k^{2}$

Step 1

Multiply $k^{2}$ by $1$ .

$5 k = k^{2}$

Step 2

Subtract $k^{2}$ from both sides of the equation.

$5 k - k^{2} = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Let $u = k$ . Substitute $u$ for all occurrences of $k$ .

$5 u - u^{2} = 0$

Step 3.2

Factor $u$ out of $5 u - u^{2}$ .

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Step 3.2.1

Factor $u$ out of $5 u$ .

$u \cdot 5 - u^{2} = 0$

Step 3.2.2

Factor $u$ out of $- u^{2}$ .

$u \cdot 5 + u (- u) = 0$

Step 3.2.3

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

$u (5 - u) = 0$

Step 3.3

Replace all occurrences of $u$ with $k$ .

$k (5 - k) = 0$

Step 4

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

$k = 0$

$5 - k = 0$

Step 5

Set $k$ equal to $0$ .

$k = 0$

Step 6

Set $5 - k$ equal to $0$ and solve for $k$ .

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Step 6.1

Set $5 - k$ equal to $0$ .

$5 - k = 0$

Step 6.2

Solve $5 - k = 0$ for $k$ .

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Step 6.2.1

Subtract $5$ from both sides of the equation.

$- k = - 5$

Step 6.2.2

Divide each term in $- k = - 5$ by $- 1$ and simplify.

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Step 6.2.2.1

Divide each term in $- k = - 5$ by $- 1$ .

$\frac{- k}{- 1} = \frac{- 5}{- 1}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{k}{1} = \frac{- 5}{- 1}$

Step 6.2.2.2.2

Divide $k$ by $1$ .

$k = \frac{- 5}{- 1}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $- 5$ by $- 1$ .

$k = 5$

Step 7

The final solution is all the values that make $k (5 - k) = 0$ true.

$k = 0, 5$

Step 8

The domain is the set of all valid $k$ values.

${k | k = 0, 5}$

Step 9