Solve Using the Quadratic Formula (4k+5)(k+1)=0

$(4 k + 5) (k + 1) = 0$

Step 1

Simplify the left side.

Simplify $(4 k + 5) (k + 1)$ .

Expand $(4 k + 5) (k + 1)$ using the FOIL Method.

Apply the distributive property.

$4 k (k + 1) + 5 (k + 1) = 0$

Apply the distributive property.

$4 k \cdot k + 4 k \cdot 1 + 5 (k + 1) = 0$

Apply the distributive property.

$4 k \cdot k + 4 k \cdot 1 + 5 k + 5 \cdot 1 = 0$

Simplify and combine like terms.

Simplify each term.

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

Move $k$ .

$4 (k \cdot k) + 4 k \cdot 1 + 5 k + 5 \cdot 1 = 0$

Multiply $k$ by $k$ .

$4 k^{2} + 4 k \cdot 1 + 5 k + 5 \cdot 1 = 0$

Multiply $4$ by $1$ .

$4 k^{2} + 4 k + 5 k + 5 \cdot 1 = 0$

Multiply $5$ by $1$ .

$4 k^{2} + 4 k + 5 k + 5 = 0$

Add $4 k$ and $5 k$ .

$4 k^{2} + 9 k + 5 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 4$ , $b = 9$ , and $c = 5$ into the quadratic formula and solve for $k$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (4 \cdot 5)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$k = \frac{- 9 \pm \sqrt{81 - 4 \cdot 4 \cdot 5}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 5$ .

Multiply $- 4$ by $4$ .

$k = \frac{- 9 \pm \sqrt{81 - 16 \cdot 5}}{2 \cdot 4}$

Multiply $- 16$ by $5$ .

$k = \frac{- 9 \pm \sqrt{81 - 80}}{2 \cdot 4}$

Subtract $80$ from $81$ .

$k = \frac{- 9 \pm \sqrt{1}}{2 \cdot 4}$

Any root of $1$ is $1$ .

$k = \frac{- 9 \pm 1}{2 \cdot 4}$

Multiply $2$ by $4$ .

$k = \frac{- 9 \pm 1}{8}$

Step 5

The final answer is the combination of both solutions.

$k = - 1, - \frac{5}{4}$

Solve Using the Quadratic Formula (4k+5)(k+1)=0

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