# How do you solve (4k+5)(k+1)=0?

Jul 6, 2016

$k = - \frac{5}{4} \mathmr{and} k = - 1$

#### Explanation:

The simple way is to realise for the right hand side of the equation to be equal to zero, one of the terms on the left hand side must be zero. ie $4 k + 5 = 0$ or $k + 1 = 0$

$\implies k = - \frac{5}{4} \mathmr{and} k = - 1$

To check our answer, we can do it the long way, expanding the brackets to obtain a quadratic and then using the quadratic formula.
You want to FOIL the brackets.

$\left(\textcolor{red}{4 k} + \textcolor{b l u e}{5}\right) \left(\textcolor{g r e e n}{k} + \textcolor{\mathmr{and} a n \ge}{1}\right) = 0$

$\textcolor{red}{4 k} \cdot \textcolor{g r e e n}{k} + \textcolor{red}{4 k} \cdot \textcolor{\mathmr{and} a n \ge}{1} + \textcolor{b l u e}{5} \cdot \textcolor{g r e e n}{k} + \textcolor{b l u e}{5} \cdot \textcolor{\mathmr{and} a n \ge}{1} = 0$

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

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

$k = \frac{- 9 \pm \sqrt{{9}^{2} - 4 \left(4\right) \left(5\right)}}{2 \left(4\right)} = \frac{- 9 \pm \sqrt{81 - 80}}{8}$

$k = \frac{- 9 \pm 1}{8} \implies k = - 1 \mathmr{and} k = - \frac{5}{4}$