# How do you solve k^ { 2} - 3k - 88= 0?

Apr 3, 2017

$k = 11$
and
$k = - 8$

#### Explanation:

${k}^{2} - 3 k - 88$

Factor

$\left(k - 11\right) \left(k + 8\right)$
Set each pair equal ($=$) to zero
$k - 11 = 0$ $\to$ $k = 11$
$k + 8 = 0$ $\to$ $k = - 8$

Apr 3, 2017

(k-11) x ( k+8) Break the trinomial into two binomials and solve for k
k = + 11 k = -8

#### Explanation:

The C value in the trinomial $A {x}^{2} + B x + C$ is negative. This means one factor of 88 must be positive and the other negative.

The B value in the the trinomial is negative this means that the negative factor of 88 must be larger than the positive factor by 3

There are several factor combinations for 88 : 1 x 88, 2 x 44 , 4 x 22 and 8 x 11. 8 x 11 has a difference of three.

So 11 must be negative since it is larger and 8 must be positive

$\left(k - 11\right) \times \left(k + 8\right) = 0$ solving for k yields

$k - 11 = 0$ add 11 to both sides

$k - 11 + 11 = 0 + 11$ which gives

$k = 11$

$k + 8 = 0$ s subtract 8 from both sides

$k + 8 - 8 = 0 = 8$ which gives

$k = - 8$ so

k = -8 and + 11