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

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Answer 1 · 2017-04-03T17:42:30

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

$k^2-3k-88$

Factor

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

Answer 2 · 2017-04-03T17:45:47

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

The C value in the trinomial $Ax^2 + Bx + C$ 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

$(k - 11 ) xx (k + 8 )= 0$ 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

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