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