How do you solve k^ { 2} - 3k - 88= 0k2−3k−88=0?

2 Answers
Chad
Apr 3, 2017

k=11k=11
and
k=-8k=−8

Explanation:

k^2-3k-88k2−3k−88

Factor

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

David Drayer
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 Ax^2 + Bx + CAx2+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(k−11)×(k+8)=0 solving for k yields

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

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

k = 11 k=11

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

k + 8 -8 = 0 =8 k+8−8=0=8 which gives

k = -8 k=−8 so

k = -8 and + 11

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