The quadratic equation (k+1)x^2+12x+(k-4)=0 has real roots. (a) Show that k^2-3k-40<=0. (b) Hence find the possible values of k.

(a) There is a quadratic equation which should e solved using the delta andthe roots formulas.

delta=a²-4ab

delta=144-4(k+1)(k-4)

delta=-4k²+12k+160

Because the problem tells us that the roots are real, the condition is that delta is 0 or greater than 0.

So, delta>=0

-4k²+12k+160>=0|:4

-k²+3k+40>=0

k²-3k-40<=0

(b) 

k²-3k-40<=0

delta=(-3)²-41(-40)

delta=169

The formula for calculating the values of a quadratic equation is k_1,2=[-b+-sqrt(delta)]/2

NB: sqrt[x] is the squared root of x.

=> k₁=(3-13)/2; k₁=-5 (1)

     k₂=(3+13)/2; k₂=8 (2)

delta>=0 (3)

(1), (2), (3)=> k=[-5,8]