Dear all,
Can anyone show me the solution?
Thanks
9768 #65
Re: 9768 #65
For 65 you have:
p(-3): 27-9a+3b=c
p(2): -8-4a-2b=c
Equating the two, it simplifies down to 7+b=a.
We also have that p'(-3)<0; so 3x^2+2ax+b<0 ==> 27-6a=b<0 ==> 27+b<6a. Substituting in for a, 27+b<6(7+b)==> -15<5b ==> -3<b.
Now lets look at -8-4a-2b=c. We have that b>-3 ==> -2b <-6 ==> -2b <-14. Similarly, a>2 ==> -4a <-8. Adding the two we have -2b-8-4a < -22 ==> c <-22. Thus -27 is the only number that satisfies.
Hopefully this helps.
p(-3): 27-9a+3b=c
p(2): -8-4a-2b=c
Equating the two, it simplifies down to 7+b=a.
We also have that p'(-3)<0; so 3x^2+2ax+b<0 ==> 27-6a=b<0 ==> 27+b<6a. Substituting in for a, 27+b<6(7+b)==> -15<5b ==> -3<b.
Now lets look at -8-4a-2b=c. We have that b>-3 ==> -2b <-6 ==> -2b <-14. Similarly, a>2 ==> -4a <-8. Adding the two we have -2b-8-4a < -22 ==> c <-22. Thus -27 is the only number that satisfies.
Hopefully this helps.
Re: 9768 #65
3rd degree polynomials with first coefficient 1, and 2 known roots
c is the y i tercept, or p(0)
Going to the factorial decomposition of p
p = (x+3)(x-2)(x-k)
Plugging 0
c = 6k
Checking that p is decreasing in -3, coming from positive values, and unbounded in -oo, we observe that the last root k is less than -3
Hence, c <-18
c is the y i tercept, or p(0)
Going to the factorial decomposition of p
p = (x+3)(x-2)(x-k)
Plugging 0
c = 6k
Checking that p is decreasing in -3, coming from positive values, and unbounded in -oo, we observe that the last root k is less than -3
Hence, c <-18