Question

Consider a quadratic formula y=ax^(2)+80x+1000 where a is an integer.
(a) If the minimum value of y is a positive integer, suggest three possible values of a.
(b) If the maximum value of u is at least 1200, suggest three possible values of a.

Answer 1

another question:
\[\log_{2} x+\log_{x} 2=\frac{ 5 }{ 2 }\]

find x.

Answer 2

because it's a minimum function, so a > 0
and because the minimum value of y is a positive it must be a definite positive

Answer 3

question 3:
If \[9^{\frac{ 1 }{ h }} = 11^{\frac{ 1 }{ k }} =99\], where h and k are not equal to 0, determine the value of h+k without finding the values of h and k.

Answer 4

I have already known a > 0 .....

Answer 5

but D < 0

Answer 6

D ??

Answer 7

lol, D=discriminant = b^2-4ac

Answer 8

.....and then?

Answer 9

D < 0
b^2 - 4ac < 0
(80)^2 - 4a(1000) < 0
6400 - 4000a < 0
a > 1.6
so, the intersection between a > 0 and a > 1.6 is a > 1.6
because a an integer number, u can take a=2,3,4,5, e.t.c

Answer 10

oh ya, I got it.
how about part b? There are two parts...

Answer 11

hmmm.... i dont understand why there is "u" there

Answer 12

@algebraic, do u know what mean u there ??

Answer 13

it should be y instead of u. sorry for that

Answer 14

ohhhh... :P

Answer 15

I have a question in part a , I substitute a=2, then the equation is : y=2x^(2)+80x+1000. How can I be factorize?

Answer 16

well, if it's a maximum function, so a < 0
and at least 1200, it means y >= 1200 (with y = -D/4a)
-(b^2 - 4ac)/4a > = 1200
(b^2 - 4ac)/4a <= 1200
((80)^2 - 4a(1000))/4a <= 1200
(6400 - 4000a)/4a <= 1200
(1600 - 1000a)/a <= 1200
(16 - 10a)/a <= 12
for zeros, a=0 or a=1.6