(Can't really explain it... hopefully these steps can help you.. maybe I'm just too lazy..)
2h^3j^-3k^4
-----------------
3jk
2h^3k^4
= -------------
3j^4k
2h^3k^3
= --------------
3j^4
So
y=ax^2+bx+c
(x,y)
sub the points and solve
(4.28,6.48)
6.48=a(4.28)^2+b(4.28)+c
(12.61,15.04)
15.04=a(12.61)^2+b(12.61)+c
well, for 3 variables, we need equations and therefor 3 points
maybe we are supposed to assume it starts at (0,0)
so then
0=a(0)^2+b(0)+c
0=c
so then
6.48=a(4.28)^2+b(4.28)
15.04=a(12.61)^2+b(12.61)
solve for a by subsitution
first equation, minut a(4.28)^2 from both sides
6.48-a(4.28)^2=b(4.28)
divide both sides by 4.28
(6.48/4.28)-4.28a=b
sub that for b in other equation
15.04=a(12.61)^2+b(12.61)
15.04=a(12.61)^2+((6.48/4.28)-4.28a)(12.61)
expand
15.04 =a(12.61)^2+(81.7128/4.28)-53.9708a
minus (81.7128/4.28) both sides
15.04-(81.7128/4.28)=a(12.61)^2-53.9708a
15.04-(81.7128/4.28)=a((12.61)^2-53.9708)
(15.04-(81.7128/4.28))/(((12.61)^2-53.9708))=a
that's the exact value of a
to find b, subsitute to get
(6.48/4.28)-4.28((15.04-(81.7128/4.28))/(((12.61)^2-53.9708)))=b
if we aprox
a≈-0.038573167896199
b≈1.6791118501845
so then the equation is
y=-0.038573167896199x²+1.6791118501845x
Answer:
i) Equation can have exactly 2 zeroes.
ii) Both the zeroes will be real and distinctive.
Step-by-step explanation:
is the given equation.
It is of the form of quadratic equation 
and highest degree of the polynomial is 2.
Now, FUNDAMENTAL THEOREM OF ALGEBRA
If P(x) is a polynomial of degree n ≥ 1, then P(x) = 0 has exactly n roots, including multiple and complex roots.
So, the equation can have exact 2 zeroes (roots).
Also, find discriminant D = 
⇒ D = 37
Here, since D > 0, So both the roots will be real and distinctive.
12^2=144
144 is divided by b which also is an odd interger, there only 2 numbers : 1 and 3
if b=3, a^2=48, then there will be no a available
if b=1 , a^2=144, then a is 12 and 12 can not be divided by 9
The answer is D
