Data Sufficiency

if ax^2 + bx^3 =5, where a and b are non-zero numbers, what is the value of a + b ?
1) ax^2 = a
2) ax^3 = b

Does anyone have answer to this question?

I get E

ax^2 + bx^3 =5 ---> x^2(a+bx) = 5 --- (1)

1) ax^2 = a ---> x^2 =1 ---> x = 1 or x= -1

substituting in (1)

x = 1 ---> a + b = 5
x = -1 ---> a -b = 5 insufficient

2) ax^3 = b ---> ax^2 = b/x

substituting in (1)
b/x + bx^3 =5 ---> b + bx^4 = 5x b and x can take different values
insufficient

1) and 2)

x = 1 --> b + b = 5 --> b = 5/2 and a = 5/2 and a+b = 5
x = -1 ---> b +b = -5 ---> b = -5/2 a = 5/2 and a - b = 5 and a + b = 0

mikeCoolBoy wrote:I get E

ax^2 + bx^3 =5 ---> x^2(a+bx) = 5 --- (1)

1) ax^2 = a ---> x^2 =1 ---> x = 1 or x= -1

substituting in (1)

x = 1 ---> a + b = 5
x = -1 ---> a -b = 5 insufficient

2) ax^3 = b ---> ax^2 = b/x

substituting in (1)
b/x + bx^3 =5 ---> b + bx^4 = 5x b and x can take different values
insufficient

1) and 2)

x = 1 --> b + b = 5 --> b = 5/2 and a = 5/2 and a+b = 5
x = -1 ---> b +b = -5 ---> b = -5/2 a = 5/2 and a - b = 5 and a + b = 0

in case 1 , can't we solve for a and b..a=5 and b=0??

tohellandback wrote:

mikeCoolBoy wrote:I get E

ax^2 + bx^3 =5 ---> x^2(a+bx) = 5 --- (1)

1) ax^2 = a ---> x^2 =1 ---> x = 1 or x= -1

substituting in (1)

x = 1 ---> a + b = 5
x = -1 ---> a -b = 5 insufficient

2) ax^3 = b ---> ax^2 = b/x

substituting in (1)
b/x + bx^3 =5 ---> b + bx^4 = 5x b and x can take different values
insufficient

1) and 2)

x = 1 --> b + b = 5 --> b = 5/2 and a = 5/2 and a+b = 5
x = -1 ---> b +b = -5 ---> b = -5/2 a = 5/2 and a - b = 5 and a + b = 0

in case 1 , can't we solve for a and b..a=5 and b=0??

the question says a and b are non zero numbers. b cannot be zero

guys, please help here..

Not sure but getting E.....

if ax^2 = a, fill into first equation

a + bx^3 = 5

if ax^3 = b, fill into first equation

a + ax^6 = 5, hence a must be positive and ax^2 must be less than 5 (but remember that x can be 1 or -1) so bx^3 must be positive as well because ax^2 + bx^3 = 5

but......b could be negative and x could be negative would would make bx^3 positive.

So we need to know whether x is positive or negative to answer this question, which we can not find out with the given information, HENCE E.