manpsingh87 wrote:ccassel wrote:Hi,
How would you explain the answer to this question?
Is it true that x>0?
(1) x^2=2x
(2) x^3=3x
cheers,
1) x^2=2x;
x^2-2x=0;
x(x-2)=0; hence x can be either 0,2; therefore 1 alone is not sufficient to answer the question.
2) x^3-3x=0;
x(x^2-3)=0; hence x can be either 0,sqrt(3),-sqrt(3) therefore 2 alone is not sufficient to answer the question.
combining 1 and 2 we have;
x^3-x^2-x=0;
x(x^2-x-1)=0; x=0 or x^2-x-1=0 as also by combining 1 and 2 we have different solutions possible, therefore answer should be
E..!!
That is not the correct way of combining both the statements.
Obviously each statement alone is not sufficient, because they are each giving more than 1 value, which may or may not be more than 0.
Combine both the statements together and check.
Take the first equation.
You get that x^2 = 2x.
Or, x(x - 2) = 0.
So, x is either 0 or 2.
Next, take the second equation.
So, x^3 - 3x = 0.
Or, x(x^2 - 3) = 0.
This gives that x = 0, sqrt3 or -sqrt3.
When we combine, we have to look for values which satisfy both the equations simultaneously.
Only 0 does that and so, x = 0.
Or x is not greater than 0.
The correct answer is therefore (C).