thangvietnam wrote:
how to pick the number ? picking number is time consuming.
any tip, trick here,
Hi,
When we look at
f(x,y)=3x^2-2xy+y^2+4 and then at the options, we find that plugging values is the best way to approach this question.
There are 3 things that should be keptp in mind while picking option values.
1.Pick integers first. They are easy to work on.
There are 3 values in the option list, which are integers.
2.Pick ‘0’ first. This will eliminate one variable completely for compuation.
3.What to choose first; x or y? One should always observe right hand side of the function. If number of terms of x is more than the number of terms of y, then plug in the option value in x first, and vice versa.We choose the values for x in the order of 0, 1, and -1 to plug in.
Now, we plug in the value of f(x, y) =8, & x=0 in the equation, and we get,
8=3.0^2-2.0.y+y^2+48=y^2+4 4=y^2 y= ±2This means for x=0, y is either 2 or -2. There is no such option available for values: 2 or-2 , hence these pair of values cannot be correct.
Now, we should try x=1. You may follow PraPon’s solution for x=1. He has done it correctly.
Hope it helps!
-Shalabh
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