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,

\(8=3.0^2-2.0.y+y^2+4\)

\(8=y^2+4\)

\(4=y^2\)

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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