A function f(x, y) is such that f(x,y)=3x^2-2xy+y^2+4. Select one

Question

A function f(x, y) is such that  f(x,y)=3x^2-2xy+y^2+4 . Select one value for x, & one value for y such that given information implies that f(x, y) = 8.
 Make only two selections, one in each column.

PrashantPonde · Accepted Answer

Answer:  x = 1  and  y= 1-\sqrt{2}

I used back-solving method to solve this problem by substituting values for x in following order  (0, 1, -1, 1-\sqrt{2})

Given that, 
 f(x,y)=3x^2-2xy+y^2+4=8 
i.e.  3x^2-2xy+y^2=4   -- To be proved

Substituting  x=0  gives  y=\pm2  which is not in the answer list.

Substitute  x=1 
 3x^2-2xy+y^2=4 
 3*1^2-2(1)y+y^2=4 
 3-2y+y^2=4 
 y^2-2y-1=0

As we know  x =  /2a  are roots for  ax^2+bx+c=0

y=  /2(1)= /2=1-\sqrt{2}

Hence Answer:  x = 1  and  y= 1-\sqrt{2}

thangvietname · Answer

how to pick the number ? picking number is time consuming.

any tip, trick here,

egmat · Answer

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= ±2

This 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

crazypriya · Answer

x=1,
y=1-root(2)

ardatt1155 · Answer

f(x,y) = 3x^2 - 2xy + y^2 + 4 = 2x^2 + (x-y)^2 + 4 
So,  f(x,y) = 8  if and only if  2x^2 + (x-y)^2 = 4  or  x^2 + ((x-y)/sqrt 2)^2 = 2

Now that you have formulated the expression this way, it is very easy to see that x=1 and y=1-sqrt(2) is the solution.

This is much faster than plugging in values.

amitpassionate · Answer

Here is another trick as number picking can be a longer process !
Solve and separate out for quadratic
we know we have to prove 3x^2-2xy+y^2=4
i will simplify this as 2x^2+(x-y)^2=4 ...... from here choosing x =1 and y = 1-root2 becomes easy

Oppenheimer1945 · Answer

Lets put an easy no, x=1 and solve the quadratic:
3-2y+y^2=4
y^2-2y-1=0
(y-1)^2=2
y=-rt(2)+1

agrasan · Answer

Simply write the equation f(x,y)=3x^2−2xy+y^2+4 as:

f(x , y) = 2*x^2 + x^2 - 2xy + y^2 + 4

f(x, y) = (x-y)^2 + 2*x^2 + 4

Now, we can try a simple case for (x-y)^2 = 2, 2*x^2 = 2 to give us total 4

We can clearly see that x = 1 and y = 1−2^0.5 would work for us

Adding 4 on top of these two, we will get 8
The learning from this question is to simplify the terms as much as possible to save time on trying options through trial and error

x	y	--------
		\(\sqrt{\frac{2}{3}}\)
		\(1-\sqrt{2}\)
		-1
		0
		1

A function f(x, y) is such that f(x,y)=3x^2-2xy+y^2+4. Select one

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A function f(x, y) is such that f(x,y)=3x^2-2xy+y^2+4. Select one

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