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Is x^2 + y^2 – xy >0?​

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Is x^2 + y^2 – xy >0?​  [#permalink]

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New post 12 Jul 2020, 10:43
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Is \(x^2 + y^2 – xy >0?​\)

​1) xy≠0​

2) xy>0​

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Is x^2 + y^2 – xy >0?​  [#permalink]

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New post 12 Jul 2020, 21:19
QuantMadeEasy wrote:
Is \(x^2 + y^2 – xy >0?​\)

​1) xy≠0​

2) xy>0​



Let us simplify the inequality..

Simplify


\(x^2 + y^2 – xy >0\).......
\(x^2+\frac{4y^2}{4}-\frac{2xy}{2}>0​\)
\(x^2+\frac{y^2}{4}+\frac{3y^2}{4}-2*x*\frac{y}{2}>0\).......
\(x^2+(\frac{y}{2})^2-2*x*\frac{y}{2}+\frac{3y^2}{4}>0\)
\((x-\frac{y}{2})^2>-\frac{3y^2}{4}\)
The LHS \((x-\frac{y}{2})^2\) will always be non-negative, and RHS will always be non-positive.
SO either
a) LHS>RHS, when atleast one of x and y is non-zero.
or
b) LHS=RHS when x and y are 0.

Requirement


Is x=y=0?

​1) xy≠0​
Case (a)
answer is YES..

2) xy>0
Case (a) again.
Again none of x and y are 0, so answer is YES.

D
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Re: Is x^2 + y^2 – xy >0?​  [#permalink]

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New post 13 Jul 2020, 05:05
QuantMadeEasy wrote:
Is \(x^2 + y^2 – xy >0?​\)

​1) xy≠0​

2) xy>0​


Imagine you have a right triangle, with short sides of lengths x and y. We'd find the hypotenuse using Pythagoras; it is √(x^2 + y^2) long. That hypotenuse is longer than x and longer than y. So if we make a rectangle using the sides x and y, its area, xy, will automatically be smaller than the area of the square we'd get by using the hypotenuse as one side (because each side of the rectangle is shorter than a side of the square). But that square's area is x^2 + y^2.

So it's always true, when x and y can be lengths, that x^2 + y^2 > xy. That inequality is also clearly true when xy is negative, since then the right side is negative and the left is positive (and when both x and y are negative, that's just like when they're both positive). The only situation where the inequality in the question can be false is when xy = 0. Since each Statement rules out that possibility, the answer is D.
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Re: Is x^2 + y^2 – xy >0?​  [#permalink]

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New post 13 Jul 2020, 08:46
1
Is x2+y2–xy>0?

​1) xy≠0​
if xy is negative then TRUE x2+y2-xy>0
if xy is positive then x^2 + y^2 would be greater than xy

case1: x and y equal; it means x^2+x^2 -x^2 = +ve
case1: unequal +ve nummber: it means one number is smaller than other ==> square of bugger number would always be greater than multiplication of one small and one bigger number , hence always+ve

SUFFICIENT

2) xy>0​
As explained above

SUFFICIENT

Hence D
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Re: Is x^2 + y^2 – xy >0?​  [#permalink]

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New post 13 Jul 2020, 10:09
Quote:
Is x^2 + y^2 – xy >0?

Step 1: Understanding the question​
Rephrasing the question by multiplying 2 on both sides of the question​
Is \( 2x^2 + 2y^2 – 2xy >0? ​\)
Is \(x^2+ y^2 – 2xy + x^2+ y^2 > 0 ?​\)
Is \((x-y)^2 + x^2 +y^2 > 0?​\)
\((x-y)^2 + x^2 +y^2\) is greater than zero only when x and y are not zero, otherwise the expression will be equal to zero.​

Quote:
(1) Xy≠0​

Step 2: Understanding (1)
Neither x nor y is zero. Sufficient​

Quote:
(2)xy>0​

Step 3: Understanding (2)
It means that both x and y have same signs ie both negative or both positive, which also means neither of them is zero. Sufficient​

(D) is correct​
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Re: Is x^2 + y^2 – xy >0?​   [#permalink] 13 Jul 2020, 10:09

Is x^2 + y^2 – xy >0?​

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