Question

PLEASE HELP!!!!

Suppose X and Y represent two different school populations where X > Y and X and Y must be greater than 0. Which of the following expressions is the largest? Explain why.

X^2 + Y^2
X^2 − Y^2
2(X + Y)
(X + Y)^2

Answer 1

@pooja195 @goformit100

Answer 2

may i help u

Answer 3

yes please

Answer 4

ok the answer is the second equation

Answer 5

can you explain how you did that?

Answer 6

cuz like the equation x=y it is going to equal 2 always that is the rule got it

Answer 7

wait, what?

Answer 8

uggh the answer is 2nd one cuz of how the equation is set up

Answer 9

@tom982 ?

Answer 10

Firstly, begin by expanding the parenthesis:
\[x^2+y^2\]\[x^2-y^2\]\[2(x+y)=2x+2y\]\[(x+y)^2=x^2+2xy+y^2\]
Let's consider the first one. It's clearly bigger than the second one as y^2 is positive so we can rule out the second one. Now we have:\[x^2+y^2\]\[2x+2y\]\[(x+y)^2=x^2+2xy+y^2\]Compare the first one with the second one, we can see that the first one is only bigger when x or y is greater than 2 - we can't really tell much from this, so let's move on. Now consider the last one, it's clearly bigger than the first one as it has a 2xy as well, which is positive, so we can rule out the first. Leaving us with:\[2(x+y)\]\[(x+y)^2=x^2+2xy+y^2\]The last bit is a bit tricky as we need to compare 2(x+y) with (x+y)^2. Break this down into two cases and test them:\[2(x+y) \lt (x+y)^2\]\[2<(x+y)\]So we can see that (x+y)^2 is greater than 2(x+y) when (x+y)>2. Next case:
\[2(x+y) \gt (x+y)^2\]\[2 \gt x+y\]So 2(x+y) is greater than (x+y)^2 only when (x+y) is less than 2 - this is to say that either x=0, y=1 or x=1, y=0 which can't be the case as the question says that the population is non-zero.
Hence (x+y)^2 is the largest :)

Answer 11

thank you so much @tom982

Answer 12

No problem, hope it made sense!