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## Comment on

FOIL Method for Expanding## For the last ques from Math

Is x^2 > y^2?

(1) x + y = 2

(2) x > y

How is the OA C? my explanation:

-Assume 1 and 2 are not suff on their own

-1+2)

x= 2, y= 0 (satisfying 1 and 2) --> Yes, 4 > 0

x= 2, y= -4 (satisfying 1 and 2) --> No, 4 is not greater than 16

Shouldn't the answer be E???

## x= 2, y= -4 does not satisfy

x= 2, y= -4 does not satisfy statement 1

## Brent, since statement 2

## "...since statement 2 clearly

"...since statement 2 clearly states that x > y, it should be true for x^2 > y^2..."

That's not true. Here's a counter-example:

x = -3 and y = 1

Here, x < y, but it is NOT the case that x² < y².

If x = -3, then x² = (-3)² = 9

If y = 1, then y² = (1)² = 1

So, x² is greater than y²

Cheers,

Brent

## Hi Brent,

Im having trouble understanding the base concept behind this question, do you have another method of explanation other than the one helpfully provided in the comments.

https://gmatclub.com/forum/for-integers-a-and-x-which-of-the-following-values-of-a-guarantees-229421.html

## Question link: https:/

Question link: https://gmatclub.com/forum/for-integers-a-and-x-which-of-the-following-v...

We're told that 4x² + ax + 16 is a perfect square.

Since this expression (4x² + ax + 16) resembles a quadratic expression, we should recall the two special products (which happen to be perfect squares):

x² + 2xy + y² = (x + y)²

x² - 2xy + y² = (x - y)²

So, as we can see, x² + 2xy + y² and x² - 2xy + y² are both perfect squares, because we can rewrite both of them as (x + y)² and (x - y)²

So, 4x² + ax + 16 = (something + something else)²

OR

4x² + ax + 16 = (something - something else)²

Let's examine each case

CASE A: 4x² + ax + 16 = (something + something else)²

Notice that 4x² = (2x)² and 16 = 4²

So, we can write: 4x² + ax + 16 = (2x + 4)²

If we EXPAND and simplify the right side, we get: 4x² + ax + 16 = 4x² + 16x + 16

So, in this case, a = 16

CASE B: 4x² + ax + 16 = (something - something else)²

Notice that 4x² = (2x)² and 16 = 4²

So, we can write: 4x² + ax + 16 = (2x - 4)²

If we EXPAND and simplify the right side, we get: 4x² + ax + 16 = 4x² - 16x + 16

So, in this case, a = -16

So, if a = 16, then 4x² + ax + 16 is a perfect square

And, if a = -16, then 4x² + ax + 16 is a perfect square

So, there are TWO values that make 4x² + ax + 16 a perfect square

Check the answer choices....answer = A

Cheers,

Brent

## https://gmatclub.com/forum

i did this question like this a = 0

then b = 4

simplifying 2x^2 -8x

now by using options x=-1,0,2,3,5 at x=2 expression is smallest hence c

sir in this type of questions can we assume value?

## Question link: https:/

Question link: https://gmatclub.com/forum/m19-184198.html

For this question, the best option is to test values, as you have done.

Cheers,

Brent

## Brent need help please!

Question link: https://gmatclub.com/forum/is-x-2-y-226126.html

Is x² > y²?

(1) x + y = 2

(2) x > y

I first simplified the target question by taking roots both sides, so is x>y?

(1) x + y = 2, this condition can be satisfied when,

x=1,y=1.....x=y

x=0,y=2.....x<y

x=3,y=-1.....x>y

we cant determine if x>y...Not sufficient

(2) x > y....this is what we need, so sufficient, answer is B

I don't understand why the above approach in the forum is complicated and the answer comes to C??!!

## Hi Gerrard,

Hi Gerrard,

The issue with your solution lies at the very beginning, when you rephrased the target question as "Is x > y?"

Asking "Is x> y?" is not the same as asking "Is x² > y²?".

For example, if x = 1 and y = -2, then it is true that x > y?

So, the answer to your REPHRASED target question is "YES, x is greater than y"

However, if x = 1 and y = -2, then it is NOT true that x² > y²?

So, the answer to the ORIGINAL target question is "NO, x² is NOT greater than y²"

In general, we cannot say that √(x²) = x

For example, if x = -3, we get: √(-3²) = -3

Simplify to get: √9 = -3 (not true)

Does that help?

Cheers,

Brent

## Understood.Thats great Brent!

## Hi Brent.

On this question:

https://gmatclub.com/forum/if-x-2-1-x-2-4-what-is-the-value-of-x-4-1-x-175113.html

In your answer:

"(x² + 1/x²)² = 4²

So, (x² + 1/x²)(x² + 1/x²) = 16

Expand to get: x⁴ + 1 + 1 + 1/x⁴ = 16

Simplify: x⁴ + 1/x⁴ = 14"

How would I know to subtract the two ones in x⁴ + 1 + 1 + 1/x⁴ = 16 from the 16? How did you know to bring the two ones over but to leave the x⁴ + 1/x⁴ ? Why not bring the 1 over from 1/x⁴?

## Question link: https:/

Question link: https://gmatclub.com/forum/if-x-2-1-x-2-4-what-is-the-value-of-x-4-1-x-1...

It's important to remember that our goal is to determine the value of x⁴ + 1/x⁴

So, once we have x⁴ + 1 + 1 + 1/x⁴ = 16, we should recognize that, except for the two 1's, the left side of the equation has the x⁴ + 1/x⁴ we need.

So, once we subtract 2 from both sides, we have x⁴ + 1/x⁴, which is what we're trying to determine the value of.

Does that help?

Cheers,

Brent

## Hi Brent,

Is x² > y²?

(1) x + y = 2

(2) x > y

Simplify the question to x² - y² > 0 or (x+y)(x-y) > 0

Stmt 1. 2(x-y)>0 or divide both sides by 2 to get x-y>0 or x>y, which is not sufficient.

Stmt 2. x>y, not sufficient.

Together, the statements provide the same info so E.

Where have I gone wrong?

Thanks

## You did a great job

You did a great job rephrasing the target question to get: "Is (x+y)(x-y) > 0?"

However, statement 1 (x + y = 2) does not translate into 2(x-y) > 0

Here's my full solution: https://www.beatthegmat.com/is-x-2-y-2-t296831.html

## Need help with the questions

For statement 1, I solved x^2 > x as follows:

x (x - 1 ) > 0

x > 0 or x > 0

However, the solution to statement 1 is posted as x < 0 or X > 1.

Could you pls explain?

Would also be helpful if you can explain how statements 1 and 2 work together to determine answer as C.

Many thanks!

## Question link: https:/

Question link: https://gmatclub.com/forum/is-x-2-2-x-2-1-x-2-x-2-1-x-87443.html

You're correct to say that: x² - x > 0

And it's also true that: x(x - 1) > 0

The critical points here are x = 0 and x = 1

So let's examine the following three regions:

Region i: x < 0

If x < 0, then x is negative and (x-1) is negative, which means x(x - 1) > 0. Perfect!

So, values of x in the region x < 0 ARE solutions did the given inequality

Region ii: 0 < x < 1

If 0 < x < 1, then x is positive and (x-1) is negative, which means x(x - 1) < 0. No good.

So, values of x in the region 0 < x < 1 are NOT solutions to the given inequality.

Region iii: 1 < x

If 1 < x, then x is positive and (x-1) is positive, which means x(x - 1) > 0. Perfect!

So, values of x in the region 1 < x ARE solutions to the given inequality.

Here's my full solution to the question: https://gmatclub.com/forum/is-x-2-2-x-2-1-x-2-x-2-1-x-87443-20.html#p249...