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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...