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Comment on Factoring - Quadratics
what is the trick in
Yes, you need to know how to
Yes, you need to know how to work with positive and negative numbers.
For example, if you want to factor x² - 2x - 15, then you're looking for two value such that their product is -15 and their sum is -2.
What are those two numbers? Well, they are -5 and 3 (positive 3)
So, we get: x² - 2x - 15 = (x - 5)(x + 3)
Does that help?
Hello Brent,
For the following question, can you please advise how to find S+T from (S+T)^2 when we have inequality sign.
http://gre.myprepclub.com/forum/manhattan-prep-gre-algebra-question-s-2-t-2-1-2st-2339.html
Thanks :)
You bet.
You bet.
Here's my response: http://gre.myprepclub.com/forum/manhattan-prep-gre-algebra-question-s-2-...
Cheers,
Brent
For the comparison question A
Unfortunately, that strategy
Unfortunately, that strategy won't always work for Quantitative Comparison questions. It's just a coincidence that it worked this time.
That said, how did you solve the equation x^2 + 1 = 0?
Good Morning,
This question: http://gre.myprepclub.com/forum/x-4-x-5508.html
can we just plug in numbers for x? I looked at your explanation on the link but to me, it just seemed easier to plug in numbers. Is that wrong for this type of question?
Question link: http:/
Question link: http://gre.myprepclub.com/forum/x-4-x-5508.html
We can certainly plug in numbers, but doing so won't yield a definitive answer.
Sure, we might plug in 4(or more) different values for x and, each time we do so, we'll find that Quantity B is greater than Quantity A. HOWEVER, we will still have doubts about whether we have failed to test a certain x-value that shows us that Quantity B is not greater than Quantity A.
I discuss this drawback starting at 2:50 in the following video: https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...
Thanks!
Hi.. can you provide an
actually i am not sure if i am able to understand the question itself
Hi Ravi,
Hi Ravi,
Here's my step-by-step solution: https://gre.myprepclub.com/forum/the-integer-v-is-greater-than-1-if-v-is...
Cheers,
Brent
Quantity A: X(4-X)
Quantity B: 6
When in doubt, plug in numbers. I plugged in 2, 3, 6 and 100 for x in quantity A. Like a doomed marriage, all efforts led to the same dead end, as it is structured for unavoidable negativity.
Quantity B will always be greater than A.
Greetings to all.
Question link: https:/
Question link: https://gre.myprepclub.com/forum/x-4-x-5508.html
I love your comments, Omar. Very funny!!
Yes, plugging in different values for x will certainly give you a strong idea of which quantity is greater.
Great work!
Cheers,
Brent
Quantity A: x(4-x)
Quantity B: 6
Hi Brent,
For the above QC, is the following approach correct?
x(4-x)
4x-x^2 i.e -x^2+4x
Subtracting by -4, therefore
quantity A: -x^2+4x-4
quantity B: 2
-(x^2-4x+4)
-(x-2)^2
-(some +ve or -ve number)^2
after squaring it becomes positive
-(positive number)
negative number
since quantity A will always be negative, quantity B will be greater.
That's a perfect approach!!
That's a perfect approach!!
Nice work.
Cheers,
Brent
Hi Brent!
For this question https://gre.myprepclub.com/forum/x-1-y-12436.html
x = 1-y
A: x^2 +2xy+y^2
B: x+y
I solved for the value of X+Y by adding Y to X=1-Y and got X+Y=1.
I then knew that x^2 +2xy+y^2 factored out to be (x+Y)^2. I replaced X+Y with 1 to get 1^2, which made answer A and B equal. Is this a valid approach?
That's a perfectly valid
That's a perfectly valid approach. Nice work!
Cheers,
Brent
Regarding this question:
When you get to the -1(x-2)^2 + 4 vs 6 part, can I just subtract 4 from both sides to get
-1(x-2)^2 vs 2
and then I could say the left side will always be negative and since the right side is always positive, then the answer is B.
Question link: https:/
Question link: https://gre.myprepclub.com/forum/x-4-x-5508.html
That's perfect reasoning. Nice work!
I'm a little confused about
I thought the product of the square of the first and last value will equal the middle value. Wouldn't it be 7x?
Thanks
Special Product: a² + 2ab +
Special Product: a² + 2ab + b² = (a + b)(a + b)
Notice that the middle term (2ab) is TWICE the product of a and b.
Also, the first term (a²) is the square of a.
Likewise, the last term (b²) is the square of b.
Given: 49x² + 14x + 1
The first term (49x²) is the square of 7x.
The last term (1) is the square of 1.
The middle term (14x) is TWICE the product of 7x and 1.
So, we can be certain that 49x² + 14x + 1 is a Special Product
As such, we can factor it as follows:
49x² + 14x + 1 = (7x)² + (2)(7x)(1) + (1)²
= (7x + 1)(7x + 1)
Does that help?
https://gmatclub.com/forum/if
for this problem why cant we take the equation and simplifiy it to 2x^2-8x+20, simplify that further to x^2-4x+10 then we know that the parabola is pointing upwards. So cant we find the vertice using x= -b/2a, plug in x into the equation and get y as the minimum, knowing that the vertice is the minimum point of the equation
Question link: https:/
Question link: https://gmatclub.com/forum/if-y-x-5-2-x-1-2-6-then-y-is-least-when-x-202...
That totally works.
So, with x² - 4x + 10, a = 1, b = -4, so -b/2a = -(-4)/(2)(1) = 2
x = 2 is the answer.
Aside: If I encountered but this question on test day, I'd probably just test the 5 answer choices.
IMPORTANT: If (something)² <
Since we now know that (s + t)² < 1, we can say that -1 < s + t < 1
Hey Brent do we take sq root to make it s + t from (s+t)^2
Something like that.
Something like that.
To be more precise, we can use a property that says √(k²) = |k|
So we can take: (s + t)² < 1
Take the square root of both sides to get: |s + t| < 1
Simplify to get: -1 < s + t < 1