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a = 5b^2 – 10b + 7

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Carcass
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a = 5b^2 – 10b + 7 [#permalink] New post   21 Jun 2017, 13:08
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a = \(5b^2\) – 10b + 7

Quantity A
Quantity B
a
b




A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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A
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GREMasterBlaster
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   01 Mar 2018, 05:25
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It's a good question.

First need to simplify and then plugin as below:

Quantity A: \(5b^2\) – 10b + 7
Quantity B: b

So, we can add 10b on both Quantities, because the inequality will not change i.e:

Quantity A: \(5b^2\) + 7
Quantity B: 11b

Now, don't plugin -ve values, because Quantity A will always greater (being positive). Also in 0, Quantity A will be greater.
Finally, just check for positive values. You will see Again, Quantity A always greater.
For positive values, just plugin one value from (0 - 1) and other one from any value greater than 1.

Choice Choice A is correct.
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YMAkib
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   04 Feb 2018, 01:19
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a = 5b^2 – 10b + 7
let b=0, so a=7
b=1, a= 2
b=-1, a=22
So Q.A is greater.
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   18 Feb 2018, 06:06
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a = 5b^2 - 10b + 7
Subtract b from both sides
a - b = 5b^2 - 11b + 7
Since the coefficient of b^2 is positive the RHS is >0.

So a-b>0 --> a>b.

Answer: A
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matayo
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   18 Feb 2018, 23:56
gremather wrote:
a = 5b^2 - 10b + 7
Subtract b from both sides
a - b = 5b^2 - 11b + 7
Since the coefficient of b^2 is positive the RHS is >0.

So a-b>0 --> a>b.

Answer: A


great approach, didn't think of that
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mosshiur
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   28 Feb 2020, 21:16
Here in one explanation, it is mentioned that, since CO-efficient of b^2 is positive, therefore RHS is positive. Why?

Posted from my mobile device Image
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   10 Mar 2020, 07:41
gremather wrote:
a = 5b^2 - 10b + 7
Subtract b from both sides
a - b = 5b^2 - 11b + 7
Since the coefficient of b^2 is positive the RHS is >0.

So a-b>0 --> a>b.

Answer: A


This is a false approach. Let a = 5b^2 - 10b

In that case, a-b = 5b^2 - 11b. Coefficient of b^2 is positive, but the RHS isn't always positive.

Consider b = 1, the RHS equals -6. The only thing you can say for certain when b^2 has a positive coefficient is that the RHS will be positive as b tends to infinity.
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   10 Mar 2020, 07:45
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mosshiur wrote:
Here in one explanation, it is mentioned that, since CO-efficient of b^2 is positive, therefore RHS is positive. Why?



It's not true. Read my other comment. The actual way to solve this problem is by finding the global minima of a-b = 5b^2 - 11b + 7, which can be found by differentiating the RHS wrt b and noting that the minima is a positive value, thus a-b is always positive.
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   17 Apr 2020, 19:16
What does RHS stand for?
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   18 Apr 2020, 04:57
FunkyMonk0 wrote:
What does RHS stand for?


Right hand side
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   27 Apr 2020, 11:01
gremather wrote:
a = 5b^2 - 10b + 7
Subtract b from both sides
a - b = 5b^2 - 11b + 7
Since the coefficient of b^2 is positive the RHS is >0.

So a-b>0 --> a>b.

Answer: A



Hello,

I didnt understad after you -b from both sides, kindly explain in detail, I was not able to follow
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   27 Apr 2020, 12:11
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Put this way

\(5b^2-10b+7\)

Now simplify

\(b^2-5b+7\)

\(b(b-5)+7\)

Basically \(b=0\)
OR
\(b=5\)

Substitute

If \(b=0\) in the equation \(b(b-5)+7\), then \(a=7\)

If \(b=5\), \(a\) is still \(= 7\)

A is always greater than b

Hope now is more clear
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   23 Sep 2020, 12:14
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Another method

a ? b
5b^2 -10b +7 ? b
5b^2-11b+7 ? 0

First derivative of the eqn on LHS will be
10b-11 = 0.........=> b = 11/10=1.1 gives the min or max for the function.
Second derivative is
10 which implies at b=1.1 the function has the minimum value

put b=1.1 in the eqn we will get a + number therefore A>B.....which is the answer.

Cheers
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Ezio23
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Re: a = 5b^2 – 10b + 7 [#permalink] New post   19 Oct 2020, 22:05
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The given equation is an upward facing parabola. The minimum value of this parabola occurs at b=1 -> where the value of equation is 2. From the point the quadratic equation grows exponentially and quantity A will always be greater.
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Re: a = 5b^2 10b + 7 [#permalink] New post   12 Aug 2022, 23:18
Carcass how did you reach the step of simplification of 5b^2 - 10b + 7 to b^2 -5b +7? Could you please walk through the question once again and also, if not too inconvenient, share similar questions for practice?
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Re: a = 5b^2 10b + 7 [#permalink] New post   13 Aug 2022, 02:39
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iASindia wrote:
Carcass how did you reach the step of simplification of 5b^2 - 10b + 7 to b^2 -5b +7? Could you please walk through the question once again and also, if not too inconvenient, share similar questions for practice?



basically it is a binomial expression + 7.

Here more https://gre.myprepclub.com/forum/gre-math- ... tml#p81624

If you have a look under the quick reply under every discussion, you will have always similar topic to practice


Attachment:
screenshot.791.jpg
screenshot.791.jpg [ 127.49 KiB | Viewed 3206 times ]


Or you can use the tag algebra to find tons of questions that are similar and NOT https://gre.myprepclub.com/forum/search.ph ... &tag_id=10
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Re: a = 5b^2 10b + 7 [#permalink] New post   04 Jul 2024, 05:12
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