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Re: a = 5b^2 – 10b + 7 [#permalink]
6
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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Re: a = 5b^2 – 10b + 7 [#permalink]
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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Re: a = 5b^2 – 10b + 7 [#permalink]
Here in one explanation, it is mentioned that, since CO-efficient of b^2 is positive, therefore RHS is positive. Why?

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Re: a = 5b^2 – 10b + 7 [#permalink]
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]
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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]
What does RHS stand for?
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Re: a = 5b^2 – 10b + 7 [#permalink]
FunkyMonk0 wrote:
What does RHS stand for?


Right hand side
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Re: a = 5b^2 – 10b + 7 [#permalink]
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]
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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]
1
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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Re: a = 5b^2 – 10b + 7 [#permalink]
1
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]
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]
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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:
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screenshot.791.jpg [ 127.49 KiB | Viewed 3076 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]
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Re: a = 5b^2 10b + 7 [#permalink]
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