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quadratic tough

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bhandariavi
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quadratic tough [#permalink] New post   20 Mar 2011, 17:12
00:00
A
B
C
D
E

Difficulty:

75% (hard)

Question Stats:

50% (02:59) correct 50% (01:53) wrong based on 6 sessions
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For a quadratic equation of the form x2 +bx + c, describe all x for which f(x + 3) = f(-2x)?

A) x= -1
B) x= b+3
C) x= Sqroot(b^2 - 3b+c)
D) x > b ^ 2
E ) x < b
plz show your calculation.



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B
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B
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Re: quadratic tough [#permalink] New post   20 Mar 2011, 19:39
I tried to factor it some extent and made a guess by eliminating others and got B as answer.

(x+3)^2 + b(x+2) + c = (-2x)^2 + b(-2x) + c

=> x^2 + 6x + 9 + bx + 2b = 4x^2 - 2bx

=> 3x^2 - 3bx - 9 - 2b = 0

=> 3x^2 -3bx -(9+2b) = 0

=> 3x^2 - 3bx - 9 = 2b

=> 3(x^2 - bx - 3) = 2b

A) x= -1 - b does not cancel out
B) x= b+3 - this is the only probable choice remaning.
C) x= Sqroot(b^2 - 3b+c) - c cancels out above
D) x > b ^ 2 - remove answers with b^2 (as it does not come during simplification.
E ) x < b - remove answers with b and without equality sign.
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 00:10
bhandariavi wrote:
For a quadratic equation of the form x2 +bx + c, describe all x for which f(x + 3) = f(-2x)?

A) x= -1
B) x= b+3
C) x= Sqroot(b^2 - 3b+c)
D) x > b ^ 2
E ) x < b
plz show your calculation.


The question doesn't seem right to me, as x can take two values and both the options A and B are correct.

Clearly x=-1 is right as that reduces both f(x + 3) and f(-2x) to f(2).

Lets do the algebra to solve for x:

We want to solve for x such that f(x + 3) = f(-2x)

or \((x+3)^2 + b*(x+3) +c = (-2x)^2-2x*b+c\)
or \(-3x^2 + x*(3b+6)+3b+9 = 0\)
or \((x+1)(3b+9-3x) = 0\)

Hence, x = -1 or x = b+3

So, both A and B are correct.

I don't know why x=-1 has been discarded in the OA
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 08:47
I doubt this question ,x has two solution

x = -1 or x = b+3

So both A and B are correct...

bhandariavi what is the source...
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 09:12
Grockit.com is the source. I dont know why it has two solutions.
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 09:32
beyondgmatscore !
How did you move from step 2 to 3. Could you please show the simplification? Is there any shortcut?
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 09:55
Onell wrote:
I doubt this question ,x has two solution

x = -1 or x = b+3

So both A and B are correct...

bhandariavi what is the source...


True!!!

The only reason I think why because x=-1 is a subset of x=b+3

x=b+3=-1 for b=-4

Thus, x=b+3 describe ALL x for the given condition.
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 16:44
Could anyone tell me how beyondgmatscore move from step 2 to 3??????????????
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 22:40
bhandariavi wrote:
beyondgmatscore !
How did you move from step 2 to 3. Could you please show the simplification? Is there any shortcut?


Well there is no shortcut really.

\(-3x^2 + x*(3b+6)+3b+9 = 0\)

can be written as \(-3x^2 + x*(3b+9-3)+3b+9 = 0\)

or \(-3x^2 -3x + x*(3b+9)+3b+9 = 0\)

or \(-3x*(x+1) + (3b+9)*(x+1) = 0\)
or \((x+1)(3b+9-3x) = 0\)

There is no short cut but once you notice that \(3b+6\)can be written as \(3b+9-3\), you would know what the factors would be and can directly deduce the final step, or you can go step by step as above.
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Re: quadratic tough [#permalink] New post   21 Mar 2011, 22:48
fluke wrote:
Onell wrote:
I doubt this question ,x has two solution

x = -1 or x = b+3

So both A and B are correct...

bhandariavi what is the source...


True!!!

The only reason I think why because x=-1 is a subset of x=b+3

x=b+3=-1 for b=-4

Thus, x=b+3 describe ALL x for the given condition.


Not sure if this argument holds completely. What about the situation when b=2, then x = b+3 only describes 5 as a possible root for f(x+3)=f(-2x), but even then x=-1 is still a root that is not captured by x=b+3.

Basically, our conclusion is that f(x+3) and f(-2x) would be same at two points - one point is x=-1 and this root is independent of value of b and other one is x=b+3 which is dependent on value of b. In the special case of b=-4, the equation has two equal roots. So x=b+3 doesn't describe ALL roots for any b other than -4.

I still believe that the question quality is suspect.
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Re: quadratic tough [#permalink] New post   22 Mar 2011, 06:17
thanks for your detailed simplification beyondgmatscore.



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Re: quadratic tough [#permalink]
22 Mar 2011, 06:17
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