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M30-13

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M30-13  [#permalink]

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New post 16 Sep 2014, 01:45
1
4
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

49% (02:23) correct 51% (02:51) wrong based on 39 sessions

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Re: M30-13  [#permalink]

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New post 16 Sep 2014, 01:45
2
1
Official Solution:

If \(f(x) = 5x^2 - 4\) and \(g(x) = 3x + 1\), then what is the sum of all the values of \(m\) for which \(f(m - 1) = g(m^2 + 1)\) ?

A. -10
B. -5
C. 0
D. 5
E. 10

\(f(x) = 5x^2 - 4\), then \(f(m - 1)=5(m-1)^2 - 4=5m^2-10 m+1\);
\(g(x) = 3x + 1\), then \(g(m^2 + 1)=3(m^2 + 1) + 1=3m^2+4\).

Since given that \(f(m - 1) = g(m^2 + 1)\), then \(5m^2-10 m+1=3m^2+4\)5m^2-10 m+1=3m^2+4, which after simplification gives \(2m^2-10m-3=0\).

Next, Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).


Thus according to the above \(m_1+m_2=\frac{-(-10)}{2}=5\).

Answer: D.
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Re: M30-13  [#permalink]

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New post 12 Nov 2016, 07:15
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I see what you wrote, but I don't get it. If i substract -9 from 2x²-10x-3=0, I get 2x²-10x-12=-9 i can write it like this: (2x+2)*(x-6)=-9
If I try to insert 5 here, it doesn't work.

How can this be correct then, since my way is a more standard way of doing this, which has worked for me on other GMAT questions with similar difficulty. I usually had to add 1 or 2 to both sides in most cases, but this formula does seem to work for other cases than ax²+bx+c=0.

Do you personally think studying this formula is necessary for the gmat?
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Re: M30-13  [#permalink]

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New post 22 Jan 2017, 14:56
SolidGod

Inserting 5 into an equation will not work of course, because 5 is the sum of all possible solutions for 2m^2-10m-3=0, not one particular solution to the equation.

for example, for (x-2)(x+3)=0, sum of all possible solutions is = -3+2 = -1.
if you plug-in either -3 or +2 to the equation, you will get a valid answer. However, if you plug-in -1 to the equation, this will not work. What you are essentially doing is plugging in -1 instead of individual solutions such as -3 or 2.

With that said, i think the theorm may be helpful as it can be applied to other similar topics as well.

Hope it helps!
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Re: M30-13  [#permalink]

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New post 20 Apr 2017, 18:11
I hate using the quadratic equation, but I solved it via [-(-10) +- root(10^2- 4*2*-3)]/2*2 = [10 +- root(124)]/4

root(124) ~ 11 so (10+11)/4 + (10-11)/4

(21/4) - (1/4) = 20/4; answer is 5
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Re: M30-13  [#permalink]

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New post 23 Aug 2017, 17:19
I have never heard of this theorem before and I have been studying quite a while. Is this common?
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New post 23 Aug 2017, 23:50
gmathopeful19 wrote:
I have never heard of this theorem before and I have been studying quite a while. Is this common?


It certainly can be useful if a question involves quadratics. Check the questions below:
https://gmatclub.com/forum/if-6-and-3-a ... 28213.html
https://gmatclub.com/forum/if-b-and-c-a ... 17885.html
https://gmatclub.com/forum/if-r-and-s-a ... 41018.html
https://gmatclub.com/forum/m28-184514.html
https://gmatclub.com/forum/if-f-x-5x-2- ... 74044.html
https://gmatclub.com/forum/if-the-graph ... 39828.html
https://gmatclub.com/forum/in-the-equat ... 09771.html
https://gmatclub.com/forum/if-x-2-3-is- ... 60524.html
https://gmatclub.com/forum/if-x-2-12x-k ... 55465.html
https://gmatclub.com/forum/in-the-equat ... 48766.html
https://gmatclub.com/forum/if-q-is-one- ... 41199.html
https://gmatclub.com/forum/if-f-x-5x-2- ... 85989.html
https://gmatclub.com/forum/if-4-is-one- ... 39119.html
https://gmatclub.com/forum/john-and-jan ... 06597.html

7. Algebra



For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: M30-13  [#permalink]

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New post 07 Sep 2017, 07:38
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I think the correct answer should be changed. It shows B as the correct answer now.
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New post 07 Sep 2017, 07:42
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M30-13  [#permalink]

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New post 17 Feb 2018, 04:03
Hi Bunuel,

Just to get this straight, I worked the solution up to 2m^2-10m-3

I couldn't work it out as a quadratic from here...does that mean that the use of the theorem was the only way of solving this - if so what conditions have to be met for one to use the viete theorem?

Additionally, was there a more conventional way of solving this question? If so...please how can it be solved more conventionally without reverting to the theorem.

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New post 17 Feb 2018, 04:06
ttaiwo wrote:
Hi Bunuel,

Just to get this straight, I worked the solution up to 2m^2-10m-3

I couldn't work it out as a quadratic from here...does that mean that the use of the theorem was the only way of solving this? Or was there a more conventional way of solving this question? If so...please how can it be solved more conventionally without reverting to the theorem.

Tosin


You could solve 2m^2-10m-3=0 to get two values of m, and get their sum.
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New post 03 Sep 2018, 16:14
Hi Bunuel

how to solve 2m^2 - 10m - 3 to find the two values of m
product of factors has to equal -6 while sum should equal -10

Does not seem possible at all. Please help

Thanks
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New post 03 Sep 2018, 23:56
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New post 12 Jul 2019, 14:08
I think this is a high-quality question and I agree with explanation. I didn't remember for the formula for the sum of two roots but I remembered the formula for each root lol
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Re M30-13   [#permalink] 12 Jul 2019, 14:08
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