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# M30-13

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Math Expert
Joined: 02 Sep 2009
Posts: 47110

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16 Sep 2014, 01:45
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Difficulty:

95% (hard)

Question Stats:

55% (01:19) correct 45% (02:05) wrong based on 49 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 47110

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16 Sep 2014, 01:45
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 $$m_1+m_2=\frac{-(-10)}{2}=5$$.

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Joined: 05 Oct 2016
Posts: 1

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12 Nov 2016, 07:15
1
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?
Manager
Joined: 28 Dec 2016
Posts: 88
Location: United States (IL)
Concentration: Marketing, General Management
Schools: Johnson '20 (M)
GMAT 1: 700 Q47 V38

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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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Joined: 29 Jul 2016
Posts: 7

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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
Manager
Joined: 09 Oct 2016
Posts: 89
Location: United States
GMAT 1: 740 Q49 V42
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23 Aug 2017, 17:19
I have never heard of this theorem before and I have been studying quite a while. Is this common?
Math Expert
Joined: 02 Sep 2009
Posts: 47110

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23 Aug 2017, 23:50
Intern
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41
GPA: 3.5

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07 Sep 2017, 07:38
1
I think the correct answer should be changed. It shows B as the correct answer now.
Math Expert
Joined: 02 Sep 2009
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07 Sep 2017, 07:42
Devbek wrote:
I think the correct answer should be changed. It shows B as the correct answer now.

_______________
Edited. Thank you.
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Joined: 13 Oct 2017
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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.

Tosin
Math Expert
Joined: 02 Sep 2009
Posts: 47110

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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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Re: M30-13 &nbs [#permalink] 17 Feb 2018, 04:06
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# M30-13

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