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If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all

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If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all [#permalink] New post 29 Oct 2009, 01:38
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If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all values for k such that f (k+2)=g(2k)?

[Reveal] Spoiler:
Looks pretty simple..however I got stuck mid-way.
Got till K^2-4K-65=0
Does anyone know how to proceed further??

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Last edited by Bunuel on 07 Oct 2012, 03:42, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Quadratic querie [#permalink] New post 29 Oct 2009, 02:11
do you have the answer options for this please?
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Re: Quadratic querie [#permalink] New post 29 Oct 2009, 02:28
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tejal777 wrote:
If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all values for k such that f (k+2)=g(2k) ?

Looks pretty simple..however I got stuck mid-way.
Got till K^2-4K-65=0
Does anyone know how to proceed further??


You've done everything right:

5*(k+2)^2=(2k)^2+12(2k)+85 --> k^2-4k-65=0.

Viete's formula for the roots x_1 and x_2 of equation ax^2+bx+c=0:

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

So in our case the roots k_1+k_2=\frac{-(-4)}{1}=4
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Re: Quadratic querie [#permalink] New post 29 Oct 2009, 02:39
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Bunuel wrote:
tejal777 wrote:
If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all values for k such that f (k+2)=g(2k) ?

Looks pretty simple..however I got stuck mid-way.
Got till K^2-4K-65=0
Does anyone know how to proceed further??


You've done everything right:

5*(k+2)^2=(2k)^2+12(2k)+85 --> k^2-4k-65=0.

Viete's formula for the roots x1 and x2 of equation ax^2+bx+c=0:

x1+x2=\frac{-b}{a} AND x1*x2=\frac{c}{a}

So in our case the roots k1+k2=\frac{-(-4)}{1}=4



Wow! math buster :)
don't have it in GMAT notes... added now, thnx and +1
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Re: Quadratic querie [#permalink] New post 29 Oct 2009, 03:36
after simplification,we get the quadratic eqn

k^2-4k-65= 0

sum of all values of k will be equal to the sum of the roots of the above q.eqn = -b/a = 4

I will go with 4
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Re: Quadratic querie [#permalink] New post 29 Oct 2009, 04:47
I tried solving this by quadratic equation formula and got 2 answers - 2+ or -\sqrt{65}.

What could be the other two possible values of k?
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Re: Quadratic querie [#permalink] New post 09 May 2011, 04:40
5(k+2)^2 = 4k^2 + 24k + 85

=> 5(k^2 + 4k + 4) = 4k^2 + 24k + 85

=> 5k^2 + 20k + 20 = 4k^2 + 24k + 85

=> k^2 - 4k - 65 = 0

No need to solve this, we need sum of two values of k:

So sum of roots = -b/a = -(-4) = 4
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Re: Quadratic querie [#permalink] New post 09 May 2011, 13:39
f(x)=5x^2
g(x)=x^2+12x+85
f(k+2)=5 (k+2)^2= 5(k^2+4k+4)=5k^2+20k+20

g(2k)= (2k)^2+ 12(2k)+85=4k^2+24k+85

5^k2+20k+20=4k^2+24k+85
k^2-4k-65=0

Using the quadratic equation (-b +/-\sqrt{b2-4ac}/2a we find the roots as follows

-(-4)+/-\sqrt{4^2 - 4(1)(-65)}/2(1)
So the first root is
4+\sqrt{16+260}/2 = 4+\sqrt{276}/2= 4+2\sqrt{69}/2
The second root is
4-\sqrt{16+260}/2 = 4-\sqrt{276}/2= 4-2\sqrt{69}/2

The sum of the roots is
4+2\sqrt{69}/2 + 4-2\sqrt{69}/2 = 8/2= 4
The answer should be 4 :-D
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Re: Quadratic querie [#permalink] New post 09 May 2011, 15:59
sum of two roots of a quadratic equation is -b/a


now we have k^2-4k-65

=> - b/a = 4
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Re: Quadratic querie [#permalink] New post 24 Nov 2011, 22:42
I also got 4 :)
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Re: Quadratic querie [#permalink] New post 07 Oct 2012, 00:38
How do we get 4??

-b/a?? what is concept behind it??

k2-4k-65=0?? i cant solve after that..

plz do explain..Thanks in advance
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Re: Quadratic querie [#permalink] New post 07 Oct 2012, 03:48
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sanjoo wrote:
How do we get 4??

-b/a?? what is concept behind it??

k2-4k-65=0?? i cant solve after that..

plz do explain..Thanks in advance


It's called Viete's theorem, which 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}.

Now, if we apply this to k^2-4k-65=0, we'll get that the sum of the roots must be k_1+k_2=\frac{-(-4)}{1}=4.

Hope it's clear.
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Re: If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all [#permalink] New post 07 Oct 2012, 09:59
Thanks Alot BUNUEL..

i didnt see formula like this in my life :( new things very hard to digest :(..
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Re: If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all [#permalink] New post 10 Dec 2012, 04:13
f(k+2) = 5(k+2)^2 = 5(k^2 + 4k + 4) = 5k^2 + 20k + 20
g(2k) = 4k^2 + 24k + 85

5k^2 + 20k + 20 = 4k^2 + 24k + 85
k^2 - 4k -65 = 0

Trick: Sum of all roots: k1 + k2 = -b/a

k1 + k2 = - \frac{-4}{1} = 4

Answer: 4
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Re: If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all [#permalink] New post 10 Jan 2014, 13:57
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Re: If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all [#permalink] New post 23 Apr 2014, 05:57
Could we try posting some answer choices for this one?

I suggest the following

A) -4
B) 4
C) 65
D) -8
E) 8

Thanks!
Cheers
J :)
Re: If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all   [#permalink] 23 Apr 2014, 05:57
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