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If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all [#permalink]
 29 Oct 2009, 02:38
Question Stats:
0% (00:00) correct
100% (00:00) wrong based on 1 sessions
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?? _________________
countdown-beginshas-ended-85483-40.html#p649902
Last edited by Bunuel on 07 Oct 2012, 04:42, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Quadratic querie [#permalink]
29 Oct 2009, 03:11
do you have the answer options for this please?
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Re: Quadratic querie [#permalink]
29 Oct 2009, 03:28
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Re: Quadratic querie [#permalink]
29 Oct 2009, 03:39
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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}=4Wow! math buster  don't have it in GMAT notes... added now, thnx and +1
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Re: Quadratic querie [#permalink]
29 Oct 2009, 04:36
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]
29 Oct 2009, 05:47
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]
09 May 2011, 05:40
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]
09 May 2011, 14:39
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 
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Re: Quadratic querie [#permalink]
09 May 2011, 16:59
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]
24 Nov 2011, 23:42
I also got 4
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Re: Quadratic querie [#permalink]
07 Oct 2012, 01:38
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]
07 Oct 2012, 04:48
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Re: If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all [#permalink]
07 Oct 2012, 10:59
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]
10 Dec 2012, 05:13
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]
10 Dec 2012, 05:13
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