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If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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29 Oct 2009, 02:38
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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)? A. 8 B. 4 C. 4 D. 8 E. 65 Looks pretty simple..however I got stuck midway. Got till \(K^24K65=0\) Does anyone know how to proceed further??
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Last edited by Bunuel on 04 Mar 2018, 07:33, edited 2 times in total.
Renamed the topic and edited the question.



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If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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29 Oct 2009, 03:28
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Re: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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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 midway. Got till \(K^24K65=0\) Does anyone know how to proceed further?? You've done everything right: \(5*(k+2)^2=(2k)^2+12(2k)+85\) > \(k^24k65=0\). 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: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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29 Oct 2009, 04:36
after simplification,we get the quadratic eqn
k^24k65= 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: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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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: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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09 May 2011, 05:40
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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: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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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^24k65=0 Using the quadratic equation (b +/\sqrt{b24ac}/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= 42\sqrt{69}/2 The sum of the roots is 4+2\sqrt{69}/2 + 42\sqrt{69}/2 = 8/2= 4 The answer should be 4
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Re: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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09 May 2011, 16:59
sum of two roots of a quadratic equation is b/a
now we have \(k^24k65\)
=>  b/a = 4



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Re: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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07 Oct 2012, 01:38
How do we get 4?? b/a?? what is concept behind it?? k24k65=0?? i cant solve after that.. plz do explain..Thanks in advance
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Re: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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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 value [#permalink]
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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 value [#permalink]
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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 value [#permalink]
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04 Mar 2018, 07:25
Bunuel , Can we have more this kinda questions of functions ?
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Re: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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04 Mar 2018, 07:35



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Re: If f(x) = 5x^2 and g(x) = x^2 + 12x + 85, what is the sum of all value [#permalink]
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04 Mar 2018, 07:46
Thanks a lot Bunuel !
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