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A given quadratic equation ax2 – 52x + 24 = 0 has 13/6 as sum

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A given quadratic equation ax2 – 52x + 24 = 0 has 13/6 as sum [#permalink]

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New post 14 Jul 2017, 08:34
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Question Stats:

67% (01:32) correct 33% (01:19) wrong based on 75 sessions

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A given quadratic equation \(ax^2 – 52x + 24 = 0\) has \(\frac{13}{6}\) as sum of the roots. Find the product of the roots.

(A) \(24\)
(B) \(1\)
(C) \(12\)
(D) \(\frac{24}{5}\)
(E) \(2\)

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Re: A given quadratic equation ax2 – 52x + 24 = 0 has 13/6 as sum [#permalink]

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New post 14 Jul 2017, 09:09
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Let the roots of the equation be x and y

Now sum of the roots = x+y = - (-52)/a

Product of the roots = x*y = 24/a

Given sum of the roots = 13/6 = 52/a => a = 52*6/13

Product of the roots = x*y = 24/a = (24 * 13)/(52 * 6) = 1

Answer is B
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Re: A given quadratic equation ax2 – 52x + 24 = 0 has 13/6 as sum [#permalink]

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New post 14 Jul 2017, 09:57
Sum of the roots of a quadratic equation \(= \frac{b}{a} = \frac{-(-52)}{a} = \frac{13}{6}\), hence, \(a = 24.\)

Product of the roots \(= \frac{c}{a} = \frac{24}{24} = 1.\) Ans - B.
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Re: A given quadratic equation ax2 – 52x + 24 = 0 has 13/6 as sum [#permalink]

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New post 01 Aug 2017, 09:22
TimeTraveller wrote:
Sum of the roots of a quadratic equation \(= \frac{b}{a} = \frac{-(-52)}{a} = \frac{13}{6}\), hence, \(a = 24.\)

Product of the roots \(= \frac{c}{a} = \frac{24}{24} = 1.\) Ans - B.


Hi,
I can't understand why you put a minus in front of -54; can you please clarify?

Thanks a lot,
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Re: A given quadratic equation ax2 – 52x + 24 = 0 has 13/6 as sum [#permalink]

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New post 02 Aug 2017, 11:04
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Heres the answer to your query
In a quadratic equation "ax^2+bx+c"

Sum of roots = -b/a
&
Product of roots = c/a

Hope this helps u. :)
Re: A given quadratic equation ax2 – 52x + 24 = 0 has 13/6 as sum   [#permalink] 02 Aug 2017, 11:04
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