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15 Mar 2021, 00:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
58% (01:24) correct
42%
(02:15)
wrong
based on 180
sessions
History
Date
Time
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What is the value of K for which the roots of the quadratic equation \(5x^2 + 13x + K = 0\) are reciprocals to each other?
A. -5
B. -3
C. 1
D. 3
E. 5
A. -5
B. -3
C. 1
D. 3
E. 5
15 Mar 2021, 05:12
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Bunuel
For a quadratic equation \(ax^2+bx+c = 0\),
The product of Roots \(α*β= \frac{c}{a}\)
Comparing \(5x^2 + 13x + K = 0\) with \(ax^2+bx+c = 0\),
a = 5, b = 13, c = K
Given, \(α = \frac{1}{β}\)
i.e. \(α*β = 1\)
But, \(α*β = \frac{c}{a} = \frac{K}{5}\)
Therefore, \(α*β = \frac{K}{5} = 1\)
i.e.
\(K = 5\)
Answer: Option E
General Discussion
Updated on: 15 Mar 2021, 18:28
Originally posted by MarmikUpadhyay on 15 Mar 2021, 01:22.
Last edited by MarmikUpadhyay on 15 Mar 2021, 18:28, edited 1 time in total.
Last edited by MarmikUpadhyay on 15 Mar 2021, 18:28, edited 1 time in total.
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Given equation: \(5x^2 + 13x + K = 0\)
Roots of the equation by Quadratic formula:
x = \(\frac{(-13 + \sqrt{13 * 13 - 4 * 5 * K})}{(2 * 5)}\)
and x = \(\frac{(-13 - \sqrt{13 * 13 - 4 * 5 * K})}{(2 * 5)}\)
Both roots are reciprocal of each other:
=> \(\frac{(-13 + \sqrt{169 - 20 * K})}{(10)}\) = \(\frac{(10)}{(-13 - \sqrt{169 - 20 * K})}\)
=> 10 * 10 = (-13 + \(\sqrt{169 - 20 * K}\)) * (-13 - \(\sqrt{169 - 20 * K}\))
Using [ \((a - b) * (a + b) = a^2 - b^2\) ]
100 = (-13) * (-13) - (169 -20 * K)
=> 100 = 169 - (169 - 20 * K) = 20 * K
=> K = 100/20 = 5
Alternatively,
Product of roots = \(\frac{K}{5}\)
Roots are reciprocal of each other, so product of roots = 1
=> \(\frac{K}{5}\) = 1
=> K = 5
So, correct answer is option E.
Roots of the equation by Quadratic formula:
x = \(\frac{(-13 + \sqrt{13 * 13 - 4 * 5 * K})}{(2 * 5)}\)
and x = \(\frac{(-13 - \sqrt{13 * 13 - 4 * 5 * K})}{(2 * 5)}\)
Both roots are reciprocal of each other:
=> \(\frac{(-13 + \sqrt{169 - 20 * K})}{(10)}\) = \(\frac{(10)}{(-13 - \sqrt{169 - 20 * K})}\)
=> 10 * 10 = (-13 + \(\sqrt{169 - 20 * K}\)) * (-13 - \(\sqrt{169 - 20 * K}\))
Using [ \((a - b) * (a + b) = a^2 - b^2\) ]
100 = (-13) * (-13) - (169 -20 * K)
=> 100 = 169 - (169 - 20 * K) = 20 * K
=> K = 100/20 = 5
Alternatively,
Product of roots = \(\frac{K}{5}\)
Roots are reciprocal of each other, so product of roots = 1
=> \(\frac{K}{5}\) = 1
=> K = 5
So, correct answer is option E.
