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If one of the roots of the equation

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soumya1989
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If one of the roots of the equation [#permalink] New post   27 Mar 2016, 23:13
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If one of the roots of the equation 2x2 + (3k + 4)x + (9k2–3k– 1) = 0 is twice the other, then which of the following can be a value of ‘k’?

A. −23

B. 23

C. −13

D. 13

E. None of the above
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C
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soumya1989
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Re: If one of the roots of the equation [#permalink] New post   28 Mar 2016, 00:30
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Explanation


Let one of the roots of the given equation be ‘a’. Then, the other root will be ‘2a’.
∴ Sum of the roots = a + 2a = -(3k+4)/2
or, a = -(3k+4)/6 ----- (i)


∴ Product of the roots = a*2a = 2a2= (9k2−3k−1)/2
or, a2 = (9k2−3k−1)/4 ---- (ii)

Combining (i) and (ii),
⇒ 72k2−51k−25 = 0
⇒(3k + 1)(24k − 25) = 0
Therefore, k = -1/3, 25/24.
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JelalHossain
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Re: If one of the roots of the equation [#permalink] New post   04 Jun 2019, 01:02
I did the backward calculation by taking the possible values of K. I tried to find whether the reformed equation ( in line with the value of K) gives two roots or not and the roots corroborate to the conditions or not. Eventually , whiling taking K=-1/3, It matched the conditions. Thus, I got C.
But this was too time-consuming.
Can any of you please give an easy and quicker solution ?
Regards
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pranab223
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Re: If one of the roots of the equation [#permalink] New post   12 Jun 2019, 05:30
1
JelalHossain wrote:
I did the backward calculation by taking the possible values of K. I tried to find whether the reformed equation ( in line with the value of K) gives two roots or not and the roots corroborate to the conditions or not. Eventually , whiling taking K=-1/3, It matched the conditions. Thus, I got C.
But this was too time-consuming.
Can any of you please give an easy and quicker solution ?
Regards



The solution provided above is less time consuming.

Whenever u receive a complex quadratic equation "ax2+bx+C " and need to find the roots, the best way is to know

1. Sum of the roots = −ba

2. product of the roots = ca
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gmatclubot
Re: If one of the roots of the equation [#permalink]
12 Jun 2019, 05:30
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