Latest Campus Aptitude Question SOLUTION: For what value of “k” will the equation (2kx2 + 5kx +2)=0 have equal roots? A. 2/7 B. 9/4 C. 16/25 D. 7/1 Read more at http://www.m4maths.com/155096-For-what-value-of-k-will-th

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(#M40155235) CAMPUS Category question determine k Keep an EYE

For what value of “k” will the equation (2kx2 + 5kx +2)=0 have equal roots?
A. 2/7 B. 9/4 C. 16/25 D. 7/1
Read more at http://www.m4maths.com/155096-For-what-value-of-k-will-the-equation-2kx2-5kx-2-0-have-equal-roots-A-2-7-B-9.html#Z17J8yZGQ65s1SSx.99

Asked In Campus ami (8 years ago)

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shreya  (7 years ago)

D=b^2-4ac
for real and equal roots D=0
here a= 2k b=5k c=2
D=(5k)^2-(4*2k*2)=0
k=16/25

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Devendra Marghade  (8 years ago)

C. 16/25

Quadratic equation 2kx^2+5kx+2=0 is of the form ax^2+bx+c=0 so here a=2k, b=5k, c=2
If r1 & r2 are the roots of the equation and r1=r2, then Sum of the roots=r1+r2=2r1=-b/a=-5k/2k=-5/2 ---(i) and
Product of the roots=r1*r2=r1^2=c/a=2/(2k) =1/k---(ii)
From (i), r1=-5/4, substituting this value in (ii), (-5/4)^2=1/k, So k=16/25

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