Latest Elitmus Aptitude Question SOLUTION: Q. ax^2+bx+c = 0 has two roots x1 and x2. If mode x1 = mode x2, then Option (a) a>c (b) b>c (c) c>a (d) none

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(#M40032649) ELITMUS Numerical Ability question Keep an EYE

Q. ax^2+bx+c = 0 has two roots x1 and x2. If mode x1 = mode x2, then
Option
(a) a>c
(b) b>c
(c) c>a
(d) none

Asked In Elitmus sid (10 years ago)

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RAKESH  (10 years ago)

mod x1 = mod x2 => x1 & x2 have same value but opposite in sign.

in this case b=0
eqn is ax^2+c = 0 => x=sqrt(-c/a)
for x to be real (-c/a) should +ve => c & a are of opposite sign
so c>a or a>c

for ex x^2-4=0 => x1=-2,x2=2 & modx1=modx2=2 [a>c]
or may -x^2+4=0 => x1=-2,x2=2 & modx1=modx2=2 [c>a]

(d) none

Is this solution Helpfull? Yes (29) | No (3)  

Mainak Das  (10 years ago)

c. c>a is the ans

Is this solution Helpfull? Yes (1) | No (4)  

vipin kumar  (10 years ago)

since x1*x2=c/a
x1*x2= +VE ( x1=|x2| )
c/a= +ve Const.
so c=a(Const)
hence c>a .......... similarly

x1+x2= const= b/a
b= a*const
so b>a
but we have c>a in option ...... it is the ans..

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ashes  (10 years ago)

i'm not sure but mod x1=mod x2 when b=0;then ax^2+c=0,x=underroot(-c/a),for this (-c/a) should be positive ,for this either c or a can be negative not both(for real roots)...so ans. is none

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Neeraj kumar jha  (10 years ago)

|x1| = |x2|, means either both are same or both have opposite sign. Lets consder the first case. (x-2)^2 = x^2 - 2x + 4, a=1,b=-2,c=4. so if both root are same then. a is always 1,b is negative and c is always positive. so c>a.

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Avigyan  (10 years ago)

is it a ?? a>c ?

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