Syntel
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Numerical Ability
Quadratic Equations
a+b+c=0, then roots of ax^2+bx+c = 0 is
Option
a) imaginary
b) real
c) coincidental
d) zero
Read Solution (Total 8)
-
- Ans is Real Values.
Roots of the quadratic equation are
x = [-b +/- sqrt(b^2 - 4ac)] / 2a ................... [1]
Since, a+b+c = 0 ==> b = -(a+c)
b^2 = (a+c)^2 = a^2+c^2 +2ac
b^2 -4ac = a^2+c^2-2ac = (a-c)^2 ........... [2]
From [1] and [2], we get,
x = [a+c +/- (a-c)] / 2a
Solving above, X roots are 1 and c/a
ROOTs are Real - 12 years agoHelpfull: Yes(19) No(1)
- Roots are real
- 11 years agoHelpfull: Yes(2) No(0)
- b=-(a+c)
ax^2+bx+c
ax^2-ax-cx+C
ax(x-1)-c(x-1)
(ax-c)(x-1)
x=c/a,1 real roots
- 10 years agoHelpfull: Yes(1) No(0)
- Take c=0,a=1,b=-1
so x^2-x=0
x=+1,-1
so real
- 10 years agoHelpfull: Yes(1) No(0)
- x=1 only is the solution as 1 is a real number ans is real
- 11 years agoHelpfull: Yes(0) No(0)
- x=1 so roots are real
- 11 years agoHelpfull: Yes(0) No(0)
- 2.real
a+b+c=0
root
alfa=-b/a,
beta=c/a
so,its roots should be real - 10 years agoHelpfull: Yes(0) No(0)
- the roots will be either 1 or c/a (use the formula to calculate roots)
hence ans=b i.e, real roots
- 10 years agoHelpfull: Yes(0) No(0)
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