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If the roots of a1x2 +b1x +c1=0 are m1,m2 and the roots of a2x2 +b2x +c2=0 are n1,n2 such that m1 n1 = m2n2 =1, then
a) a1/ a2= b1/ b2= c1/ c2
b) a1 a2= b1 b2= c1 c2
c) a1/ c2= b1/ b2= c1/ a2
Dear Pooja,
correct option is c)
solution :
it is given that the roots of the two equations are reciprocals
n1 = 1/ m1 and n2 = 1/ m2
m1 and m2 are the roots of
a1x2 +b1x +c1=0
m1 + m2 = -(b1/a1) m1*m2= c1/a1
and n1 and n2 are roots of a2x2 +b2x +c2=0 -------1
n1+n2= 1/m1+1/m2 = (m1+m2)/(m1*m2)=-(b1/c1) ----- sum of roots
n1*n2= 1/(m1*m2)=a1/c1-----------product of roots
now we will form the equation as x2 +(sum of roots)x +product of roots =0
x2 + (b1/c1) x +a1/c1 =0
multiply by c1 we get
c1x2 + b1x +a1 =0 -------2 this is same as the equation 1 now comparing we get
we get c1 = a2 , b1 = b2 , a1 = c2
from this we get c1/a2 = 1 b1/b2= 1 a1/c2 =1 thus equating all three we get
a1/ c2= b1/ b2= c1/ a2 so option c
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