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If the roots of a 1 x 2 +b 1 x +c 1 =0 are m 1, m 2 and the roots of a 2 x 2 +b 2 x +c 2 =0 are n 1 ,n 2 such that m 1 n 1 = m 2 n 2 =1, then a) a 1/ a 2 = b 1/ b 2= c 1/ c 2 b) a 1 a 2= b 1 b 2= c 1 c 2 c) a 1/ c 2= b 1/ b 2= c 1/ a 2


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


Grade:12

1 Answers

jitender lakhanpal
62 Points
9 years ago

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  ,    b= b2    ,  a= 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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